what is conditional problem solving explain with an example

Conditional Probability Explained (with Formulas and Real-life Examples)

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Have you ever wondered what the likelihood of getting your dream job or winning the lottery is? At first glance, the idea seems completely out of reach, because these instances are influenced by many different factors. This is where probability comes in. While it may not help you win the lottery, it helps you calculate the odds for a variety of different scenarios – in fact, you’ve probably used it multiple times just today.

In less casual terms, probability is used by many industry professionals for its ability to resolve complex real-world problems in a more quantifiable way. Whether it’s through the different types of distribution , expected values, or mathematical modeling methods like the Monte Carlo simulation, mastering this theory will help you make a giant leap towards a successful data science career .

In this tutorial, we will introduce you to the key concept of conditional probability and how it applies to real-life instances. You will also learn how to:

• Distinguish between independent and dependent events
• Interpret the conditional probability formula
• Apply the law of total probability

What Is Conditional Probability?

Essentially, conditional probability is the likelihood of an event occurring, assuming a different one has already happened. Otherwise said, there must be some sort of relationship with the past. Moreover, its formula, which we will expand on in this tutorial, is based on the Bayes’ Theorem .

But first, in order to fully understand how to calculate conditional probability, we must look at the events that can affect it.

What Are Independent and Dependent Events?

There are two types of events that can influence conditional probability:

• Independent

It’s important to know the differences in order to successfully solve a problem. In fact, we use conditional probability to distinguish between the events.

What Is an Independent Event?

We call the theoretical probability that remains unaffected by other occurrences an independent event.

The easiest example is a coin flip – you always have a 50% chance of landing on your desired side, regardless of the result of the previous throw. In this case, we have two events – A and B. If A is flipping heads and B represents getting the same on the previous try, the probability doesn’t change – it remains 0.5. Therefore, we say that:

\[P(A) = P(A|B) = 0.5\]

What this means is that the probability of the coin landing on heads in a new flip is unconditional with respect to previous flips.

In cases where any two events are independent, the probability of their intersection is the product of the individual probabilities:

\[P(A \cap B) = P(A) \times P(B)\]

What Is a Dependent Event?

The probabilities of dependent events vary as conditions change.

For instance, what is the probability of drawing the Queen of Spades? Normally, we have exactly one favorable outcome and 52 elements in the sample space, so the result is:

\[P(A) =  \frac {1}{52}\]

Now, imagine we know that we drew a spade . The new sample space contains all the 13 cards from the suit only and the probability becomes:

\[P(A|B) =  \frac {1}{13}\]

The two events are, therefore, dependent.

Alternatively, our sample can consist of only four cards if we know that the one we have drawn is a queen instead of a spade. Thus, the probability of drawing the Queen of Spades becomes:

\[P(A|B) =  \frac {1}{4}\]

That is because the probability of getting our desired card alters if we know it is a queen. In other words, A and C are also dependent.

With this example, you could clearly see how the probability of an event changes depending on the information we have.

The Conditional Probability Formula

By definition, the conditional probability equals the probability of the intersection of events A and B over the probability of event B occurring:

\[P(A|B) =  \frac {P (A \cap B)}{P (B)}\]

This holds true only if the probability of $B > 0$. This is logically so because if we have $P(B) = 0$, then the event would never occur. Thus, $P(A|B)$ would not be interpretable.

First, to satisfy the conditional probability formula, we need both events B and A to occur simultaneously. This suggests that the intersection of A and B would consist of all our favorable outcomes.

Second, the conditional probability requires that event B occurs, so the sample space would simply be all outcomes where event B is satisfied.

It’s important to remember that the order in which we write the elements for the conditional probability is crucial. $P(A|B)$ is not the same as $P(B|A)$, even if the numeric values are equal.

To illustrate, let’s explore the characteristics of Hamilton College’s class of 2018:

• 5% percent of the students who got a degree in Economics graduated with honors
• At the same time, 5% of all students who graduated with honors completed a concentration in Economics

These two statements might have the same conditional probability, but they hold completely different meanings.

In particular, the first suggests that only four of the 80 Economics majors graduated with a distinction:

\[P(H|E) =  \frac {4}{80}\]

Meanwhile, the second statement shows that four out of the 80 students who graduated with high grades completed a degree in Economics:

\[P(E|H) =  \frac {4}{80}\]

Conditional Probability in Real Life: An Example

Many scientific papers rely on conducting experiments or surveys. They often provide summarized statistics we use to analyze and interpret how certain factors affect one another.

Imagine you conducted a survey where you asked 100 men and women of all ages if they eat meat and obtain the following results:

We see 15 of the 47 women that participated are vegetarian, as well as 29 out of the 53 men.

Now, if A represents being vegetarian and B represents being a woman, then $P(A|B)$ and $P(B|A)$ express different events.

The likelihood of a woman being vegetarian is:

\[P(A|B) = \frac {15}{47} \]

Meanwhile, the likelihood of a vegetarian being a woman is:

\[P(B|A) = \frac {15}{44} \]

Based on these outcomes, we can conclude that it is more likely for a vegetarian to be female, than for a woman not to eat meat. This goes to show that in probability theory things are never straightforward.

The Law of Total Probability

Now that you understand the distinctions between the different conditional probabilities of two events, we can introduce an important concept – the law of total probability.

Let’s say A is the union of some infinitely many events:

\[A = B_1 \cup B_2 \cup~...~\cup B_n\]

This law dictates that the probability of A is:

\[P(A) = P(A|B_1) \times P(B_1) + P(A|B_2) \times P(B_2)...\]

So, going back to our survey example, the probability of being vegetarian equals

\[ P(vegetarian) = P(vegetarian|male) \times P(male) \ + P(vegetarian|female) \times P(female) \]

If we plug in values from the results table, we get that the probability of being vegetarian equals:

\[\frac {29}{53} \times \frac {53}{100} + \frac {15}{47} \times \frac {47}{100} = 0.44\]

Therefore, according to the survey, there is a 44% chance of someone being vegetarian.

Conditional Probability: Next Steps

As you can see, there are many benefits to learning how to apply probability in order to solve real-life problems. In fact, the theory is used in branches such as finance , business analytics , healthcare , and many more. In other words, it is indeed an essential skill for everyone looking to work with data.

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Conditional Probability

The conditional probability , as its name suggests, is the probability of happening an event that is based upon a condition. For example, assume that the probability of a boy playing tennis in the evening is 95% (0.95) whereas the probability that he plays given that it is a rainy day is less which is 10% (0.1). Then the former case is just normal probability whereas the latter case is the conditional probability. In this example, we represent the two probabilities as P(Play tennis) = 0.95 and P(Play tennis | Rainy day) = 0.1.

Let us learn more about conditional probability along with its formula, examples, and practice questions.

What Is Conditional Probability?

Conditional probability is one of the important concepts in probability and statistics . The "probability of A given B" (or) the "probability of A with respect to the condition B" is denoted by the conditional probability P(A | B) (or) P (A / B) (or) P\(_B\)(A). Thus, P(A | B) represents the probability of A which happens after event B has happened already. the probability of an event may alter if there is a condition given.

Definition of Conditional Probability

If A and B are two events associated with the same sample space of a random experiment, the conditional probability of event A given that B has occurred is given by P(A/B) = P( A ∩ B)/ P (B) , provided P(B) ≠ 0.

Let us understand conditional probability with an example. Let us find the conditional probability of getting at least two tails given that it is a head on the first toss when 3 coins are tossed. The sample space, S (the list of all outcomes) when 3 coins are tossed is given as follows:

Let us assume the two events A and B as follows:

• A = the event of getting at least two tails
• B = the event of getting a head on the first toss

Then, A = {HTT, THT, TTH, TTT} and B = {HHH, HHT, HTH, HTT}.

Then P(A) = 4/8 = 1/2 and P(B) = 4/8 = 1/2.

We have to find the probability of getting at least two tails given that it is a head on the first toss. It means, out of all elements of B, we have to choose only the ones with two tails. We can see that among the elements of B, there is only one element (which is HTT) with two tails. Thus, the required probability is P(A | B) = 1/4 (only 1 outcome of B is favorable to A out of 4 outcomes of B).

Conditional Probability Formula

In the above example, we have got P(A | B) = 1/4, here 1 represents the element HTT which is present both in "A and B" and 4 represents the total number of elements in B. Using this, we can derive the formula of conditional probability as follows.

P(A | B) = P(A ∩ B) / P(B) (Note that P(B) ≠ 0 here)

Similarly, we can define P(B | A) as follows:

P(B | A) = P(A ∩ B) / P(A) (Note that P(A) ≠ 0 here)

These formulas are also known as the "Kolmogorov definition" of conditional probability.

• P(A | B) = The probability of A given B (or) the probability of A which happens after B
• P(B | A) = The probability of B given A (or) the probability of B which happens after A
• P(A ∩ B) = The probability of happening of both A and B
• P(A) = The probability of A
• P(B) = The probability of B

Derivation of Conditional Probability

Note that the elements of B which favor the event A are the common elements of A and B. i.e. the sample points of A ∩ B.

Thus P(A/B) = Number of events favorable to A ∩ B ÷ Number of events favorable to B.

P(A/B) = \(\dfrac{\dfrac{n(A ∩ B)}{n(S)}}{\dfrac{n(B)}{n(S)}}\)

Thus P(A | B) = P(A ∩ B) / P(B)

Properties of Conditional Probability

Here are some properties of conditional probability along with their proofs (derivations) which we may need to use while solving the problems. All these properties depend on the conditional probability formula (which is mentioned in the previous section).

Let S be the sample space of an experiment and A be any event. Then P(S | A) = P(A | A) = 1.

By the formula of conditional probability,

P(S | A) = P(S ∩ A) / P(A) = P(A) / P(A) = 1

P(A | A) = P(A ∩ A) / P(A) = P(A) / P(A) = 1

Hence property 1 is proved.

Let S be the sample space of an experiment and A and B be any two events. Let E be any other event such that P(E) ≠ 0. Then P((A ⋃ B) | E) = P(A | E) + P(B | E) - P((A ∩ B) | E).

P((A ⋃ B) | E) = [P((A ⋃ B) ∩ E)] / P(E)

= [ P(A ∩ E) ⋃ P(B ∩ E) ] / P(E) (using a property of sets )

= [P(A ∩ E) + P(B ∩ E) - P(A ∩ B ∩ E)] / P(E) (using addition theorem of probability )

= P(A ∩ E) / P(E) + P(B ∩ E) / P(E) - P(A ∩ B ∩ E) / P(E)

= P(A | E) + P(B | E) - P((A ∩ B) | E) (By conditional probability formula)

Hence property 2 is proved.

P(A' | B) = 1 - P(A | B), where A' is the complement of the set A.

By Property 1, we have P(S | B) = 1.

We know that S = A ⋃ A'. Thus by the above property,

P( A ⋃ A' | B) = 1

Since A and A' are disjoint events,

P(A | B) + P(A' | B) = 1

• P(A' | B) = 1 - P(A | B)

Hence property 3 is proved.

Dependent and Independent Events

The definition of independent and dependent events is connected to conditional probability. Let us see the definitions of independent and dependent events along with their formulas.

• Dependent Events

Dependent events , as the name suggests, are any two events in which the happening of one event depends on the happening of the other event.

• If A depends on B, then the probability of A is P(A | B).
• If B depends on A, then the probability of B is P(B | A).

By the conditional probability formulas,

P(A | B) = P(A ∩ B) / P(B) ⇒ P(A ∩ B) = P(A | B) · P(B)

P(B | A) = P(A ∩ B) / P(A) ⇒ P(A ∩ B) = P(B | A) · P(A)

Thus, two event A and B are said to be dependent events if one of the conditions is satisfied.

• P(A ∩ B) = P(A | B) · P(B) (or)
• P(A ∩ B) = P(B | A) · P(A)
• Independent Events

Independent events , as the name suggests, are any two events in which the happening of one event does not depend on the happening of the other event. i.e., if A and B are independent then P(A | B) = P(A) and P(B | A) = P(B). Thus, to get the formula of independent events, we just need to replace P(A | B) with P(A) (or P(B | A) with P(B)) in one of the above (dependent events) formulas. Hence, two events are said to be independent if

P(A ∩ B) = P(A) · P(B)

This is also called as multiplication rule of probability.

☛ Also Check:

• Probability
• Addition Theorem of Probability

Important Notes:

• The probability of A given B is called the conditional probability and it is calculated using the formula P(A | B) = P(A ∩ B) / P(B).
• The events that are part of conditional probability are dependent events. For example, if we have P(A | B) anywhere in the problem, then it means that A and B are dependent.
• If two events A and B are independent, then P(A | B) = P(A) and P(B | A) = P(B).
• For any two events A and B, P(A ∩ B) = P(A) · P(B). This is called the multiplication theorem of probability.

Examples of Conditional Probability

Example 1: The table below shows the occurrence of diabetes in 100 people. Let D and N be the events where a randomly selected person "has diabetes" and "not overweight". Then find P(D | N).

From the given table, P(N) = (5+45) / 100 = 50/100.

P(D ∩ N) = 5/100.

By the conditional probability formula,

P(D | N) = P(D ∩ N) / P(N)

= (5/100) / (50/100)

Answer: P(D | N) = 1/10.

Example 2: The probability that it will be sunny on Friday is 4/5. The probability that an ice cream shop will sell ice creams on a sunny Friday is 2/3 and the probability that the ice cream shop sells ice creams on a non-sunny Friday is 1/3. Then find the probability that it will be sunny and the ice cream shop sells the ice creams on Friday.

Let us assume that the probabilities for a Friday to be sunny and for the ice cream shop to sell ice creams be S and I respectively. Then,

P(S) = 4/5.

P(I | S) = 2/3.

P(I | S') = 1/3.

We have to find P(S ∩ I).

We can see that S and I are dependent events. By using the dependent events' formula of conditional probability,

P(S ∩ I) = P(I | S) · P(S) = (2/3) · (4/5) = 8/15.

Answer: The required probability = 8/15.

Example 3: If a fair die is rolled twice, observe the numbers that face up. Find the conditional probability that the sum of the numbers is 7, given that the first number is 2.

Let us determine the sample space of rolling a die twice. S = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6)(4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) }

Considering events A and B as given: we have

A : the sum of the numbers is 7. Thus set A = {(1,6),(2,5), (3,4), (4,3), (5,2),(6,1) }

B: the first number is 2. Thus set B = {(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)}

A ∩ B: {(2,5)}

By the conditional probability, we know that P(A ) = P(A ∩ B) / P(B)

P(A ) = \(\dfrac{\dfrac{1}{36}}{\dfrac{6}{36}}\)

P(A ) = 1/6

Answer: The conditional probability that the sum of the numbers is 7, given that the first number is 2 is 1/6

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Practice Questions on Conditional Probability

Faqs on conditional probability.

