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

Conditional probability is one type of probability in which the possibility of an event depends upon the existence of a previous event. 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.

Let us understand the concept with few examples.

Example 1 We flip two coins. What is the probability that both are heads, given that at least one of them is heads? We are given that there is at least one head.. So there are three possibilities HH, HT, TH. Only one outcome has both coins as heads: HH. So our answer is = 1/3

Example 2 We roll a six-sided die. What is the probability that the roll is a 6, given that the outcome is an even number The possible outcomes when rolling a die are: 1, 2, 3, 4, 5, 6. The even numbers are: 2, 4, 6 (3 outcomes). Only one of these is a 6. So our answer = 1/3

Table of Content

Conditional Probability Formula

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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.

Rolling a Dice with 3 in the first Roll and 9 as Sum

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.

A King Given a Face Card Suppose you have a deck of 52 cards. What is the probability of drawing a king, given that the card is a face card (king, queen, or jack)?

Total number of face cards = 12 (4 kings, 4 queens, 4 jacks). Number of kings = 4. P(King ∩ Face Card ) = 4/52 since all kings are face cards P(Face Card) = 12/52

P(King | Face Card) = P(King ∩ Face Card ) / P(Face Card) = 4 / 12 = 1/3

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 of Independent Events

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:

Tossing a Coin

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.

Drawing Cards

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.

How to Apply the Multiplication Rule?

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?

Resources related to Conditional Probability:

Probability Class 12 Notes Probability Class 12 NCERT Solutions

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.

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

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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3.2: Problems on Conditional Probability

• Last updated
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• Page ID 10875

• Paul Pfeiffer
• Rice University

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Exercise \(\PageIndex{1}\)

Given the following data:

\(P(A) = 0.55\), \(P(AB) = 0.30\), \(P(BC) = 0.20\), \(P(A^c \cup BC) = 0.55\), \(P(A^c BC^c) = 0.15\)

Determine, if possible, the conditional probability \(P(A^c|B) = P(A^cB)/P(B)\).

Exercise \(\PageIndex{2}\)

In Exercise 11 from "Problems on Minterm Analysis," we have the following data: A survey of a represenative group of students yields the following information:

• 52 percent are male
• 85 percent live on campus
• 78 percent are male or are active in intramural sports (or both)
• 30 percent live on campus but are not active in sports
• 32 percent are male, live on campus, and are active in sports
• 8 percent are male and live off campus
• 17 percent are male students inactive in sports

Let A = male, B = on campus, C = active in sports.

• A student is selected at random. He is male and lives on campus. What is the (conditional) probability that he is active in sports?
• A student selected is active in sports. What is the(conditional) probability that she is a female who lives on campus?

Exercise \(\PageIndex{3}\)

In a certain population, the probability a woman lives to at least seventy years is 0.70 and is 0.55 that she will live to at least eighty years. If a woman is seventy years old, what is the conditional probability she will survive to eighty years? Note that if \(A \subset B\) then \(P(AB) = P(A)\).

Let \(A=\) event she lives to seventy and \(B=\) event she lives to eighty. Since \(B \subset A\), \(P(B|A) = P(AB)/P(A) = P(B)/P(A) = 55/70\).

Exercise \(\PageIndex{4}\)

From 100 cards numbered 00, 01, 02, \(\cdot\cdot\cdot\), 99, one card is drawn. Suppose A i is the event the sum of the two digits on a card is \(i\), \(0 \le i \le 18\), and \(B_j\) is the event the product of the two digits is \(j\). Determine \(P(A_i|B_0)\) for each possible \(i\).

\(B_0\) is the event one of the first ten is draw. \(A_i B_0\) is the event that the card with numbers \(0i\) is drawn. \(P(a_i|B_0) = (1/100)/(1/10) = 1/10\) for each \(i\), 0 through 9.

Exercise \(\PageIndex{5}\)

Two fair dice are rolled.

• What is the (conditional) probability that one turns up two spots, given they show different numbers?
• What is the (conditional) probability that the first turns up six, given that the sum is \(k\), for each \(k\) from two through 12?
• What is the (conditional) probability that at least one turns up six, given that the sum is \(k\), for each \(k\) from two through 12?

a. There are \(6 \times 5\) ways to choose all different. There are \(2 \times 5\) ways that they are different and one turns up two spots. The conditional probability is 2/6.

b. Let \(A_6\) = event first is a six and \(S_k = \) event the sum is \(k\). Now \(A_6S_k = \emptyset\) for \(k \le 6\). A table of sums shows \(P(A_6S_k) = 1/36\) and \(P(S_k) = 6/36, 5/36, 4/36, 3/36, 2/36, 1/36\) for \(k = 7\) through 12, respectively. Hence \(P(A_6|S_k) = 1/6, 1/5. 1/4, 1/3. 1/2, 1\), respectively.

c. If \(AB_6\) is the event at least one is a six, then \(AB_6S_k) = 2/36\) for \(k = 7\) through 11 and \(P(AB_6S_12) = 1/36\). Thus, the conditional probabilities are 2/6, 2/5, 2/4, 2/3, 1, 1, respectively.