The conditional probability is the probability of happening of an event of A given that another event B has already occurred. It is denoted by P(A | B) and it is calculated by the formula P(A | B) = P(A ∩ B) / P(B).

What Is Conditional Probability Formula?

The conditional probability of A given B is given as P(A | B) = P(A ∩ B) / P(B) and the conditional probability of B given A is P(B | A) = P(A ∩ B) / P(A).

What Are the Properties of Conditional Probability?

Here are the important properties of conditional probability. In all the properties, assume that S is the sample space and A, B, and E are the events.

• P(S | A) = P(A | A) = 1.
• P((A ⋃ B) | E) = P(A | E) + P(B | E) - P((A ∩ B) | E)

Which Example Does Best describe Conditional Probability?

Assume that there are 100 blood donors available in a hospital. Among them, a non-diabetic person has to be chosen given that his blood group is O + . This situation best describes conditional probability. If N and O are the events of selecting a non-diabetic person and a person with the blood group O + respectively, then the conditional probability representing the above situation is P(N | O) and is calculated using the formula, P(N | O) = P(N ∩ O) / P(O).

Which Theorem Best Explains Conditional Probability and Independence?

The multiplication theorem of probability (which is derived from conditional probability) best describes the independent events. According to this, two events A and B are said to be independent if P(A ∩ B) = P(A) · P(B).

How To Read Conditional Probability P(A | B)?

The conditional probability P(A | B) is read as "the probability of A given B" (or) "the probability of A after B has happened". P(A | B) can also be written as P(A/B) (or) P\(_B\)(A).

Why Is Conditional Probability Important?

The conditional probability is important when we have to find the probability of an event that depends on another event. If event A depends on another event B (i.e., event A happens after B has happened), then the probability of event A is denoted by the conditional probability P(A | B) and is calculated using the formula P(A/B) = P(A ∩ B)/P(B).

• Math Article

Conditional Probability

Conditional probability is known as the possibility of an event or outcome happening, based on the existence of a previous event or outcome. It is calculated by multiplying the probability of the preceding event by the renewed probability of the succeeding, or conditional, event.

Here the concept of the independent event and dependent event occurs. Imagine a student who takes leave from school twice a week, excluding Sunday. If it is known that he will be absent from school on Tuesday then what are the chances that he will also take a leave on Saturday in the same week? It is observed that in problems where the occurrence of one event affects the happening of the following event, these cases of probability are known as conditional probability.

Table of Contents:

• Marginal probability
• Joint probability
• Conditional probability and Bayes theorem

The probability of occurrence of any event A when another event B in relation to A has already occurred is known as conditional probability. It is depicted by P(A|B).

As depicted by the above diagram, sample space is given by S, and there are two events A and B. In a situation where event B has already occurred, then our sample space S naturally gets reduced to B because now the chances of occurrence of an event will lie inside B.

As we have to figure out the chances of occurrence of event A, only a portion common to both A and B is enough to represent the probability of occurrence of A, when B has already occurred. The common portion of the events is depicted by the intersection of both the events A and B, i.e. A ∩ B.

This explains the concept of conditional probability problems, i.e. occurrence of any event when another event in relation to has already occurred.

What is Marginal probability?

Marginal probability is the probability of an event happening, such as (p(A)), and it can be mentioned as an unconditional probability. It does not depend on the occurrence of another event. For example, the likelihood that a card is drawn from a deck of cards is black (P(black) = 0.5), and the probability that a card is drawn is 7 (P(7)=1/13), both are independent events since the outcome of another event does not condition the result of one event.

What is Joint Probability?

A joint probability is the probability of event A and event B happening, P(A and B). It is the likelihood of the intersection of two or more events. The probability of the intersection of A and B is written as P(A ∩ B). For example, the likelihood that a card is black and seven is equal to P(Black and Seven) = 2/52 = 1/26. (There are two Black-7 in a deck of 52: the 7 of clubs and the 4 of spades).

Also, read:

When the intersection of two events happen, then the formula for conditional probability for the occurrence of two events is given by;

Where P(A|B) represents the probability of occurrence of A given B has occurred.

N(A ∩ B) is the number of elements common to both A and B.

N(B) is the number of elements in B, and it cannot be equal to zero.

Let N represent the total number of elements in the sample space.

Since N(A ∩ B)/N and N(B)/N denotes the ratio of the number of favourable outcomes to the total number of outcomes; therefore, it indicates the probability.

Therefore, N(A ∩ B)/N can be written as P(A ∩ B) and N(B)/N as P(B).

⇒ P(A|B) = P(A ∩ B)/P(B)

Therefore, P(A ∩ B) = P(B) P(A|B) if P(B) ≠ 0

= P(A) P(B|A) if P(A) ≠ 0

Similarly, the probability of occurrence of B when A has already occurred is given by,

P(B|A) = P(B ∩ A)/P(A)

To have a better insight, let us practice some conditional probability examples.

Conditional Probability and Bayes Theorem

Bayes’ theorem defines the probability of occurrence of an event associated with any condition. It is considered for the case of conditional probability. Also, this is known as the formula for the likelihood of “causes”.

A tree diagram will help you in understanding the probability of different events in different cases.

Property 1: Let E and F be events of a sample space S of an experiment, then we have:

P(S|F) = P(F|F) = 1.

Property 2: If A and B are any two events of a sample space S and F is an event of S such that P(F) ≠ 0, then;

P((A ∪ B)|F) = P(A|F) + P(B|F) – P((A ∩ B)|F)

Property 3: P(A′|B) = 1 − P(A|B)

Problems and Solutions

Example 1: Two dies are thrown simultaneously, and the sum of the numbers obtained is found to be 7. What is the probability that the number 3 has appeared at least once?

Solution: The sample space S would consist of all the numbers possible by the combination of two dies. Therefore S consists of 6 × 6, i.e. 36 events.

Event A indicates the combination in which 3 has appeared at least once.

Event B indicates the combination of the numbers which sum up to 7.

A = {(3, 1), (3, 2), (3, 3)(3, 4)(3, 5)(3, 6)(1, 3)(2, 3)(4, 3)(5, 3)(6, 3)}

B = {(1, 6)(2, 5)(3, 4)(4, 3)(5, 2)(6, 1)}

P(A) = 11/36

P(B) = 6/36

P(A ∩ B) = 2/36

Applying the conditional probability formula we get,

P(A|B) = P(A∩B)/P(B) = (2/36)/(6/36) = ⅓

Example 2: In a group of 100 computer buyers, 40 bought CPU, 30 purchased monitor, and 20 purchased CPU and monitors. If a computer buyer chose at random and bought a CPU, what is the probability they also bought a Monitor?

Solution: As per the first event, 40 out of 100 bought CPU,

So, P(A) = 40% or 0.4

Now, according to the question, 20 buyers purchased both CPU and monitors. So, this is the intersection of the happening of two events. Hence,

P(A∩B) = 20% or 0.2

By the formula of conditional probability we know;

P(B|A) = P(A∩B)/P(B)

P(B|A) = 0.2/0.4 = 2/4 = ½ = 0.5

The probability that a buyer bought a monitor, given that they purchased a CPU, is 50%.

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7.9 Conditional Probability and the Multiplication Rule

Learning objectives.

After completing this section, you should be able to:

• Calculate conditional probabilities.
• Apply the Multiplication Rule for Probability to compute probabilities.

Back in Example 7.18 , we constructed the following table ( Figure 7.38 ) to help us find the probabilities associated with rolling two standard 6-sided dice:

For example, 3 of these 36 equally likely outcomes correspond to rolling a sum of 10, so the probability of rolling a 10 is 3 36 = 1 12 3 36 = 1 12 . However, if you choose to roll the dice one at a time, the probability of rolling a 10 will change after the first die comes to rest. For example, if the first die shows a 5, then the probability of rolling a sum of 10 has jumped to 1 6 1 6 —the event will occur if the second die also shows a 5, which is 1 of 6 equally likely outcomes for the second die. If instead the first die shows a 3, then the probability of rolling a sum of 10 drops to 0—there are no outcomes for the second die that will give us a sum of 10.

Understanding how probabilities can shift as we learn new information is critical in the analysis of our second type of compound events: those built with “and.” This section will explain how to compute probabilities of those compound events.

Conditional Probabilities

When we analyze experiments with multiple stages, we often update the probabilities of the possible final outcomes or the later stages of the experiment based on the results of one or more of the initial stages. These updated probabilities are called conditional probabilities .

In other words, if O O is a possible outcome of the first stage in a multistage experiment, then the probability of an event E E conditional on O O (denoted P ( E | O ) P ( E | O ) , read “the probability of E E given O O ”) is the updated probability of E E under the assumption that O O occurred.

In the example that opened this section, we might consider rolling two dice as a multistage experiment: rolling one, then the other. If we define E E to be the event “roll a sum of 10,” O O to be the event “first die shows 5,” and Q Q to be the event “first die shows 3,” then we computed P ( E ) = 1 12 P ( E ) = 1 12 , P ( E | O ) = 1 6 P ( E | O ) = 1 6 , and P ( E | Q ) = 0 P ( E | Q ) = 0 .

Example 7.31

Computing conditional probabilities.

• April is playing a coin-flipping game with Ben. She will flip a coin 3 times. If the coin lands on heads more than tails, April wins; if it lands on tails more than heads, Ben wins. Let A A be the event “April wins,” H H be “first flip is heads,” and T T be “first flip is tails.” Compute P ( A ) P ( A ) , P ( A | H ) P ( A | H ) , and P ( A | T ) P ( A | T ) .
• You are about to draw 2 cards without replacement from a deck containing only these 10 cards: A ♡ A ♡ , A ♠ A ♠ , A ♣ A ♣ , A ♢ A ♢ , K ♠ K ♠ , K ♣ K ♣ , Q ♡ Q ♡ , Q ♠ Q ♠ , J ♡ J ♡ , J ♠ J ♠ . We’ll define the following events: F F is “both cards are the same rank,” A A is “first card is an ace,” and K K is “first card is a king.” Compute P ( F | A ) P ( F | A ) and P ( F | K ) P ( F | K ) .
• Jim’s sock drawer contains 5 black socks and 3 blue socks. To avoid waking his partner, Jim doesn’t want to turn the lights on, so he puts on 2 socks at random. Let M M be the event “Jim’s 2 socks match,” let K K be the event “the sock on Jim’s left foot is black,” and let L L be the event “the sock on Jim’s left foot is blue.” Compute P ( M ) P ( M ) , P ( M | K ) P ( M | K ) , and P ( M | L ) P ( M | L ) .

Step 1. The sample space is {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}. The event A A consists of the first 4 of those outcomes: HHH, HHT, HTH, and THH. Thus, P ( A ) = 4 8 = 1 2 P ( A ) = 4 8 = 1 2 .

Step 2. Now, let’s compute P ( A | H ) P ( A | H ) . We are assuming the result of the first flip is heads. That leaves us with 4 possible outcomes: HHH, HHT, HTH, and HTT. Of those, April wins 3 (HHH, HHT, HTH) and loses one (HTT). So, P ( A | H ) = 3 4 P ( A | H ) = 3 4 .

Step 3. If the result of the first flip is instead tails, the 4 possible outcomes are THH, THT, TTH, and TTT. Of those, April wins 1 (THH) and loses 3 (THT, TTH, TTT). So, P ( A | T ) = 1 4 P ( A | T ) = 1 4 .

Step 1. If the event A A happens, then 1 of the 4 aces is drawn first; the remaining cards in the deck are 3 aces, 2 kings, 2 queens, and 2 jacks. In order for the event F F to occur, the second card drawn has to be an ace. Since there are 3 aces among the remaining 9 cards, P ( F | A ) = 3 9 = 1 3 P ( F | A ) = 3 9 = 1 3 .

Step 2. If the event K K happens instead, then the first card drawn is a king. That leaves 4 aces, 1 king, 2 queens, and 2 jacks in the deck. Under the assumption that the first card is a king, the event F F will occur only if the second card is also a king. Since only one of the remaining 9 cards is a king, we have P ( F | K ) = 1 9 P ( F | K ) = 1 9 .

Step 1. We can view the event M M as a compound event using “or”: both socks are blue or both socks are black. Let’s compute the probability that both socks are blue using combinations. We’re choosing 2 socks from a group of 8; 3 of the 8 are blue. So, P ( both socks blue ) = 3 C 2 8 C 2 = 3 28 P ( both socks blue ) = 3 C 2 8 C 2 = 3 28 . Similarly, P ( both socks black ) = 5 C 2 8 C 2 = 10 28 P ( both socks black ) = 5 C 2 8 C 2 = 10 28 . Therefore, since these events are mutually exclusive, we can use the Addition Rule: P ( M ) = P ( both socks blue ) + P ( both socks black ) = 3 28 + 10 28 = 13 28 P ( M ) = P ( both socks blue ) + P ( both socks black ) = 3 28 + 10 28 = 13 28 .

Step 2. If the sock on Jim’s left foot is black (i.e., K K occurred), then there are 4 remaining black socks of the 7 in the drawer. So, P ( M | K ) = 4 7 P ( M | K ) = 4 7 .

Step 3. If the sock on Jim’s left foot is blue ( L L occurred), then there are 2 blue socks among the 7 remaining in the drawer. So, P ( M | L ) = 2 7 P ( M | L ) = 2 7 .

Your Turn 7.31

In Tree Diagrams, Tables, and Outcomes , we introduced the concept of dependence between stages of a multistage experiment. We stated at the time that two stages were dependent if the result of one stage affects the other stage. We explained that dependence in terms of the sample space, but sometimes that dependence can be a little more subtle; it’s more properly understood in terms of conditional probabilities. Two stages of an experiment are dependent if P ( E | F ) ≠ P ( E | F ′ ) P ( E | F ) ≠ P ( E | F ′ ) for some outcome of the second stage E E and outcome of the first stage F F .

Protecting Bombers in World War II

In his book How Not to Be Wrong , Jordan Ellenberg recounts this anecdote: During World War II, the American military wanted to add additional armor plating to bomber aircraft, in order to reduce the chances that they get shot down. So, they collected data on planes after returning from missions. The data showed that the fuselage, wings, and fuel system had many more bullet holes (per unit area) than the engine compartments, so the military brass wanted to add additional armor to the parts of the plane that were hit most often. Luckily, before they added the armor to the planes, they asked for a second opinion. Abraham Wald, a Jewish mathematician who had fled the rising Nazi regime, pointed out that it was far more important that the armor plating be added to areas where there were fewer bullet holes. Why? The planes they were studying had already completed their missions, so the military was essentially looking at conditional probabilities: the probability of suffering a bullet strike, given that the plane made it back safely. More bullet holes in an area on the plane indicated that was a region that wasn’t as important for the plane’s survival!

Compound Events Using “And” and the Multiplication Rule

For multistage experiments, the outcomes of the experiment as a whole are often stated in terms of the outcomes of the individual stages. Commonly, those statements are joined with “and.” For example, in the sock drawer example just above, one outcome might be “the left sock is black and the right sock is blue.” As with “or” compound events, these probabilities can be computed with basic arithmetic.

Multiplication Rule for Probability: If E E and F F are events associated with the first and second stages of an experiment, then P ( E and ⁢ F ) = P ( E ) × P ( F | E ) P ( E and ⁢ F ) = P ( E ) × P ( F | E ) .

In The Addition Rule for Probability , we considered probabilities of events connected with “and” in the statement of the Inclusion/Exclusion Principle. These two scenarios are different; in the statement of the Inclusion/Exclusion Principle, the events connected with “and” are both events associated with the same single-stage experiment (or the same stage of a multistage experiment). In the Multiplication Rule, we’re looking at events associated with different stages of a multistage experiment.

Example 7.32

Using the multiplication rule for probability.

You are president of a club with 10 members: 4 seniors, 3 juniors, 2 sophomores, and 1 first-year. You need to choose 2 members to represent the club on 2 college committees. The first person selected will be on the Club Awards Committee and the second will be on the New Club Orientation Committee. The same person cannot be selected for both. You decide to select these representatives at random.

• What is the probability that a senior is chosen for both positions?
• What is the probability that a junior is chosen first and a sophomore is chosen second?
• What is the probability that a sophomore is chosen first and a senior is chosen second?
• We need the probability that a senior is chosen first and a senior is chosen second. These are two stages of a multistage experiment, so we’ll apply the Multiplication Rule for Probability: P ( senior chosen first and senior chosen second ) = P ( senior chosen first ) × P ( senior chosen second | senior chosen first ) . P ( senior chosen first and senior chosen second ) = P ( senior chosen first ) × P ( senior chosen second | senior chosen first ) . Since there are 4 seniors among the 10 members, P ( senior chosen first ) = 4 10 = 2 5 P ( senior chosen first ) = 4 10 = 2 5 . Next, assuming a senior is chosen first, there are 3 seniors among the 9 remaining members. So, P ( senior chosen second | senior chosen first ) = 3 9 = 1 3 P ( senior chosen second | senior chosen first ) = 3 9 = 1 3 . Putting this all together, we get P ( senior chosen first and senior chosen second ) = 2 5 × 1 3 = 2 15 P ( senior chosen first and senior chosen second ) = 2 5 × 1 3 = 2 15 .
• There are 3 juniors among the 10 members, so P ( junior chosen first ) = 3 10 P ( junior chosen first ) = 3 10 . Assuming a junior is chosen first, there are 2 sophomores among the remaining 9 members, so P ( sophomore chosen second | junior chosen first ) = 2 9 P ( sophomore chosen second | junior chosen first ) = 2 9 . Thus, using the Multiplication Rule for Probability, we have P ( junior chosen first and sophomore chosen second ) = 3 10 × 2 9 = 1 15 P ( junior chosen first and sophomore chosen second ) = 3 10 × 2 9 = 1 15 .
• The probability that a sophomore is chosen first is 2 10 = 1 5 2 10 = 1 5 , and the probability that a senior is chosen second given that a sophomore was chosen first is 4 9 4 9 . Thus, using the Multiplication Rule for Probability, we have: P ( sophomore chosen first and senior chosen second ) = 1 5 × 4 9 = 4 45 P ( sophomore chosen first and senior chosen second ) = 1 5 × 4 9 = 4 45 .

Your Turn 7.32

Work it out, the birthday problem.

One of the most famous problems in probability theory is the Birthday Problem, which has to do with shared birthdays in a large group. To make the analysis easier, we’ll ignore leap days, and assume that the probability of being born on any given date is 1 365 1 365 . Now, if you have 366 people in a room, we’re guaranteed to have at least one pair of people who share a single birthday. Imagine filling the room by first admitting someone born on January 1, then someone born on January 2, and so on… The 365th person admitted would be born on December 31. If you add one more person to the room, that person’s birthday would have to match someone else’s.

Let’s look at the other end of the spectrum. If you choose two people at random, what is the probability that they share a birthday? As with many probability questions, this is best addressed by find out the probability that they do not share a birthday. The first person’s birthday can be anything (probability 1), and the second person’s birthday can be anything other than the first person’s birthday (probability 364 365 364 365 ). The probability that they have different birthdays is 1 × 364 365 = 364 365 1 × 364 365 = 364 365 . So, the probability that they share a birthday is 1 − 364 365 = 1 365 1 − 364 365 = 1 365 .

What if we have three people? The probability that they all have different birthdays can be obtained by extending our previous calculation: The probability that two people have different birthdays is 364 365 364 365 , so if we add a third to the mix, the probability that they have a different birthday from the other two is 363 365 363 365 . So, the probability that all three have different birthdays is 364 365 × 363 365 ≈ 0.9918 364 365 × 363 365 ≈ 0.9918 , and thus the probability that there’s a shared birthday in the group is 1 − 0.9918 ≈ 0.0082 1 − 0.9918 ≈ 0.0082 .

The big question is this: How many people do we need in the room to have the probability of a shared birthday greater than 1 2 1 2 ? Make a guess, then with a partner keep adding hypothetical people to the group and computing probabilities until you get there!

It is often useful to combine the rules we’ve seen so far with the techniques we used for finding sample spaces. In particular, trees can be helpful when we want to identify the probabilities of every possible outcome in a multistage experiment. The next example will illustrate this.

Example 7.33

Using tree diagrams to help find probabilities.

The board game Clue uses a deck of 21 cards: 6 suspects, 6 weapons, and 9 rooms. Suppose you are about to draw 2 cards from this deck. There are 6 possible outcomes for the draw: 2 suspects, 2 weapons, 2 rooms, 1 suspect and 1 weapon, 1 suspect and 1 room, or 1 weapon and 1 room. What are the probabilities for each of these outcomes?

Step 1: Let’s start by building a tree diagram that illustrates both stages of this experiment. Let’s use S, W, and R to indicate drawing a suspect, weapon, and room, respectively ( Figure 7.39 ).

Step 2: We want to start computing probabilities, starting with the first stage. The probability that the first card is a suspect is 6 21 = 2 7 6 21 = 2 7 . The probability that the first card is a weapon is the same: 2 7 2 7 . Finally, the probability that the first card is a room is 9 21 = 3 7 9 21 = 3 7 .

Step 3: Let’s incorporate those probabilities into our tree: label the edges going into each of the nodes representing the first-stage outcomes with the corresponding probabilities ( Figure 7.40 ).

Note that the sum of the probabilities coming out of the initial node is 1; this should always be the case for the probabilities coming out of any node!

Step 4: Let’s look at the case where the first card is a suspect. There are 3 edges emanating from that node (leading to the outcomes SS, SW, and SR). We’ll label those edges with the appropriate conditional probabilities, under the assumption that the first card is a suspect. First, there are 5 remaining suspect cards among the 20 left in the deck, so P ( second is suspect | first is suspect ) = 5 20 = 1 4 P ( second is suspect | first is suspect ) = 5 20 = 1 4 . Using similar reasoning, we can compute P ( second is weapon | first is suspect ) = 6 20 = 3 10 P ( second is weapon | first is suspect ) = 6 20 = 3 10 and P ( second is room | first is suspect ) = 9 20 P ( second is room | first is suspect ) = 9 20 .

Step 5: Checking our work, we see that the sum of these 3 probabilities is again equal to 1. Let’s add those to our tree ( Figure 7.41 ).

Step 6: Let’s continue filling in the conditional probabilities at the other nodes, always checking to make sure the sum of the probabilities coming out of any node is equal to 1 ( Figure 7.42 ).

Step 7: We can compute the probability of landing on any final node by multiplying the probabilities along the path we would take to get there. For example, the probability of drawing a suspect first and a weapon second (i.e., ending up on the node labeled “SW”) is 2 7 × 3 10 = 3 35 2 7 × 3 10 = 3 35 , as illustrated in Figure 7.43 .

Step 8: Let’s fill in the rest of the probabilities ( Figure 7.44 ).

Step 9: A helpful feature of tree diagrams is that the final outcomes are always mutually exclusive, so the Addition Rule can be directly applied. For example, the probability of drawing one suspect and one room (in any order) would be P ( S R ) + P ( R S ) = 9 70 + 9 70 = 9 35 P ( S R ) + P ( R S ) = 9 70 + 9 70 = 9 35 . We can find the probabilities of the other outcomes in a similar fashion, as shown in the following table:

Checking once again, the sum of these 6 probabilities is 1, as expected.

Your Turn 7.33

The monty hall problem.

On the original version of the game show Let’s Make a Deal , originally hosted by Monty Hall and now hosted by Wayne Brady, one contestant was chosen to play a game for the grand prize of the day (often a car). Here’s how it worked: On the stage were three areas concealed by numbered curtains. The car was hidden behind one of the curtains; the other two curtains hid worthless prizes (called “Zonks” on the show). The contestant would guess which curtain concealed the car. To build tension, Monty would then reveal what was behind one of the other curtains, which was always one of the Zonks (Since Monty knew where the car was hidden, he always had at least one Zonk curtain that hadn’t been chosen that he could reveal). Monty then turned to the contestant and asked: “Do you want to stick with your original choice, or do you want to switch your choice to the other curtain?” What should the contestant do? Does it matter?

With a partner or in a small group, simulate this game. You can do that with a small candy (the prize) hidden under one of three cups, or with three playing cards (just decide ahead of time which card represents the “Grand Prize”). One person plays the host, who knows where the prize is hidden. Another person plays the contestant and tries to guess where the prize is hidden. After the guess is made, the host should reveal a losing option that wasn’t chosen by the contestant. The contestant then has the option to stick with the original choice or switch to the other, unrevealed option. Play about 20 rounds, taking turns in each role and making sure that both contestant strategies (stick or switch) are used equally often. After each round, make a note of whether the contestant chose “stick” or “switch” and whether the contestant won or lost. Find the empirical probability of winning under each strategy. Then, see if you can use tree diagrams to verify your findings.

Check Your Understanding

Section 7.9 exercises.

• P ( tile shows A )
• P ( tile shows A | tile shows a vowel )
• P ( tile shows a vowel )
• P ( tile shows a vowel | tile shows a letter that comes after M alphabetically )

In the following exercises deal with the game “Punch a Bunch,” which appears on the TV game show The Price Is Right . In this game, contestants have a chance to punch through up to 4 paper circles on a board; behind each circle is a card with a dollar amount printed on it. There are 50 of these circles; the dollar amounts are given in this table:

Contestants are shown their selected dollar amounts one at a time, in the order selected. After each is revealed, the contestant is given the option of taking that amount of money or throwing it away in favor of the next amount. (You can watch the game being played in the video Playing “Punch a Bunch.” ) Jeremy is playing “Punch a Bunch” and gets 2 punches.

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Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Access for free at https://openstax.org/books/contemporary-mathematics/pages/1-introduction

• Authors: Donna Kirk
• Publisher/website: OpenStax
• Book title: Contemporary Mathematics
• Publication date: Mar 22, 2023
• Location: Houston, Texas
• Book URL: https://openstax.org/books/contemporary-mathematics/pages/1-introduction
• Section URL: https://openstax.org/books/contemporary-mathematics/pages/7-9-conditional-probability-and-the-multiplication-rule

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Conditional Probability

How to handle Dependent Events

Life is full of random events! You need to get a "feel" for them to be a smart and successful person.

Independent Events

Events can be " Independent ", meaning each event is not affected by any other events.

Example: Tossing a coin.

Each toss of a coin is a perfect isolated thing.

What it did in the past will not affect the current toss.

The chance is simply 1-in-2, or 50%, just like ANY toss of the coin.

So each toss is an Independent Event .

Dependent Events

But events can also be "dependent" ... which means they can be affected by previous events ...

Example: Marbles in a Bag

2 blue and 3 red marbles are in a bag.

What are the chances of getting a blue marble?

The chance is 2 in 5

But after taking one out the chances change!

So the next time:

This is because we are removing marbles from the bag.

So the next event depends on what happened in the previous event, and is called dependent .

Replacement

Note: if we replace the marbles in the bag each time, then the chances do not change and the events are independent :

• With Replacement: the events are Independent (the chances don't change)
• Without Replacement: the events are Dependent (the chances change)

Dependent events are what we look at here.

Tree Diagram

A Tree Diagram is a wonderful way to picture what is going on, so let's build one for our marbles example.

There is a 2/5 chance of pulling out a Blue marble, and a 3/5 chance for Red:

We can go one step further and see what happens when we pick a second marble:

If a blue marble was selected first there is now a 1/4 chance of getting a blue marble and a 3/4 chance of getting a red marble.

If a red marble was selected first there is now a 2/4 chance of getting a blue marble and a 2/4 chance of getting a red marble.

Now we can answer questions like "What are the chances of drawing 2 blue marbles?"

Answer: it is a 2/5 chance followed by a 1/4 chance :

Did you see how we multiplied the chances? And got 1/10 as a result.

The chances of drawing 2 blue marbles is 1/10

We love notation in mathematics! It means we can then use the power of algebra to play around with the ideas. So here is the notation for probability:

P(A) means "Probability Of Event A"

In our marbles example Event A is "get a Blue Marble first" with a probability of 2/5:

And Event B is "get a Blue Marble second" ... but for that we have 2 choices:

• If we got a Blue Marble first the chance is now 1/4
• If we got a Red Marble first the chance is now 2/4

So we have to say which one we want , and use the symbol "|" to mean "given":

P(B|A) means "Event B given Event A"

In other words, event A has already happened, now what is the chance of event B?

P(B|A) is also called the "Conditional Probability" of B given A.

And in our case:

P(B|A) = 1/4

So the probability of getting 2 blue marbles is:

And we write it as

"Probability of event A and event B equals the probability of event A times the probability of event B given event A "

Let's do the next example using only notation:

Example: Drawing 2 Kings from a Deck

Event A is drawing a King first, and Event B is drawing a King second.

For the first card the chance of drawing a King is 4 out of 52 (there are 4 Kings in a deck of 52 cards):

P(A) = 4/52

But after removing a King from the deck the probability of the 2nd card drawn is less likely to be a King (only 3 of the 51 cards left are Kings):

P(B|A) = 3/51

P(A and B) = P(A) x P(B|A) =(4/52)x (3/51) = 12/2652 = 1/221

So the chance of getting 2 Kings is 1 in 221, or about 0.5%

Finding Hidden Data

Using Algebra we can also "change the subject" of the formula, like this:

And we have another useful formula:

"The probability of event B given event A equals the probability of event A and event B divided by the probability of event A "

Example: Ice Cream

70% of your friends like Chocolate, and 35% like Chocolate AND like Strawberry.

What percent of those who like Chocolate also like Strawberry?

P(Strawberry|Chocolate) = P(Chocolate and Strawberry) / P(Chocolate)

50% of your friends who like Chocolate also like Strawberry

Big Example: Soccer Game

You are off to soccer, and want to be the Goalkeeper, but that depends who is the Coach today:

• with Coach Sam the probability of being Goalkeeper is 0.5
• with Coach Alex the probability of being Goalkeeper is 0.3

Sam is Coach more often ... about 6 out of every 10 games (a probability of 0.6 ).