Exercise \(\PageIndex{6}\)

Four persons are to be selected from a group of 12 people, 7 of whom are women.

• What is the probability that the first and third selected are women?
• What is the probability that three of those selected are women?
• What is the (conditional) probability that the first and third selected are women, given that three of those selected are women?

\(P(W_1W_3) = P(W_1W_2W_3) + P(W_1W_2^c W_3) = \dfrac{7}{12} \cdot \dfrac{6}{11} \cdot \dfrac{5}{10} + \dfrac{7}{12} \cdot \dfrac{5}{11} \cdot \dfrac{6}{10} = \dfrac{7}{22}\)

Exercise \(\PageIndex{7}\)

Twenty percent of the paintings in a gallery are not originals. A collector buys a painting. He has probability 0.10 of buying a fake for an original but never rejects an original as a fake, What is the (conditional) probability the painting he purchases is an original?

Let \(B=\) the event the collector buys, and \(G=\) the event the painting is original. Assume \(P(B|G) = 1\) and \(P(B|G^c) = 0.1\). If \(P(G) = 0.8\), then

\(P(G|B) = \dfrac{P(GB)}{P(B)} = \dfrac{P(B|G) P(G)}{P(B|G)P(G) + P(B|G^c)P(G^c)} = \dfrac{0.8}{0.8 + 0.1 \cdot 0.2} = \dfrac{40}{41}\)

Exercise \(\PageIndex{8}\)

Five percent of the units of a certain type of equipment brought in for service have a common defect. Experience shows that 93 percent of the units with this defect exhibit a certain behavioral characteristic, while only two percent of the units which do not have this defect exhibit that characteristic. A unit is examined and found to have the characteristic symptom. What is the conditional probability that the unit has the defect, given this behavior?

Let \(D=\) the event the unit is defective and \(C=\) the event it has the characteristic. Then \(P(D) = 0.05\), \(P(C|D) = 0.93\), and \(P(C|D^c) = 0.02\).

\(P(D|C) = \dfrac{P(C|D) P(D)}{P(C|D) P(D) + P(C|D^c) P(D^c)} = \dfrac{0.93 \cdot 0.05}{0.93 \cdot 0.05 + 0.02 \cdot 0.95} = \dfrac{93}{131}\)

Exercise \(\PageIndex{9}\)

A shipment of 1000 electronic units is received. There is an equally likely probability that there are 0, 1, 2, or 3 defective units in the lot. If one is selected at random and found to be good, what is the probability of no defective units in the lot?

Let \(D_k =\) the event of \(k\) defective and \(G\) be the event a good one is chosen.

\(P(D_0|G) = \dfrac{P(G|D_0) P(D_0)}{P(G|D_0) P(D_0) + P(G|D_1) P(D_1) + P(G|D_2) P(D_2) + P(G|D_3) P(D_3)}\)

\(= \dfrac{1 \cdot 1/4}{(1/4)(1 + 999/1000 + 998/1000 + 997/1000)} = \dfrac{1000}{3994}\)

Exercise \(\PageIndex{10}\)

\(P(S_i|E_j)\)

• Suppose a person has a university education (no graduate study). What is the (conditional) probability that he or she will make $25,000 or more?
• Find the total probability that a person's income category is at least as high as his or her educational level.

a. \(P(E_3S_3) = P(S_3|E_3)P(E_3) = 0.45 \cdot 0.05 = 0.0225\)

b. \(P(S_2 \vee S_3|E_2) = 0.80 + 0.10 = 0.90\)

c. \(p = (0.85 + 0.10 + 0.05) \cdot 0.65 + (0.80 + 0.10) \cdot 0.30 + 0.45 \cdot 0.05 = 0.9425\)

Exercise \(\PageIndex{11}\)

In a survey, 85 percent of the employees say they favor a certain company policy. Previous experience indicates that 20 percent of those who do not favor the policy say that they do, out of fear of reprisal. What is the probability that an employee picked at random really does favor the company policy? It is reasonable to assume that all who favor say so.