So, what is the probability you will be a Goalkeeper today?

Let's build a tree diagram . First we show the two possible coaches: Sam or Alex:

The probability of getting Sam is 0.6, so the probability of Alex must be 0.4 (together the probability is 1)

Now, if you get Sam, there is 0.5 probability of being Goalie (and 0.5 of not being Goalie):

If you get Alex, there is 0.3 probability of being Goalie (and 0.7 not):

The tree diagram is complete, now let's calculate the overall probabilities. Remember that:

P(A and B) = P(A) x P(B|A)

Here is how to do it for the "Sam, Yes" branch:

(When we take the 0.6 chance of Sam being coach times the 0.5 chance that Sam will let you be Goalkeeper we end up with an 0.3 chance.)

But we are not done yet! We haven't included Alex as Coach:

With 0.4 chance of Alex as Coach, followed by the 0.3 chance gives 0.12

And the two "Yes" branches of the tree together make:

0.3 + 0.12 = 0.42 probability of being a Goalkeeper today

(That is a 42% chance)

One final step: complete the calculations and make sure they add to 1:

0.3 + 0.3 + 0.12 + 0.28 = 1

Yes, they add to 1 , so that looks right.

Friends and Random Numbers

Here is another quite different example of Conditional Probability.

4 friends (Alex, Blake, Chris and Dusty) each choose a random number between 1 and 5. What is the chance that any of them chose the same number?

Let's add our friends one at a time ...

First, what is the chance that Alex and Blake have the same number?

Blake compares his number to Alex's number. There is a 1 in 5 chance of a match.

As a tree diagram :

Note: "Yes" and "No" together  makes 1 (1/5 + 4/5 = 5/5 = 1)

Now, let's include Chris ...

But there are now two cases to consider:

• If Alex and Blake did match, then Chris has only one number to compare to.
• But if Alex and Blake did not match then Chris has two numbers to compare to.

And we get this:

For the top line (Alex and Blake did match) we already have a match (a chance of 1/5).

But for the "Alex and Blake did not match" there is now a 2/5 chance of Chris matching (because Chris gets to match his number against both Alex and Blake).

And we can work out the combined chance by multiplying the chances it took to get there:

Following the "No, Yes" path ... there is a 4/5 chance of No, followed by a 2/5 chance of Yes:

Following the "No, No" path ... there is a 4/5 chance of No, followed by a 3/5 chance of No:

Also notice that when we add all chances together we still get 1 (a good check that we haven't made a mistake):

(5/25) + (8/25) + (12/25) = 25/25 = 1

Now what happens when we include Dusty?

It is the same idea, just more of it:

OK, that is all 4 friends, and the "Yes" chances together make 101/125:

Answer: 101/125

But here is something interesting ... if we follow the "No" path we can skip all the other calculations and make our life easier:

The chances of not matching are:

(4/5) × (3/5) × (2/5) = 24/125

So the chances of matching are:

1 - (24/125) = 101/125

(And we didn't really need a tree diagram for that!)

And that is a popular trick in probability:

It is often easier to work out the "No" case (and subtract from 1 for the "Yes" case)

(This idea is shown in more detail at Shared Birthdays .)

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Statistics By Jim

Making statistics intuitive

Conditional Probability: Definition, Formula & Examples

By Jim Frost 4 Comments

What is Conditional Probability?

A conditional probability is the likelihood of an event occurring given that another event has already happened. Conditional probabilities allow you to evaluate how prior information affects probabilities. For example, what is the probability of A given B has occurred? When you incorporate existing facts into the calculations, it can change the likelihood of an outcome.

Typically, the problem statement for conditional probability questions assumes that the initial event occurred or indicates that an observer witnesses it. The goal is to calculate the chances of the second event under the condition that the first event occurred.

This concept might sound complicated, but it makes sense that knowing an event occurred can affect the chances of another event.

For example, if someone asks you, what is the likelihood that you’re carrying an umbrella? Wouldn’t your first question be, is it raining? Obviously, knowing whether it is raining affects the chances that you’re carrying an umbrella.

Related post : Probability Fundamentals

Conditional Probability Examples

P (A|B) denotes the conditional probability of event A occurring given that event B has occurred.

P (Cat | Open box on floor) = 0.8

This notation indicates that the chances of someone owning a cat given the presence of an open box on their floor is 0.8.

I know because I own cats and often have an empty box on my floor for them to enjoy!

For a more serious conditional probability example, consider medical testing, such as COVID tests. In this context, evaluating conditional probabilities is critical. You need to know the likelihood of obtaining a positive test result when a person has COVID and the chances of a negative result when a person does not have COVID.

For studies that assess medical tests, researchers already know whether a volunteer has COVID (or whatever disease they’re testing). They then administer a COVID test and record the results. Consequently, we’re evaluating the likelihood of a test result given the known status of a participant.

The following conditional probability notation represents these two cases:

• P (Positive COVID test | Person has COVID)
• P (Negative COVID test | Person does not have COVID)

Tests can perform well for one, both, or neither of these conditions. They need high conditional probabilities for both cases to limit the chances for false negatives and positives, respectively.

Conditional Probability Formula

Use the following conditional probability formula to find the probability of A given B:

In the conditional probability formula, the numerator of the ratio is the joint chance that A and B occur together. We need the joint likelihood in the numerator because we’re interested in the subset of cases where both events happen.

The denominator of the conditional probability formula is the likelihood that B occurs. We use that value in the denominator because the chances of event B defines the total sample space. Remember, the nature of a conditional probability is that a given event occurs, and the denominator accounts for that occurrence.

In the conditional probability formula, the numerator is a subset of the denominator. Together, the formula gives us the ratio of the chances of both events occurring relative to the likelihood that the given event occurs, which is the conditional probability!

Therefore, if the ratio equals one, event A always occurs when event B has occurred. Conversely, when the ratio equals zero, event A never occurs after B happens. For most conditional probability examples, the ratio is somewhere between 0 and 1, indicating that A sometimes occurs after B. Hence, we find the probability of A given B!

Related post : Joint Probability: Definition, Formula & Examples

Worked Conditional Probability Example

Let’s return to the conditional probability example of carrying an umbrella when it’s raining. Assume that a study in a rainy city assesses chances related to rainy days and carrying umbrellas. The researchers were out observing the weather and people.

We want to see how the chances of carrying an umbrella change depending on whether it is raining or not. That requires calculating the following two conditional probabilities:

• P (Umbrella | Rain): What is the likelihood that someone is carrying an umbrella given that it is raining.
• P (Umbrella | No Rain): What is the likelihood that someone is carrying an umbrella given that it is not raining.

To find our answers, we’ll use the following two conditional probability formulas:

The researchers find the following probabilities:

• P(Umbrella ⋂ Rain): 0.20
• P(Umbrella ⋂ No Rain): 0.40
• P(Rain): 0.25
• P(No Rain): 0.75

Calculating the Conditional Probabilities

If you look at the joint likelihoods in the first two bullets, it appears that the chance of carrying an umbrella when it’s not raining (0.40) is higher than when it is raining (0.20). That seems backward, and we’ll come back to that. For the correct answer, we need to calculate the conditional probability. Let’s plug these numbers into the conditional probability formula!

Based on the conditional probabilities, we see that people are more likely to carry an umbrella given that it’s raining (0.80) compared to when it’s not raining (0.53). That makes sense!

The joint likelihoods were misleading because they do not account for the fact that days with no rain are three times as likely as rainy days (0.75 vs. 0.25)! In this study, you’re more likely to see people carrying umbrellas when there’s no rain because there are many more days with no rain. The conditional probabilities consider that fact.

This example of conditional probabilities involves dependent events. We know that’s the case because when you change the initial event, the chances of the second event change. Consequently, the likelihood of the second event depends on the first event.

Now, let’s see what happens when we look at independent events.

Related post : Venn diagrams can display probabilities effectively

Example of Conditional Probabilities with Independent Events

When you assess conditional probabilities of independent events, the following is true:

P (A|B) = P (A)

What does that mean?

The probability of A given that B occurred equals the likelihood of A. In other words, whether B occurs or not has no effect on the chances of A. That makes sense because the events are independent ! There’s also a mathematical proof for it, which I won’t cover.

Imagine we’re playing a game. For each turn, you roll two dice, but you roll them one at a time. You want to roll two sixes. The dice rolls are independent events because the outcome of the first roll does not affect the second roll.

Learn more about Independent Events: Definition & Probability .

Calculations

We’ll use the conditional probability formula to find the likelihood of rolling a second six given that the first roll was a six. For this example, I’ll denote the first six as 6 1 and the second as 6 2 .

Because these are independent events, we can use the multiplication rule to calculate the joint probability of P (P 2 ⋂ P 1 ).  Each roll has a 1/6 = 0.167 chance of getting a six.

Consequently, the joint likelihood of two sixes is: P (6 2 ⋂ P 1 ) = 0.167 * 0.167 = 0.028

And the chance of rolling a six on the first throw: P (6 1 ) = 0.167

By entering these values into the conditional probability formula, we know that the chances of rolling a 2 nd six given that the first roll was a six is the following:

No surprise. The chance of throwing that 2 nd six is still 1/6. The first six doesn’t affect the likelihood at all, which makes sense for independent events!

You can also calculate conditional probabilities and other types using contingency tables. To learn about that, read my post Using Contingency Tables to Calculate Probabilities .

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Reader Interactions

January 11, 2022 at 1:38 am

Appreciate the explanation. I do have a question which I think conditional probability is the best solution but I could be wrong.

For example: there are 20 envelopes. 10 have money and 10 do not. What is the probability of selecting an envelope with money before selecting 3 envelopes with no money? I am needing to find out the probability of selecting 1 thru 10 with money before getting 3 with no money.

Clear as mud?

October 19, 2021 at 10:52 am

I think you mean 6 not P on this statement

“Consequently, the joint probability of two sixes is: P (62 ⋂ 61) = 0.167 * 0.167 = 0.028”

October 20, 2021 at 12:58 am

What I wrote is correct. It’s standard probability notation the signifies the joint probability of rolling two sixes. The P just means “probability.”

October 18, 2021 at 7:59 am

Very clear explanation. Thank you.

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1.4.5 Solved Problems: Conditional Probability

In die and coin problems, unless stated otherwise, it is assumed coins and dice are fair and repeated trials are independent.

You purchase a certain product. The manual states that the lifetime $T$ of the product, defined as the amount of time (in years) the product works properly until it breaks down, satisfies $$P(T \geq t)=e^{-\frac{t}{5}}, \textrm{ for all } t \geq 0.$$ For example, the probability that the product lasts more than (or equal to) $2$ years is $P(T \geq 2)=e^{-\frac{2}{5}}=0.6703$. I purchase the product and use it for two years without any problems. What is the probability that it breaks down in the third year?

• What is the probability of three heads, $HHH$?
• What is the probability that you observe exactly one heads?
• Given that you have observed at least one heads, what is the probability that you observe at least two heads?
• $P(HHH)=P(H)\cdot P(H) \cdot P(H)=0.5^3=\frac{1}{8}$.

• $A$ and $C$ are independent,
• $B$ and $C$ are independent,
• $A$ and $B$ are disjoint,
• $P(A \cup C)=\frac{2}{3}, P(B \cup C)=\frac{3}{4}, P(A \cup B\cup C)=\frac{11}{12}$

We can use the Venn diagram in Figure 1.26 to better visualize the events in this problem. We assume $P(A)=a, P(B)=b$, and $P(C)=c$. Note that the assumptions about independence and disjointness of sets are already included in the figure.

Now we can write $$P(A \cup C)= a+c-ac=\frac{2}{3};$$ $$P(B \cup C)=b+c-bc=\frac{3}{4};$$ $$P(A \cup B\cup C)=a+b+c-ac-bc=\frac{11}{12}.$$ By subtracting the third equation from the sum of the first and second equations, we immediately obtain $c=\frac{1}{2}$, which then gives $a=\frac{1}{3}$ and $b=\frac{1}{2}$.

• $A$ and $B$ are conditionally independent given $C_i$, for all $i \in \{1,2,\cdots,M\}$;
• $B$ is independent of all $C_i$'s.

• What is the probability that it's not raining and there is heavy traffic and I am not late?
• What is the probability that I am late?
• Given that I arrived late at work, what is the probability that it rained that day?

Let $R$ be the event that it's rainy, $T$ be the event that there is heavy traffic, and $L$ be the event that I am late for work. As it is seen from the problem statement, we are given conditional probabilities in a chain format. Thus, it is useful to draw a tree diagram. Figure 1.27 shows a tree diagram for this problem. In this figure, each leaf in the tree corresponds to a single outcome in the sample space. We can calculate the probabilities of each outcome in the sample space by multiplying the probabilities on the edges of the tree that lead to the corresponding outcome.

• You pick a coin at random and toss it. What is the probability that it lands heads up?
• You pick a coin at random and toss it, and get heads. What is the probability that it is the two-headed coin?

Here is another variation of the family-with-two-children problem [1] [7] . A family has two children. We ask the father, "Do you have at least one daughter named Lilia?" He replies, "Yes!" What is the probability that both children are girls? In other words, we want to find the probability that both children are girls, given that the family has at least one daughter named Lilia. Here you can assume that if a child is a girl, her name will be Lilia with probability $\alpha \ll 1$ independently from other children's names. If the child is a boy, his name will not be Lilia. Compare your result with the second part of Example 1.18 .

Let's compare the result with part (b) of Example 1.18. Amazingly, we notice that the extra information about the name of the child increases the conditional probability of $GG$ from $\frac{1}{3}$ to about $\frac{1}{2}$. How can we explain this intuitively? Here is one way to look at the problem. In part (b) of Example 1.18, we know that the family has at least one girl. Thus, the sample space reduces to three equally likely outcomes: $GG, GB, BG$, thus the conditional probability of $GG$ is one third in this case. On the other hand, in this problem, the available information is that the event $L$ has occurred. The conditional sample space here still is $GG, GB, BG$, but these events are not equally likely anymore. A family with two girls is more likely to name at least one of them Lilia than a family who has only one girl ($P(L|BG)=P(L|GB)=\alpha$, $ P(L|GG)=2 \alpha-\alpha^2$), thus in this case the conditional probability of $GG$ is higher. We would like to mention here that these problems are confusing and counterintuitive to most people. So, do not be disappointed if they seem confusing to you. We seek several goals by including such problems.

First, we would like to emphasize that we should not rely too much on our intuition when solving probability problems. Intuition is useful, but at the end, we must use laws of probability to solve problems. Second, after obtaining counterintuitive results, you are encouraged to think deeply about them to explain your confusion. This thinking process can be very helpful to improve our understanding of probability. Finally, I personally think these paradoxical-looking problems make probability more interesting.

So the answer again is different from the second part of Example 1.18. This is surprising to most people. The two problem statements look very similar but the answers are completely different. This is again similar to the previous problem (please read the explanation there). The conditional sample space here still is $GG, GB, BG$, but the point here is that these are not equally likely as in Example 1.18. The probability that a randomly chosen child from a family with two girls is a girl is one, while this probability for a family who has only one girl is $\frac{1}{2}$. Thus, intuitively, the conditional probability of the outcome $GG$ in this case is higher than $GB$ and $BG$, and thus this conditional probability must be larger than one third.

Okay, another family-with-two-children problem. Just kidding! This problem has nothing to do with the two previous problems. I toss a coin repeatedly. The coin is unfair and $P(H)=p$. The game ends the first time that two consecutive heads ($HH$) or two consecutive tails ($TT$) are observed. I win if $HH$ is observed and lose if $TT$ is observed. For example if the outcome is $HTH\underline{TT}$, I lose. On the other hand, if the outcome is $THTHT\underline{HH}$, I win. Find the probability that I win.

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Conditional Probability

Conditional probability is one type of probability in which the possibility of an event depends upon the existence of a previous event. As this type of event is very common in real life, conditional probability is often used to determine the probability of such cases.

Conditional probability is  the likelihood of an outcome occurring based on a previous outcome in similar circumstances. In probability notation, this is denoted as A given B, expressed as P(A|B), indicating that the probability of event A is dependent on the occurrence of event B.

To know about conditional probability, we need to be familiar with independent events and dependent events. Let’s understand conditional probability, and its formula with solved examples in this article.

Table of Content

What is Conditional Probability?

Conditional probability definition, conditional probability formula, how to find probability of one event given another, conditional probability of independent events, conditional probability vs joint probability vs marginal probability, conditional probability and bayes’ theorem, conditional probability examples, tossing a coin, drawing cards, properties of conditional probability , multiplication rule of probability, how to apply the multiplication rule, applications of conditional probability, conditional probability sample problems, unsolved practice problems, resources related to conditional probability:.

Conditional probability is the probability that depends on a previous result or event . Due to this fact, they help us understand how events are related to each other. Simply put, conditional probability tells us the likelihood of the occurrence of an event based on the occurrence of some previous outcome. 

With the help of conditional probability, we can tell apart dependent and independent events . When the probability of one event happening doesn’t influence the probability of any other event, then events are called independent, otherwise dependent events.

Conditional Probability is defined as the probability of any event occurring when another event has already occurred. In other words, it calculates the probability of one event happening given that a certain condition is satisfied . It is represented as P (A | B) which means the probability of A when B has already happened.

For Example, let’s consider the case of rolling two dice, sample space of this event is as follows:

{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

Now, consider an event A = getting 3 on the first die and B = getting a sum of 9.

Then the probability of getting 9 when on the first die it’s already 3 is P(B | A),

which can be calculated as follows:

All the cases for the first die as 3 are (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6).

In all of these cases, only one case has a sum of 9.

Thus, P (B | A) = 1/36.

In case, we have to find P (A | B),

All cases where the sum is 9 are (3, 6), (4, 5), (5, 4), and (6, 3).

In all of these cases, only one case has 3 on the first die i.e., (3, 6)

Thus, P(A | B) = 1/36.

As we can calculate the Conditional Probability of simple cases without any formula, just as we have seen in the above heading but for complex cases, we need a Conditional Probability Equation as we can’t possibly count all the cases for those. Let’s consider two events A and B, then the formula for conditional probability of A when B has already occurred is given by:

P(A|B) = P (A ∩ B) / P(B) Where, P (A ∩ B) represents the probability of both events A and B occurring simultaneously. P(B) represents the probability of event B occurring.

In other words, the conditional probability of A given B has already occurred is equal to the probability of the intersection of A and B divided by the probability of event B.

To calculate the conditional probability, we can use the following step-by-step method:

Step 1: I dentify the Events. Let’s call them Event A and Event B. Step 2: Determine the Probability of Event A i.e., P(A) Step 3: Determine the Probability of Event B i.e., P(B) Step 4: Determine the Probability of Event A and B i.e., P(A∩B). Step 5: Apply the Conditional Probability Formula and calculate the required probability.

When two events are independent, those conditional probability is the same as the probability of the event individually i.e., P (A | B) is the same as P(A) as there is no effect of event B on the probability of event A. For independent events, A and B, the conditional probability of A and B with respect to each other is given as follows:

• P(B|A) = P(B)
• P(A|B) = P(A)

Check, Probability Formulas

The difference between Conditional Probability, Joint Probability , and Marginal Probability is given in the following table:

Bayes’ Theorem is a fundamental concept in probability theory named after the Reverend Thomas Bayes . It provides a mathematical framework for updating beliefs or hypotheses in light of new evidence or information. This theorem is extensively used in various fields, including statistics, machine learning, and artificial intelligence.

At its core, Bayes’ Theorem enables us to calculate the probability of a hypothesis being true given observed evidence. The theorem is expressed mathematically as follows:

P(A∣B) = (P(B∣A) × P(A))​ / P(B)

• P(A∣B) is the posterior probability of hypothesis A given evidence B .
• P(B∣A) is the likelihood of observing evidence B given that hypothesis A is true.
• P(A) is the prior probability of hypothesis A before observing any evidence.
• P(B) is the probability of observing evidence B regardless of the truth of hypothesis A .

Here’s a breakdown of how Bayes’ Theorem works:

Prior Probability P(A): This represents our initial belief in the likelihood of hypothesis A being true before considering any new evidence. Likelihood P(B∣A): This indicates the probability of observing the evidence B given that hypothesis A is true. It quantifies how well the evidence supports the hypothesis. Evidence P(B): This term serves as a normalization factor and represents the total probability of observing the evidence B across all possible hypotheses. Posterior Probability P(A∣B) : This is the updated probability of hypothesis A being true after taking into account the observed evidence B . It’s what we’re ultimately interested in determining.

Bayes’ Theorem is particularly powerful because it allows us to incorporate new evidence incrementally, refining our beliefs as more data becomes available . This iterative process of updating beliefs with new evidence forms the basis of Bayesian inference, which is widely used in fields such as medical diagnosis, spam filtering, weather forecasting, and many others.

Bayes’ Theorem provides a principled approach for reasoning under uncertainty, making it a cornerstone of probabilistic reasoning and decision-making in diverse domains.

Read in Detail: Bayes’s Theorem for Conditional Probability

There are various examples of conditional probability as in real life all the events are related to each other and happening any event affects the probability of another event. For example , if it rains, the probability of road accidents increases as roads have less friction. Let’s consider some problem-based examples here:

Let’s consider two events in tossing two coins be,

• A: Getting a head on the first coin.
• B: Getting a head on the second coin.

Sample space for tossing two coins is:

S = {HH, HT, TH, TT}

Conditional probability of getting a head on the second coin (B) given that we got a head on the first coin (A) is = P(B|A)

Since the coins are independent (one coin’s outcome does not affect the other), P(B|A) = P(B) = 0.5 (50%), which is the probability of getting a head on a single coin toss.

In a deck of 52 cards where two cards are being drawn, then let’s consider the events be.

• A: Drawing a red card on the first draw, and
• B: Drawing a red card on the second draw.

Conditional probability of drawing a red card on the second draw (B) given that we drew a red card on the first draw (A) is = P(B|A) 

After drawing a red card on the first draw, there are 25 red cards and 51 cards remaining in the deck. So, P(B|A) = 25/51 ≈ 0.49 (approximately 49%).

Some of the common properties of conditional property are:

Property 1: Let’s consider an event A in any sample space S of an experiment.

P(S|A) = P(A|A) = 1

Property 2: For any two events A and B of a sample space S, and an event X such that P(X) ≠ 0,

P((A ∪ B)|X) = P(A|X) + P(B|X) – P((A ∩ B)|X)

Property 3: The order of set or events is important in conditional probability, i.e.,

P(A|B) ≠ P(B|A)

Property 4: The complement formula for probability only holds conditional probability if it is given in the context of the first argument in conditional probability i.e.,

P(A’|B)=1-P(A|B) P(A|B’) ≠ 1-P(A|B)

Property 5: For any two or three independent events, the intersection of events can be calculated using the following formula:

• For the intersection of two events A and B,

P(A ⋂ B) = P(A) P(B)

• For the intersection of three events A, B, and C,

P (A ⋂ B ⋂ C) = P(A) P(B) P(C)

Multiplication Rule of Probability , when applied in the context of conditional probability, helps us calculate the probability of the intersection of two events when the probability of one event depends on the occurrence of the other event. This rule is crucial in understanding the joint probability of events under specific conditions.

In the context of conditional probability, the Multiplication Rule is often stated as follows:

P(A∩B) = P(A) × P(B∣A)

Here’s what each term represents :

• P(A∩B) : This denotes the probability that both events A and B occur simultaneously.
• P(A) : This represents the probability of event A happening.
• P(B∣A) : This is the conditional probability of event B occurring given that event A has already occurred.

To apply the Multiplication Rule in the context of conditional probability, we can use the following steps:

• First we calculate the probability of event A occurring.
• Then, we compute the probability of event B occurring given that event A has occurred.
• Multiplying these probabilities together gives us the joint probability of both events happening under the specified conditions.
• This rule is particularly useful when dealing with events that are not independent, meaning that the occurrence of one event affects the probability of the other event.

Various applications of conditional probability are,

Finance and Risk Management

• Example: Assessing the probability of default for a borrower given certain financial indicators.
• Application: Banks and financial institutions use conditional probability to evaluate the risk associated with loans and investments.

Healthcare and Diagnostics

• Example: Determining the probability of a patient having a specific disease given the results of diagnostic tests.
• Application: Conditional probability is crucial in medical diagnoses and decision-making, helping healthcare professionals make informed decisions based on test results.

Marketing and Customer Relationship Management (CRM)

• Example: Predicting the probability of a customer making a purchase based on their past buying behavior.
• Application: Businesses use conditional probability to tailor marketing strategies, optimize customer experiences, and personalize product recommendations.

Machine Learning and Artificial Intelligence

• Example: Predicting the likelihood of a user clicking on a particular ad based on their online behavior.
• Application: Conditional probability is fundamental in machine learning algorithms for tasks such as classification, recommendation systems, and natural language processing.

Weather Forecasting

• Example: Estimating the probability of rain tomorrow given today’s weather conditions.
• Application: Meteorologists use conditional probability to make weather predictions based on historical data and current atmospheric conditions.

Question 1: A bag contains 5 red balls and 7 blue balls. Two balls are drawn without replacement. What is the probability that the second ball drawn is red, given that the first ball drawn was red?

Let the events be, Event A: The first ball drawn is red. Event B: The second ball drawn is red. P(A) = 5/12  and P(B) = 4/11 (as first ball drawn is already red, thus only 4 red balls remain in the bag) Therefore, probability of the second ball drawn being red given that the first ball drawn was red is 4/11.

Question 2: A box contains 5 green balls and 3 yellow balls. Two balls are drawn without replacement. What is the probability that both balls are green?

Let events be: Event A: The first ball drawn is green, and Event B: The second ball drawn is green. P(A) = 5/8 P(B) = 4/7 (as there are 4 green balls left out of 7) Thus, probability that both balls drawn are green is (5/8) × (4/7) = 20/56 = 5/14.

Question 3: In a bag, there are 8 red marbles, 4 blue marbles, and 3 green marbles. If one marble is randomly drawn, what is the probability that it is not blue?

Let the events be: Event A: The marble drawn is not blue, and  Event B: The marble drawn is blue. As A and B are complementary Events, we know P(A) = 1 – P(B) ⇒ P(A) = 1 – 4/15  ⇒ P(A) = (15 – 4)/15  ⇒ P(A) = 11/15 Thus, probability of drawing a marble out of bag which is not blue is 11/15.

Question 4: In a survey among a group of students, 70% play football, 60% play basketball, and 40% play both sports. If a student is chosen at random and it is known that the student plays basketball, what is the probability that the student also plays football?

Let’s assume there are 100 students in the survey. Number of students who play football = n(A) = 70 Number of students who play basketball = n(B) = 60 Number of students who play both sports = n(A ∩ B) = 40 To find the probability that a student plays football given that they play basketball, we use the conditional probability formula : P(A|B) = n(A ∩ B) / n(B) Substituting the values, we get: P(A|B) = 40 / 60 = 2/3 Therefore, probability that a randomly chosen student who plays basketball also plays football is 2/3.

Question 5: In a deck of 52 playing cards, 4 cards are drawn without replacement. What is the probability that all 4 cards are aces, given that the first card drawn is an ace?

Let the events be, Event A: The first card drawn is an ace, Event B: The second card drawn is an ace, Event C: The third card drawn is an ace, and Event D: The fourth card drawn is an ace. P(A) = 4/52 (there are 4 ace out of 52) P(B | A) = 3/51 (one is already drawn, thus 3 ace left) P(C | A and B) = 2/50 (two is already drawn, thus 2 ace left) P(D | A and B and C) = 1/49 (three is already drawn, thus 1 ace left) To find the probability that all four cards are aces, we multiply the probabilities of the individual events. P(A and B and C and D) = P(A) × P(B|A) × P(C|A and B) × P(D|A and B and C) = (4/52) × (3/51) × (2/50) × (1/49) = 1/270725 Therefore, probability that all 4 cards drawn are aces, given that the first card drawn is an ace, is 1/270725 .

Question 6: In a certain city, it rains 30% of the days. A weather forecaster correctly predicts rain 80% of the time when it actually rains, and correctly predicts no rain 90% of the time when it doesn’t rain. If the forecast predicts rain, what is the probability that it will actually rain?

Let R be the event that it rains, and F be the event that rain is forecast. P(R|F) = P(F|R) × P(R) / P(F) P(F) = P(F|R) × P(R) + P(F|not R) × P(not R) = 0.8 × 0.3 + 0.1 × 0.7 = 0.31 P(R|F) = (0.8 × 0.3) / 0.31 = 0.7059

Question 7: A fair die is rolled twice. Given that the sum of the two rolls is even, what is the probability that the first roll was an even number?

There are 18 ways to get an even sum (out of 36 total outcomes). 12 of these 18 ways have an even number on the first roll. P(First even | Sum even) = 12/18 = 2/3

Question 8: A diagnostic test for a disease has a false positive rate of 2% and a false negative rate of 3%. The disease occurs in 1% of the population. If a person tests positive, what is the probability that they actually have the disease?

Let D be the event of having the disease, and T be the event of testing positive. P(D|T) = P(T|D) × P(D) / P(T) P(T) = P(T|D) × P(D) + P(T|not D) × P(not D) = 0.97 × 0.01 + 0.02 × 0.99 = 0.0293 P(D|T) = (0.97 × 0.01) / 0.0293 = 0.3305

Question 9: In a bag, there are 4 red balls and 6 blue balls. Two balls are drawn without replacement. What is the probability that the second ball is red, given that the first ball drawn was blue?