\(P(S) = 0.85\), \(P(S|F^c) = 0.20\). Also, reasonable to assume \(P(S|F) = 1\).

\(P(S) = P(S|F) P(F) + P(S|F^c) [1 - P(F)]\) implies \(P(F) = \dfrac{P(S) - P(S|F^c)}{1 - P(S|F^c)} = \dfrac{13}{16}\)

Exercise \(\PageIndex{12}\)

\(\dfrac{D^c|T}{P(D|T)} = \dfrac{P(T|D^c)P(D^c)}{P(T|D)P(D)} = \dfrac{0.98 \cdot 0.99}{0.05 \cdot 0.01} = \dfrac{9702}{5}\)

\(P(D^c|T) = \dfrac{9702}{9707} = 1 - \dfrac{5}{9707}\)

Exercise \(\PageIndex{13}\)

Five boxes of random access memory chips have 100 units per box. They have respectively one, two, three, four, and five defective units. A box is selected at random, on an equally likely basis, and a unit is selected at random therefrom. It is defective. What are the (conditional) probabilities the unit was selected from each of the boxes?

\(H_i =\) the event from box \(i\). \(P(H_i) = 1/5\) and \(P(D|H_i) = i/100\).

\(P(H_i|D) = \dfrac{P(D|H_i) P(H_i)}{\sum P(D|H_i) P(H_j)} = i/15\), \(1 \le i \le 5\)

Exercise \(\PageIndex{14}\)

Two percent of the units received at a warehouse are defective. A nondestructive test procedure gives two percent false positive indications and five percent false negative. Units which fail to pass the inspection are sold to a salvage firm. This firm applies a corrective procedure which does not affect any good unit and which corrects 90 percent of the defective units. A customer buys a unit from the salvage firm. It is good. What is the (conditional) probability the unit was originally defective?

Let \(T\) = event test indicates defective, \(D\) = event initially defective, and \(G =\) event unit purchased is good. Data are

\(P(D) = 0.02\), \(P(T^c|D) = 0.02\), \(P(T|D^c) = 0.05\), \(P(GT^c) = 0\),

\(P(G|DT) = 0.90\), \(P(G|D^cT) = 1\)

\(P(D|G) = \dfrac{P(GD)}{P(G)}\), \(P(GD) = P(GTD) = P(D) P(T|D) P(G|TD)\)

\(P(G) = P(GT) = P(GDT) + P(GD^c T) = P(D) P(T|D) P(G|TD) + P(D^c) P(T|D^c) P(G|TD^c)\)

\(P(D|G) = \dfrac{0.02 \cdot 0.98 \cdot 0.90}{0.02 \cdot 0.98 \cdot 0.90 + 0.98 \cdot 0.05 \cdot 1.00} = \dfrac{441}{1666}\)

Exercise \(\PageIndex{15}\)

At a certain stage in a trial, the judge feels the odds are two to one the defendent is guilty. It is determined that the defendent is left handed. An investigator convinces the judge this is six times more likely if the defendent is guilty than if he were not. What is the likelihood, given this evidence, that the defendent is guilty?

Let \(G\) = event the defendent is guilty, \(L\) = the event the defendent is left handed. Prior odds: \(P(G)/P(G^c) = 2\). Result of testimony: \(P(L|G)/P(L|G^c) = 6\).

\(\dfrac{P(G|L)}{P(G^c|L)} = \dfrac{P(G)}{P(G^c)} \cdot \dfrac{P(L|G)}{P(L|G^c)} = 2 \cdot 6 = 12\)

\(P(G|L) = 12/13\)

Exercise \(\PageIndex{16}\)

Show that if \(P(A|C) > P(B|C)\) and \(P(A|C^c) > P(B|C^c)\), then \(P(A) > P(B)\). Is the converse true? Prove or give a counterexample.

\(P(A) = P(A|C) P(C) + P(A|C^c) P(C^c) > P(B|C) P(C) + P(B|C^c) P(C^c) = P(B)\).

The converse is not true. Consider \(P(C) = P(C^c) = 0.5\), \(P(A|C) = 1/4\).

\(P(A|C^c) = 3/4\), \(P(B|C) = 1/2\), and \(P(B|C^c) = 1/4\). Then

\(1/2 = P(A) = \dfrac{1}{2} (1/4 + 3/4) > \dfrac{1}{2} (1/2 + 1/4) = P(B) = 3/8\)

But \(P(A|C) < P(B|C)\).