After drawing a blue ball, there are 3 red balls and 5 blue balls left. P(Second is red | First was blue) = 3 / (3 + 5) = 3/8 = 0.444

Question 10: A software company has two development teams: Team A and Team B. Team A completes 60% of all projects, while Team B handles the rest. Team A has a 95% success rate, while Team B has an 85% success rate. If a project is successful, what is the probability that it was handled by Team A?

Let S be the event that a project is successful, and A be the event that Team A handled the project. P(A|S) = P(S|A) × P(A) / P(S) P(S) = P(S|A) × P(A) + P(S|B) × P(B) = 0.95 × 0.6 + 0.85 × 0.4 = 0.91 P(A|S) = (0.95 × 0.6) / 0.91 = 0.6316

P1. A card is drawn at random from a standard 52-card deck. Given that the card drawn is a face card (Jack, Queen, or King), what is the probability that it’s a heart?

P2. In a class of 30 students, 18 play basketball and 12 play football. If 6 students play both sports, what is the probability that a randomly selected student plays basketball, given that they play football?

P3. A bag contains 5 red marbles, 3 blue marbles, and 2 green marbles. Two marbles are drawn without replacement. What is the probability that the second marble is blue, given that the first marble drawn was red?

P4. A manufacturing process produces 5% defective items. The quality control system detects 98% of defective items and 3% of non-defective items are incorrectly identified as defective. If an item is identified as defective by the quality control system, what is the probability that it is actually defective?

P5. A family has two children. Given that at least one of the children is a boy, what is the probability that both children are boys? Assume that the probability of having a boy or a girl is equal.

P6. In a certain population, 10% of people have a particular disease. A test for this disease correctly identifies 95% of people who have the disease (true positives) and 90% of people who don’t have the disease (true negatives). If a person tests positive, what is the probability that they actually have the disease?

P7. A software has three modules: A, B, and C. The probabilities of a bug being in these modules are 0.3, 0.4, and 0.3 respectively. If a bug is found in module A, the probability of it affecting module B is 0.2 and module C is 0.1. What is the probability that a bug affects both modules A and B?

P8. In a group of 100 people, 60 speak English, 40 speak French, and 20 speak both languages. If a person is selected at random and is known to speak French, what is the probability that they also speak English?

P9. A student has to answer 3 out of 5 questions on an exam. The student knows the answers to 3 questions for sure, has a 50% chance of answering the 4th question correctly, and doesn’t know the answer to the 5th question at all. If the student answers a question correctly, what is the probability that it was one of the questions they knew for sure?

P10. A biased coin has a 60% chance of landing heads. It is flipped twice. Given that at least one of the flips resulted in heads, what is the probability that both flips resulted in heads?

Probability Class 12 Notes Probability Class 12 NCERT Solutions

Summary – Conditional Probability

Conditional probability is a fundamental concept in probability theory that quantifies the likelihood of an event occurring given that another event has already occurred. Expressed as P(A|B), it represents the probability of event A happening under the condition that event B has occurred. This concept is crucial in various fields such as statistics, finance, and machine learning, where understanding the relationship between events and their outcomes is essential for making informed decisions. Conditional probability allows us to assess the impact of one event on the occurrence of another, providing valuable insights into probabilistic relationships and dependencies.

FAQs on Conditional Probability

What is conditional probability.

Conditional probability is a measure of the probability of an event occurring given that another event has already occurred or is known to have occurred. It is denoted as P(A|B), which reads as “the probability of event A given event B” .

What is difference between conditional probability and joint probability?

Conditional probability focuses on the probability of one event given that another event has already occurred while Joint probability focuses on the probability of multiple events occurring simultaneously.

What are some real-world examples of conditional probability?

The probability of a person having a specific disease given that they tested positive. The probability of it raining today given that it is cloudy. The probability of a student passing an exam given that they studied.

How is conditional probability calculated?

Conditional Probability Equation is calculated using the formula:  P(A|B) = P(A ∩ B) / P(B)

What is the difference between conditional probability and regular probability?

Regular probability , often referred to as unconditional probability, calculates the likelihood of an event occurring without any prior information and is the basic probability of an event in isolation. Conditional probability takes into account additional information or the occurrence of another event to calculate the probability of a particular event.

Can conditional probability be greater than 1?

No, conditional probability cannot be greater than 1 as the conditional probability is a type of probability. It can only be between 0 and 1.

Can conditional probability be used in Bayesian inference?

Yes, conditional probability plays a crucial role in Bayesian inference, where prior beliefs are updated based on observed evidence using Bayes’ theorem.

What is the relationship between conditional probability and independence?

Two events A and B are considered independent if the occurrence of one event does not affect the probability of the other event. In terms of conditional probability , if events A and B are independent, then P(A|B) = P(A), and similarly, P(B|A) = P(B). In other words, the conditional probability of A given B is equal to the unconditional probability of A, and vice versa.

Can conditional probability be negative?

No, conditional probability cannot be negative as probabilities are always non-negative values between 0 and 1.

What is the interpretation of conditional probability?

Conditional probability measures the likelihood of one event occurring, given that another event has already occurred. It helps in updating probabilities based on new information.

What is conditional probability with an example?

Conditional probability , denoted P(A|B), assesses the likelihood of event A happening, given that event B has already occurred. For example, in a medical test scenario, it calculates the chance of having a disease when the test result is positive. It’s a key concept in probability theory, helping us understand real-world situations where one event depends on another.

What are the applications of conditional probability?

Conditional probability has diverse applications across fields such as f inance, medicine, genetics, marketing, weather forecasting, manufacturing, sports analytics, criminal justice, and environmental science.

What are the types of conditional probability?

Various types of conditional probability includes: Simple Conditional Probability Joint Probability Marginal Probability Conditional Probability Distribution Posterior Probability Prior Probability Sequential Conditional Probability Time Series Analysis Markov Chains

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Conditional Probability

In these lessons, we will learn what is conditional probability and how to use the formula for conditional probability.

Related Pages Dependent Events More Lessons On Probability Probability Tree Diagrams

The following diagram shows the formula for conditional probability. Scroll down the page for more examples and solutions on finding the conditional probability.

What Is Conditional Probability?

The probability of an event occurring given that another event has already occurred is called a conditional probability .

Recall that when two events, A and B, are dependent , the probability of both occurring is:

P(A and B) = P(A) × P(B given A) or P(A and B) = P(A) × P(B | A)

If we divide both sides of the equation by P(A) we get the Formula for Conditional Probability

How To Find The Conditional Probability From A Word Problem?

Step 1: Write out the Conditional Probability Formula in terms of the problem Step 2: Substitute in the values and solve.

Example: Susan took two tests. The probability of her passing both tests is 0.6. The probability of her passing the first test is 0.8. What is the probability of her passing the second test given that she has passed the first test?

Example: A bag contains red and blue marbles. Two marbles are drawn without replacement. The probability of selecting a red marble and then a blue marble is 0.28. The probability of selecting a red marble on the first draw is 0.5. What is the probability of selecting a blue marble on the second draw, given that the first marble drawn was red?

Solution: What is the probability that the total of two dice will be greater than 9, given that the first die is a 5?

Solution: Let A = first die is 5 Let B = total of two dice is greater than 9

Possible outcomes for A and B: (5, 5), (5, 6)

How To Use Real World Examples To Explain Conditional Probability?

Conditional probability is about narrowing down the set of possible circumstances so that the statistics can be measured more accurately.

How To Define Conditional Probability?

This video introduces the basic definition of conditional probability as it is defined in standard probability theory.

How To Calculate Conditional Probability?

Tutorial on how to calculate conditional probability for two events P(A), P(B), P(B|A) with two examples.

• Example 1: What is the probability of rolling a dice and its value is less than 4 knowing that the value is an odd number?
• Example 2: What is the probability of rolling a dice and its value is 1 knowing that the value is an odd number?

How To Determine The Conditional Probability From The Given Word Problems?

• You roll one 6-sided die, what is the probability of a 3 given you know the number is odd?
• At P-Town High School, the probability that a student takes Computer Programming and Spanish is 0.15. The probability that a student takes Computer Programming is 0.4. What is the probability that a student takes Spanish given that the student is taking Computer Programming?
• Here are the results of a survey completed with adult parents with children. What is the probability a person thinks college is too expensive given they have a child in college?
• Two cards are drawn without replacement in succession. What is the probability that the second card drawn is an ace, given that the first can drawn was an ace?
• Two cards are drawn without replacement. What is the probability the second card is a red face card given the first card is a ref face card?

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Conditional Statement – Definition, Truth Table, Examples, FAQs

What is a conditional statement, how to write a conditional statement, what is a biconditional statement, solved examples on conditional statements, practice problems on conditional statements, frequently asked questions about conditional statements.

A conditional statement is a statement that is written in the “If p, then q” format. Here, the statement p is called the hypothesis and q is called the conclusion. It is a fundamental concept in logic and mathematics. 

Conditional statement symbol :  p → q

A conditional statement consists of two parts.

• The “if” clause, which presents a condition or hypothesis.
• The “then” clause, which indicates the consequence or result that follows if the condition is true. 

Example : If you brush your teeth, then you won’t get cavities.

Hypothesis (Condition): If you brush your teeth

Conclusion (Consequence): then you won’t get cavities 

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Conditional Statement: Definition

A conditional statement is characterized by the presence of “if” as an antecedent and “then” as a consequent. A conditional statement, also known as an “if-then” statement consists of two parts:

• The “if” clause (hypothesis): This part presents a condition, situation, or assertion. It is the initial condition that is being considered.
• The “then” clause (conclusion): This part indicates the consequence, result, or action that will occur if the condition presented in the “if” clause is true or satisfied. 

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Representation of Conditional Statement

The conditional statement of the form ‘If p, then q” is represented as p → q. 

It is pronounced as “p implies q.”

Different ways to express a conditional statement are:

• p implies q
• p is sufficient for q
• q is necessary for p

Parts of a Conditional Statement

There are two parts of conditional statements, hypothesis and conclusion. The hypothesis or condition will begin with the “if” part, and the conclusion or action will begin with the “then” part. A conditional statement is also called “implication.”

Conditional Statements Examples:

Example 1: If it is Sunday, then you can go to play. 

Hypothesis: If it is Sunday

Conclusion: then you can go to play. 

Example 2: If you eat all vegetables, then you can have the dessert.

Condition: If you eat all vegetables

Conclusion: then you can have the dessert 

To form a conditional statement, follow these concise steps:

Step 1 : Identify the condition (antecedent or “if” part) and the consequence (consequent or “then” part) of the statement.

Step 2 : Use the “if… then…” structure to connect the condition and consequence.

Step 3 : Ensure the statement expresses a logical relationship where the condition leads to the consequence.

Example 1 : “If you study (condition), then you will pass the exam (consequence).” 

This conditional statement asserts that studying leads to passing the exam. If you study (condition is true), then you will pass the exam (consequence is also true).

Example 2 : If you arrange the numbers from smallest to largest, then you will have an ascending order.

Hypothesis: If you arrange the numbers from smallest to largest

Conclusion: then you will have an ascending order

Truth Table for Conditional Statement

The truth table for a conditional statement is a table used in logic to explore the relationship between the truth values of two statements. It lists all possible combinations of truth values for “p” and “q” and determines whether the conditional statement is true or false for each combination. 

The truth value of p → q is false only when p is true and q is False. 

If the condition is false, the consequence doesn’t affect the truth of the conditional; it’s always true.

In all the other cases, it is true.

The truth table is helpful in the analysis of possible combinations of truth values for hypothesis or condition and conclusion or action. It is useful to understand the presence of truth or false statements. 

Converse, Inverse, and Contrapositive

The converse, inverse, and contrapositive are three related conditional statements that are derived from an original conditional statement “p → q.” 

Consider a conditional statement: If I run, then I feel great.

• Converse: 

The converse of “p → q” is “q → p.” It reverses the order of the original statement. While the original statement says “if p, then q,” the converse says “if q, then p.” 

Converse: If I feel great, then I run.

• Inverse: 

The inverse of “p → q” is “~p → ~q,” where “” denotes negation (opposite). It negates both the antecedent (p) and the consequent (q). So, if the original statement says “if p, then q,” the inverse says “if not p, then not q.”

Inverse : If I don’t run, then I don’t feel great.

• Contrapositive: 

The contrapositive of “p → q” is “~q → ~p.” It reverses the order and also negates both the statements. So, if the original statement says “if p, then q,” the contrapositive says “if not q, then not p.”

Contrapositive: If I don’t feel great, then I don’t run.

A biconditional statement is a type of compound statement in logic that expresses a bidirectional or two-way relationship between two statements. It asserts that “p” is true if and only if “q” is true, and vice versa. In symbolic notation, a biconditional statement is represented as “p ⟺ q.”

In simpler terms, a biconditional statement means that the truth of “p” and “q” are interdependent. 

If “p” is true, then “q” must also be true, and if “q” is true, then “p” must be true. Conversely, if “p” is false, then “q” must be false, and if “q” is false, then “p” must be false. 

Biconditional statements are often used to express equality, equivalence, or conditions where two statements are mutually dependent for their truth values. 

Examples : 

• I will stop my bike if and only if the traffic light is red.  
• I will stay if and only if you play my favorite song.

Facts about Conditional Statements

• The negation of a conditional statement “p → q” is expressed as “p and not q.” It is denoted as “𝑝 ∧ ∼𝑞.” 
• The conditional statement is not logically equivalent to its converse and inverse.
• The conditional statement is logically equivalent to its contrapositive. 
• Thus, we can write p → q ∼q → ∼p

In this article, we learned about the fundamentals of conditional statements in mathematical logic, including their structure, parts, truth tables, conditional logic examples, and various related concepts. Understanding conditional statements is key to logical reasoning and problem-solving. Now, let’s solve a few examples and practice MCQs for better comprehension.

Example 1: Identify the hypothesis and conclusion. 

If you sing, then I will dance.

Solution : 

Given statement: If you sing, then I will dance.

Here, the antecedent or the hypothesis is “if you sing.”

The conclusion is “then I will dance.”

Example 2: State the converse of the statement: “If the switch is off, then the machine won’t work.” 

Here, p: The switch is off

q: The machine won’t work.

The conditional statement can be denoted as p → q.

Converse of p → q is written by reversing the order of p and q in the original statement.

Converse of  p → q is q → p.

Converse of  p → q: q → p: If the machine won’t work, then the switch is off.

Example 3: What is the truth value of the given conditional statement? 

If 2+2=5 , then pigs can fly.

Solution:  

q: Pigs can fly.

The statement p is false. Now regardless of the truth value of statement q, the overall statement will be true. 

F → F = T

Hence, the truth value of the statement is true. 

Conditional Statement - Definition, Truth Table, Examples, FAQs

Attend this quiz & Test your knowledge.

What is the antecedent in the given conditional statement? If it’s sunny, then I’ll go to the beach.

A conditional statement can be expressed as, what is the converse of “a → b”, when the antecedent is true and the consequent is false, the conditional statement is.

What is the meaning of conditional statements?

Conditional statements, also known as “if-then” statements, express a cause-and-effect or logical relationship between two propositions.

When does the truth value of a conditional statement is F?

A conditional statement is considered false when the antecedent is true and the consequent is false.

What is the contrapositive of a conditional statement?

The contrapositive reverses the order of the statements and also negates both the statements. It is equivalent in truth value to the original statement.

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Bayes' Theorem and Conditional Probability

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Applied probability.

A framework for understanding the world around us, from sports to science.

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Bayes' theorem is a formula that describes how to update the probabilities of hypotheses when given evidence. It follows simply from the axioms of conditional probability , but can be used to powerfully reason about a wide range of problems involving belief updates.

Given a hypothesis \(H\) and evidence \(E\), Bayes' theorem states that the relationship between the probability of the hypothesis before getting the evidence \(P(H)\) and the probability of the hypothesis after getting the evidence \(P(H \mid E)\) is

\[P(H \mid E) = \frac{P(E \mid H)} {P(E)} P(H).\]

Many modern machine learning techniques rely on Bayes' theorem. For instance, spam filters use Bayesian updating to determine whether an email is real or spam, given the words in the email. Additionally, many specific techniques in statistics, such as calculating \(p\)-values or interpreting medical results , are best described in terms of how they contribute to updating hypotheses using Bayes' theorem.

Explaining Counterintuitive Results

Deriving bayes' theorem, visualizing bayes’ theorem, diagnosing disease, more examples.

Probability problems are notorious for yielding surprising and counterintuitive results. One famous example--or a pair of examples--is the following:

A couple has 2 children and the older child is a boy. If the probabilities of having a boy or a girl are both 50%, what's the probability that the couple has two boys? We already know that the older child is a boy. The probability of two boys is equivalent to the probability that the younger child is a boy, which is \(50\%\). A couple has two children, of which at least one is a boy. If the probabilities of having a boy or a girl are both \(50\%\), what is the probability that the couple has two boys? At first glance, this appears to be asking the same question. We might reason as follows: “We know that one is a boy, so the only question is whether the other one is a boy, and the chances of that being the case are \(50\%\). So again, the answer is \(50\%\).” This makes perfect sense. It also happens to be incorrect.

Bayes' theorem centers on relating different conditional probabilities . A conditional probability is an expression of how probable one event is given that some other event occurred (a fixed value). For instance, "what is the probability that the sidewalk is wet?" will have a different answer than "what is the probability that the sidewalk is wet given that it rained earlier?"

For a joint probability distribution over events \(A\) and \(B\), \(P(A \cap B)\), the conditional probability of \(A\) given \(B\) is defined as

\[P(A\mid B) = \frac{P(A\cap B)}{P(B)}.\]

In the sidewalk example, where \(A\) is "the sidewalk is wet" and \(B\) is "it rained earlier," this expression reads as "the probability the sidewalk is wet given that it rained earlier is equal to the probability that the sidewalk is wet and it rains over the probability that it rains."

Note that \(P(A \cap B)\) is the probability of both \(A\) and \(B\) occurring, which is the same as the probability of \(A\) occurring times the probability that \(B\) occurs given that \(A\) occurred: \(P(B \mid A) \times P(A).\) Using the same reasoning, \(P(A \cap B)\) is also the probability that \(B\) occurs times the probability that \(A\) occurs given that \(B\) occurs: \(P(A \mid B) \times P(B)\). The fact that these two expressions are equal leads to Bayes' Theorem. Expressed mathematically, this is:

\[\begin{align} P(A \mid B) &= \frac{P(A\cap B)}{P(B)}, \text{ if } P(B) \neq 0, \\ P(B \mid A) &= \frac{P(B\cap A)}{P(A)}, \text{ if } P(A) \neq 0, \\ \Rightarrow P(A\cap B) &= P(A\mid B)\times P(B)=P(B\mid A)\times P(A), \\ \Rightarrow P(A \mid B) &= \frac{P(B \mid A) \times P(A)} {P(B)}, \text{ if } P(B) \neq 0. \end{align}\]

Notice that our result for dependent events and for Bayes’ theorem are both valid when the events are independent. In these instances, \(P(A \mid B) = P(A)\) and \(P(B \mid A) = P(B)\), so the expressions simplify.

Bayes' Theorem \[P(A \mid B) = \frac{P(B \mid A)} {P(B)} P(A)\]

While this is an equation that applies to any probability distribution over events \(A\) and \(B\), it has a particularly nice interpretation in the case where \(A\) represents a hypothesis \(H\) and \(B\) represents some observed evidence \(E\). In this case, the formula can be written as

\[P(H \mid E) = \frac{P(E \mid H)}{P(E)} P(H).\]

This relates the probability of the hypothesis before getting the evidence \(P(H)\), to the probability of the hypothesis after getting the evidence, \(P(H \mid E)\). For this reason, \(P(H)\) is called the prior probability , while \(P(H \mid E)\) is called the posterior probability . The factor that relates the two, \(\frac{P(E \mid H)}{P(E)}\), is called the likelihood ratio . Using these terms, Bayes' theorem can be rephrased as "the posterior probability equals the prior probability times the likelihood ratio."

If a single card is drawn from a standard deck of playing cards, the probability that the card is a king is 4/52, since there are 4 kings in a standard deck of 52 cards. Rewording this, if \(\text{King}\) is the event "this card is a king," the prior probability \(P(\text{King}) = \frac{4}{52} = \frac{1}{13}.\) If evidence is provided (for instance, someone looks at the card) that the single card is a face card, then the posterior probability \(P(\text{King} \mid \text{Face})\) can be calculated using Bayes' theorem: \[P(\text{King} \mid \text{Face}) = \frac{P(\text{Face} \mid \text{King})}{P(\text{Face})} P(\text{King}).\] Since every King is also a face card, \(P(\text{Face} \mid \text{King}) = 1\). Since there are 3 face cards in each suit (Jack, Queen, King) , the probability of a face card is \(P(\text{Face}) = \frac{3}{13}\). Combining these gives a likelihood ratio of \(\frac{1}{\hspace{2mm} \frac3{13}\hspace{2mm} } = \frac{13}{3}\). Using Bayes' theorem gives \(P(\text{King} \mid \text{Face}) = \frac{13}{3} \frac{1}{13} = \frac{1}{3}\). \(_\square\)

You randomly choose a treasure chest to open, and then randomly choose a coin from that treasure chest. If the coin you choose is gold, then what is the probability that you chose chest A?

Bayes' theorem clarifies the two-children problem from the first section:

1. A couple has two children, the older of which is a boy. What is the probability that they have two boys? 2. A couple has two children, one of which is a boy. What is the probability that they have two boys? \[\] Define three events, \(A\), \(B\), and \(C\), as follows: \[ \begin{align} A & = \mbox{ both children are boys}\\ B & = \mbox{ the older child is a boy}\\ C & = \mbox{ one of their children is a boy.} \end{align}\] Question 1 is asking for \(P(A \mid B)\), and Question 2 is asking for \(P(A \mid C)\). The first is computed using the simpler version of Bayes’ theorem: \[P(A \mid B) = \frac{P(A)P(B \mid A)}{P(B)} = \frac{ \frac{1}{4}\cdot 1 }{\frac{1}{2}} = \frac{1}{2}.\] To find \(P(A \mid C)\), we must determine \(P(C)\), the prior probability that the couple has at least one boy. This is equal to \(1 - P(\mbox{both children are girls}) = 1 - \frac{1}{4}=\frac{3}{4}\). Therefore the desired probability is \[P(A \mid C) = \frac{P(A)P(C \mid A)}{P(C)} = \frac{\frac{1}{4}\cdot 1}{\frac{3}{4}} = \frac{1}{3}. \ _\square \] For a similarly paradoxical problem, see the Monty Hall problem .

Venn diagrams are particularly useful for visualizing Bayes' theorem, since both the diagrams and the theorem are about looking at the intersections of different spaces of events.

A disease is present in 5 out of 100 people, and a test that is 90% accurate (meaning that the test produces the correct result in 90% of cases) is administered to 100 people. If one person in the group tests positive, what is the probability that this one person has the disease?

The intuitive answer is that the one person is 90% likely to have the disease. But we can visualize this to show that it’s not accurate. First, draw the total population and the 5 people who have the disease:

The circle A represents 5 out 100, or 5% of the larger universe of 100 people.

Next, overlay a circle to represent the people who get a positive result on the test. We know that 90% of those with the disease will get a positive result, so need to cover 90% of circle A, but we also know that 10% of the population who does not have the disease will get a positive result, so we need to cover 10% of the non-disease carrying population (the total universe of 100 less circle A).

Circle B is covering a substantial portion of the total population. It actually covers more area than the total portion of the population with the disease. This is because 14 out of the total population of 100 (90% of the 5 people with the disease + 10% of the 95 people without the disease) will receive a positive result. Even though this is a test with 90% accuracy, this visualization shows that any one patient who tests positive (Circle B) for the disease only has a 32.14% (4.5 in 14) chance of actually having the disease.

Main article: Bayesian theory in science and math

Bayes’ theorem can show the likelihood of getting false positives in scientific studies. An in-depth look at this can be found in Bayesian theory in science and math .

Many medical diagnostic tests are said to be \(X\)% accurate, for instance 99% accurate, referring specifically to the probability that the test result is correct given your condition (or lack thereof). This is not the same as the posterior probability of having the disease given the result of the test. To see this in action, consider the following problem.

The world had been harmed by a widespread Z-virus, which already turned 10% of the world's population into zombies.

The scientists then invented a test kit with the sensitivity of 90% and specificity of 70%: 90% of the infected people will be tested positive while 70% of the non-infected will be tested negative.

If the test kit showed a positive result, what would be the probability that the tested subject was truly zombie?

If the solution is in a form of \(\frac{a}{b}\), where \(a\) and \(b\) are coprime positive integers, submit your answer as \(a+b\).

A disease test is advertised as being 99% accurate: if you have the disease, you will test positive 99% of the time, and if you don't have the disease, you will test negative 99% of the time.

If 1% of all people have this disease and you test positive, what is the probability that you actually have the disease?

Balls numbered 1 through 20 are placed in a bag. Three balls are drawn out of the bag without replacement. What is the probability that all the balls have odd numbers on them? In this situation, the events are not independent. There will be a \(\frac{10}{20} = \frac{1}{2}\) chance that any particular ball is odd. However, the probability that all the balls are odd is not \(\frac{1}{8}\). We do have that the probability that the first ball is odd is \(\frac{1}{2}.\) For the second ball, given that the first ball was odd, there are only 9 odd numbered balls that could be drawn from a total of 19 balls, so the probability is \(\frac{9}{19}\). For the third ball, since the first two are both odd, there are 8 odd numbered balls that could be drawn from a total of 18 remaining balls. So the probability is \(\frac{8}{18}\). So the probability that all 3 balls are odd numbered is \(\frac{10}{20} \times \frac{9}{19} \times \frac{8}{18} = \frac{2}{19}.\) Notice that \(\frac{2}{19} \approx 0.105\), whereas \(\frac{1}{8} = 0.125.\) \(_\square\)

A family has two children. Given that one of the children is a boy, what is the probability that both children are boys? We assume that the probability of a child being a boy or girl is \(\frac{1}{2}\). We solve this using Bayes’ theorem. We let \(B\) be the event that the family has one child who is a boy. We let \(A\) be the event that both children are boys. We want to find \(P(A \mid B) = \frac{P(B \mid A) \times P(A)}{P(B)}\). We can easily see that \(P(B \mid A) = 1\). We also note that \(P(A) = \frac{1}{4}\) and \(P(B) = \frac{3}{4}\). So \(P(A \mid B) = \frac{1 \times \frac{1}{4}}{\frac{3}{4}} = \frac{1}{3} \). \(_\square\)

A family has two children. Given that one of the children is a boy, and that he was born on a Tuesday, what is the probability that both children are boys? Your first instinct to this question might be to answer \(\frac{1}{3}\), since this is obviously the same question as the previous one. Knowing the day of the week a child is born on can’t possibly give you additional information, right? Let’s assume that the probability of being born on a particular day of the week is \(\frac{1}{7}\) and is independent of whether the child is a boy or a girl. We let \(B\) be the event that the family has one child who is a boy born on Tuesday and \(A\) be the event that both children are boys, and apply Bayes’ Theorem. We notice right away that \(P(B \mid A)\) is no longer equal to one. Given that there are 7 days of the week, there are 49 possible combinations for the days of the week the two boys were born on, and 13 of these have a boy who was born on a Tuesday, so \(P( B \mid A) = \frac{13}{49}\). \(P(A)\) remains unchanged at \(\frac{1}{4}\). To calculate \(P(B)\), we note that there are \(14^2\ = 196\) possible ways to select the gender and the day of the week the child was born on. Of these, there are \(13^2 = 169\) ways which do not have a boy born on Tuesday, and \(196 - 169 = 27\) which do, so \(P(B) = \frac{27}{196}\). This gives is that \(P(A \mid B) = \frac{ \frac{13}{49} \times \frac{1}{4}} {\frac{27}{196}} = \frac{13}{27}\). \(_\square\) Note: This answer is certainly not \(\frac{1}{3}\), and is actually much closer to \(\frac{1}{2}\).

Zeb's coin box contains 8 fair, standard coins (heads and tails) and 1 coin which has heads on both sides. He selects a coin randomly and flips it 4 times, getting all heads. If he flips this coin again, what is the probability it will be heads? (The answer value will be from 0 to 1, not as a percentage.)

There are 10 boxes containing blue and red balls.

The number of blue balls in the \(n^\text{th}\) box is given by \(B(n) = 2^n\). The number of red balls in the \(n^\text{th}\) box is given by \(R(n) = 1024 - B(n)\).

A box is picked at random, and a ball is chosen randomly from that box. If the ball is blue, and the probability that the \(10^\text{th}\) box was picked can be expressed as \( \frac ab\), where \(a\) and \(b\) are coprime positive integers, find \(a+b\).

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What Is Conditional Probability

• How It Works
• Joint and Marginal Probability
• Bayes' Theorem

The Bottom Line

• Fundamental Analysis

Conditional Probability: Formula and Real-Life Examples

Investopedia contributors come from a range of backgrounds, and over 25 years there have been thousands of expert writers and editors who have contributed.

Thomas J Catalano is a CFP and Registered Investment Adviser with the state of South Carolina, where he launched his own financial advisory firm in 2018. Thomas' experience gives him expertise in a variety of areas including investments, retirement, insurance, and financial planning.

Investopedia / Madelyn Goodnight

Condition probability measures the likelihood of an event or outcome happening based on the occurrence of some earlier event.

Conditional probability is a principle in probability theory. It relates to the probability that a certain event will occur based on the fact that a previous event has already occurred.