Exercise \(\PageIndex{17}\)

Since \(P(\cdot |B)\) is a probability measure for a given \(B\), we must have \(P(A|B) + P(A^c|B) = 1\). Construct an example to show that in general \(P(A|B) + P(A|B^c) \ne 1\).

Suppose \(A \subset B\) with \(P(A) < P(B)\). Then \(P(A|B) = P(A)/P(B) < 1\) and \(P(A|B^c) = 0\) so the sum is less than one.

Exercise \(\PageIndex{18}\)

Use property ( CP4 ) to show

a. \(P(A|B) > P(A)\) iff \(P(A|B^c) < P(A)\)

b. \(P(A^c|B) > P(A^c)\) iff \(P(A|B) < P(A)\)

c. \(P(A|B) > P(A)\) iff \(P(A^c|B^c) > P(A^c)\)

a. \(P(A|B) > P(A)\) iff \(P(AB) > P(A) P(B)\) iff \(P(AB^c) < P(A) P(B^c)\) iff \(P(A|B^c) < P(A)\)

b. \(P(A^c|B) > P(A^c)\) iff \(P(A^c B) > P(A^c) P(B)\) iff \(P(AB) < P(A) P(B)\) iff \(P(A|B) < P(A)\)

c. \(P(A|B) > P(A)\) iff \(P(AB) > P(A) P(B)\) iff \(P(A^c B^c) > P(A^c) P(B^c)\) iff \(P(A^c|B^c) > P(A^c)\)

Exercise \(\PageIndex{19}\)

Show that \(P(A|B) \ge (P(A) + P(B) - 1)/P(B)\).

\(1 \ge P(A \cup B) = P(A) + P(B) - P(AB) = P(A) + P(B) - P(A|B) P(B)\). Simple algebra gives the desired result.

Exercise \(\PageIndex{20}\)

Show that \(P(A|B) = P(A|BC) P(C|B) + P(A|BC^c) P(C^c|B)\).

\(P(A|B) = \dfrac{P(AB)}{P(B)} = \dfrac{P(ABC) + P(ABC^c)}{P(B)}\)

\(= \dfrac{P(A|BC) P(BC) + P(A|BC^c) P(BC^c)}{P(B)} = P(A|BC) P(C|B) + P(A|BC^c) P(C^c|B)\)

Exercise \(\PageIndex{21}\)

\(P(A|B) = 1\), \(P(A|B^c) = 1/n\), \(P(B) = p\)

\(P(B|A) = \dfrac{P(A|B) P(B)}{P(A|B) P(B) +P(A|B^c) P(B^c)} = \dfrac{p}{p + \dfrac{1}{n} (1 - p)} = \dfrac{np}{(n - 1) p + 1}\)

\(\dfrac{P(B|A)}{P(B)} = \dfrac{n}{np + 1 - p}\) increases from 1 to \(1/p\) as \(n \to \infty\)

Exercise \(\PageIndex{22}\)

Polya's urn scheme for a contagious disease . An urn contains initially \(b\) black balls and \(r\) red balls \((r + b = n)\). A ball is drawn on an equally likely basis from among those in the urn, then replaced along with \(c\) additional balls of the same color. The process is repeated. There are \(n\) balls on the first choice, \(n + c\) balls on the second choice, etc. Let \(B_k\) be the event of a black ball on the \(k\)th draw and \(R_k\) be the event of a red ball on the \(k\)th draw. Determine

a. \(P(B_2|R_1)\) b. \(P(B_1B_2)\) c. \(P(R_2)\) d. \(P(B_1|R_2)\)

a. \(P(B_2|R_1) = \dfrac{b}{n + c}\)

b. \(P(B_1B_2) = P(B_2) P(B_2|B_1) = \dfrac{b}{n} \cdot \dfrac{b + c}{n + c}\)

c. \(P(R_2) P(R_2|R_1) P(R_1) + P(R_2|B_1) P(B_1)\)

\(= \dfrac{r + c}{n + c} \cdot \dfrac{r}{n} + \dfrac{r}{n + c} \cdot \dfrac{b}{n} = \dfrac{r(r + c + b)}{n(n + c)}\)

d. \(P(B_1|R_2) = \dfrac{P(R_2|B_1) P(B_1)}{P(R_2)}\) with \(P(R_2|B_1) P(B_1) = \dfrac{r}{n + c} \cdot \dfrac{b}{n}\). Using (c), we have

\(P(B_1|R_2) = \dfrac{b}{r + b + c} = \dfrac{b}{n + c}\)