It involves two or more events that are not independent, and asks, "If we know A has happened, what's the chance of B also happening?" Conditional probability is calculated by multiplying the probability of the preceding event by the updated probability of the succeeding, or conditional, event.

Key Takeaways

• Conditional probability refers to the chances that some outcome (A) occurs given that another event (B) has already occurred.
• In probability, this is written as A given B, or as this formula: P(A|B), where the probability of A happening depends on that of B happening.
• Conditional probability can be contrasted with unconditional probability.
• Probabilities are classified as conditional, marginal (the base probability without any dependence on another event), or joint (the probability of two events occurring together).
• Bayes' theorem is a mathematical formula that can calculate conditional probabilities dealing with uncertain events.

Understanding Conditional Probability

Conditional probability measures the likelihood of a certain outcome (A), based on the occurrence of some earlier event (B).

Two events are said to be independent if one event occurring does not affect the probability that the other event will occur. However, if one event occurring (or not occurring) does affect the likelihood that the other event will happen, the two events are said to be dependent.

An example of dependent events is a company's stock price increasing after the company reports higher-than-expected earnings.

If events are independent, then the probability of event B occurring is not contingent on what happens with event A. For example, an increase in Apple's shares has nothing to do with a drop in wheat prices.

Conditional probability is often written as the "probability of A given B" and notated as P(A|B).

Other Types of Probabilities

• Conditional probability can be contrasted with unconditional probability . The latter is also called marginal probability, which measures the chance of a single event without depending on any other. In contrast, conditional probability determines the likelihood of one event given that another event has occurred, linking them.
• Independent probability doesn’t have that interconnectedness and instead looks at the probability of some event in isolation because it’s believed to be independent.
• A joint probability is the likelihood of two events occurring together. From these ideas, one can get to Bayes’ Theorem, which provides a way to flip conditional probabilities mathematically. If you know the chance of event B happening given event A, you can reverse-calculate the conditional probability of A given B.

Overall, while marginal and joint probabilities measure individual and paired events, conditional probability can measure precedence and dependence between events.

Conditional probability is used in a variety of fields, such as insurance , economics, politics, and different areas of mathematics.

Conditional Probability Formula

P ( B ∣ A ) = P ( A a n d B ) / P ( A ) P(B|A) = P(A and B) / P(A) P ( B ∣ A ) = P ( A an d B ) / P ( A )

P ( B ∣ A ) = P ( A ∩ B ) / P ( A ) P(B|A) = P(A∩B) / P(A) P ( B ∣ A ) = P ( A ∩ B ) / P ( A )

Where the letters are for the following:

P = Probability

A = Event A

B = Event B

Unconditional probability, also known as marginal probability, measures the chance of something happening while ignoring any knowledge of previous or external events. Since this probability also ignores new information, it remains constant.

Examples of Conditional Probability

Example 1: marbles in a bag.

An example of conditional probability using marbles is illustrated below. The steps are as follows:

Step 1 : Understand the scenario

Initially, you're given a bag with six red marbles, three blue marbles, and one green marble. Thus, there are 10 marbles in the bag.

Step 2 : Identify the events

Two events are defined:

• Event A: Drawing a red marble from the bag
• Event B: Drawing a marble that is not green

Step 3 : Calculate the probability of event B: P(B)

Event B is drawing a marble that is not green. There are 10 marbles altogether, nine of which are not green: the six red and three blue marbles.

P ( B ) = ( N u m b e r o f m a r b l e s t h a t a r e n o t g r e e n ) / ( T o t a l n u m b e r o f m a r b l e s ) = 9 / 10 P(B) = (Number of marbles that are not green)/(Total number of marbles) = 9/10 P ( B ) = ( N u mb ero f ma r b l es t ha t a re n o t g ree n ) / ( T o t a l n u mb ero f ma r b l es ) = 9/10

Step 4 : Identify the intersection of events A and B: P(A∩B)

The intersection of events A and B involves drawing a red marble that is also not green. Since all red marbles are not green, the intersection is simple: the event of drawing a red marble.

Step 5 : Calculate the probability of the intersection of events A and B: P(A∩B)

P ( A ∩ B ) = ( N u m b e r o f r e d m a r b l e s ) / ( T o t a l n u m b e r o f m a r b l e s ) = 6 / 10 = 3 / 5 P(A∩B) = (Number of red marbles)/(Total number of marbles) = 6/10 = 3/5 P ( A ∩ B ) = ( N u mb ero f re d ma r b l es ) / ( T o t a l n u mb ero f ma r b l es ) = 6/10 = 3/5

Step 6 : Calculate the conditional probability: P(A|B)

Using the conditional probability formula, P(A|B), that is, the probability of drawing a red marble given that the marble drawn is not green, the probability is calculated.

P ( A ∣ B ) = P ( A ∩ B ) / P ( B ) = ( 3 / 5 ) / ( 9 / 10 ) = 2 / 3 P(A|B) = P(A∩B)/P(B) = (3/5)/(9/10) = 2/3 P ( A ∣ B ) = P ( A ∩ B ) / P ( B ) = ( 3/5 ) / ( 9/10 ) = 2/3

Result: The conditional probability of drawing a red marble given that the marble drawn is not green, is 2/3.

Example 2: Rolling a Fair Die

Let's consider another example of conditional probability using a fair die. The steps are as follows:

You have a fair six-sided die. You want to determine the probability of rolling an even number, given that the number rolled is greater than four.

Step 2: Identify the events

The possible outcomes (sample space) for a six-sided die are the numbers one through six. From this list, you can define the two events:

• Event A: Rolling an even number. Event A would mean rolling {2,4,6}.
• Event B: Rolling a number greater than four. Event B would mean rolling {5,6}.

Step 3 : Calculate the probability of each event

The probability of each event can be calculated by dividing the number of favorable outcomes (the ones you're looking for) by the total number of outcomes in the sample space.

P(A) is the probability of rolling an even number. There are three even numbers {2,4,6} out of the six possible outcomes. Thus, P(A) = 3/6 = 1/2.

P(B) is the probability of rolling a number greater than four. Two numbers are greater than four {5,6} out of the six possible outcomes. Thus, P(B) = 4/6 = 2/3.

Step 4 : Identify the intersection of events A and B

The intersection of events A and B includes the outcomes that satisfy both conditions simultaneously. In this case, that means rolling a number that is even and also greater than four. The only outcome that does both is rolling a six.

Step 5 : Calculate the probability of the intersection of events A and B

We'll spell this out, even if it's easy, given the above, because other examples might prove more difficult: P(A∩B) is the probability of rolling six since six is the only outcome that is both even and greater than six. There is one outcome out of six possibilities. So P(A∩B) = 1/6.

Step 6 : Calculate the conditional probability: P(B|A)

The formula for conditional probability is as follows:

When the values are substituted into the formula, here is the result:

P ( B ∣ A ) = ( 1 / 6 ) / ( 1 / 2 ) = 1 / 3 P(B|A) = (1/6)/(1/2) = 1/3 P ( B ∣ A ) = ( 1/6 ) / ( 1/2 ) = 1/3

Result: This means that given the die rolled is even, the probability that this number is also greater than four is 1/3.

Example 3: Multiple Conditional Probabilities

Another scenario involves a student applying for admission to a college who hopes to get a scholarship and a stipend for books, meals, and housing. The steps to determine the conditional probability of getting a stipend and the scholarship are as follows:

First, the student wants to know the likelihood of being accepted to the university. Then, if accepted, the student would like to receive an academic scholarship. Moreover, if possible, the student would also like to receive a stipend for books, meals, and housing if they get the scholarship.

There are three events:

• Event A: Being accepted to the university.
• Event B: Receiving a scholarship upon acceptance
• Event C: Receiving a stipend for books, meals, and housing upon receiving a scholarship

Step 3 : Calculate the probability of being accepted (event A)

The university accepts 100 out of every 1,000 applicants who have applications similar to the student's. Thus, the probability of a student being accepted is P(A) = 100/1000 = 0.10 or 10%.

Step 4 : Determine the probability of receiving a scholarship once accepted: P(B|A)

It's known that out of the students accepted, 10 out of every 500 receive a scholarship. Thus the probability of receiving a scholarship given acceptance is as follows:

P ( B ∣ A ) = 10 / 500 = 0.02 = 2 P(B|A) = 10/500 = 0.02 = 2% P ( B ∣ A ) = 10/500 = 0.02 = 2 %

Step 5 : Calculate the probability of being accepted and receiving a scholarship

To calculate the probability of being accepted and also receiving a scholarship, the likelihood of acceptance is multiplied by the conditional probability of receiving a scholarship given acceptance.

P ( A ∩ B ) = P ( A ) × P ( B ∣ A ) = 0.1 × 0.02 = 0.002 = 0.2 P(A∩B)=P(A)×P(B∣A)=0.1×0.02=0.002=0.2 P ( A ∩ B ) = P ( A ) × P ( B ∣ A ) = 0.1 × 0.02 = 0.002 = 0.2 %

Step 6 : Determine the probability of receiving a stipend having gotten a scholarship: P(C|B)

It's also known that among the scholarship recipients, 50% receive a stipend for books, meals, and housing. Thus, P(C|B) = 0.5 = 50%.

Step 7 : Calculate the probability of being accepted, receiving a scholarship, and receiving a stipend

To calculate the probability of a student being accepted, receiving a scholarship, and then also receiving a stipend, the probabilities of the events are multiplied.

P ( A ∩ B ∩ C ) = P ( A ) × P ( B ∣ A ) × P ( C ∣ B ) = 0.1 × 0.02 × 0.5 = 0.001 = 0.1 P(A∩B∩C)=P(A)×P(B∣A)×P(C∣B)=0.1×0.02×0.5=0.001=0.1% P ( A ∩ B ∩ C ) = P ( A ) × P ( B ∣ A ) × P ( C ∣ B ) = 0.1 × 0.02 × 0.5 = 0.001 = 0.1 %

This step-by-step breakdown illustrates how the probabilities for each scenario are calculated using basic probability formulas and conditional probability.

Conditional Probability vs. Joint Probability and Marginal Probability

Let's now differentiate calculating conditional probability from other kinds of probability.

Conditional Probability

The example this time is a regular deck of cards. Two events are defined:

• Event A: Drawing a four
• Event B: Drawing a red card

A standard deck has 52 cards divided into four suits (hearts, diamonds, clubs, and spades). Hearts and diamonds are red, and clubs and spades are black. Each suit has 13 cards: Ace, then two through 10, and then the face cards Jack, Queen, and King.

The deck contains 26 red cards, 13 hearts, and 13 diamonds. Thus, the probability of drawing a red card is P(B) = 26/52 = 1/2.

Within the red cards are a four of hearts and a four of diamonds. Therefore, if a red card has to be drawn, a subset of the deck that includes only these 26 red cards needs to be considered.

Given that a red card has been drawn, the probability of it being a four is calculated as follows:

P ( A ∣ B ) = ( N u m b e r o f r e d f o u r s ) / ( T o t a l n u m b e r o f r e d c a r d s ) = 2 / 26 = 1 / 13 P(A|B) = (Number of red fours)/(Total number of red cards) = 2/26 = 1/13 P ( A ∣ B ) = ( N u mb ero f re df o u rs ) / ( T o t a l n u mb ero f re d c a r d s ) = 2/26 = 1/13

Marginal Probability

The marginal probability, P(A), is the probability of an event A happening on its own. It does not consider the occurrence of any other event.

Since event A is drawing a four, P(A) is calculated by dividing the number of fours by the total number of cards in the deck.

P ( A ) = ( N u m b e r o f f o u r s i n t h e d e c k ) / ( T o t a l n u m b e r o f c a r d s i n d e c k ) = 4 / 52 = 1 / 13 P(A) = (Number of fours in the deck)/(Total number of cards in deck) = 4/52 = 1/13 P ( A ) = ( N u mb ero ff o u rs in t h e d ec k ) / ( T o t a l n u mb ero f c a r d s in d ec k ) = 4/52 = 1/13

Joint Probability

Joint probability is the likelihood of two or more events happening at the same time. This is denoted as P(A∩B), the probability of events A and B occurring.

Assuming that the previous events are the same, that is, event A is the occurrence of drawing a card that is a four and event B is drawing a red card, we can find the joint probability of drawing a card that is both a four and red.

There are two cards that meet both criteria, the four of hearts and the four of diamonds. Thus, the joint probability of drawing a card that is both a four and red is calculated as follows:

P ( A ∩ B ) = ( N u m b e r o f r e d f o u r s ) / ( T o t a l n u m b e r o f c a r d s ) = 2 / 52 = 1 / 26 P(A∩B) = (Number of red fours)/(Total number of cards) = 2/52 = 1/26 P ( A ∩ B ) = ( N u mb ero f re df o u rs ) / ( T o t a l n u mb ero f c a r d s ) = 2/52 = 1/26

Bayes' Theorem and Conditional Probability

Bayes’ theorem is used to calculate conditional probabilities when dealing with uncertain events. In investing, this allows you to update your probability estimates of a market outcome when you get new relevant data.

For example, suppose you wanted to know the probability that the S&P 500 would return a positive percentage this year, given initial gross domestic product (GDP) figures. In that case, you’d start with Bayes’ theorem, considering the index’s historical return rates to get an initial estimate of projected economic expansion.

You would then revise this first probability using the latest GDP estimates. This would provide more refined probability assessments incorporating all evidence as the year progresses.

While a bit complex mathematically, Bayes’ theorem is quite logical. If an investor discovers new economic information relevant to potential market returns, it makes sense to integrate this data to get a more precise calculation.

The 18th-century English minister Thomas Bayes devised this statistical technique, which remains central in financial modeling and other fields requiring predictions under uncertain conditions.

Bayes' theorem is well suited to and widely used in machine learning.

What Is a Conditional Probability Calculator?

A conditional probability calculator is an online tool that calculates conditional probability. It provides the probability of the first and second events occurring. A conditional probability calculator saves the user from doing the mathematics manually.

What Is the Difference Between Probability and Conditional Probability?

Probability looks at the likelihood of one event occurring. Conditional probability looks at two events occurring in relation to one another. More specifically, it looks at the probability of a second event occurring based on the probability of the first event occurring.

What Is Prior Probability?

Prior probability is the probability of an event occurring before any data has been gathered. It is the probability as determined by a prior belief. Prior probability is a part of Bayesian statistical inference since you can revise these beliefs and arrive mathematically at a posterior probability .

What Is Compound Probability?

Compound probability looks to determine the likelihood of two independent events occurring. Compound probability multiplies the probability of the first event by the probability of the second event. The most common example is a coin flipped twice and finding if the second result will be the same as the first.

Conditional probability examines the likelihood of an event occurring based on the likelihood of a preceding event occurring. The second event is dependent on the first event.

For example, we might want to know the probability that some stock will go up if the index for its sector is on the rise. The conditional probability calculation takes into account both, how likely the first event is (the stock rising in price), as well as how much the two events overlap.

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