`P(A|B)`
- Introduction
- Understanding Conditional Probability \( P(A|B) \)
- Understanding the \( P(A|B) \) Formula
- Special Cases of Conditional Probability \( P(A|B) \)
- Bayes Theorem
- Applications of Conditional Probability \( P(A|B) \)
- Solved Examples
- Practice Problems
- Frequently Asked Questions
Introduction
Conditional probability, often denoted as \( P(A|B) \) and read as probability of `A` given `B`, is a concept in probability theory where the likelihood of one event occurring depends on the occurrence of another event. It can be understood as the probability of event `A` given that event `B` has already happened. Conditional probability is frequently encountered in real-life situations, such as weather forecasting or medical diagnosis, where the outcome of one event influences the likelihood of another. Understanding conditional probability is essential and it is often utilized to solve complex problems involving dependent events.
Additionally, the Bayes’ theorem is a fundamental concept related to conditional probability, providing a framework for updating probabilities based on new information. In this article, we will see various examples and applications to better conditional probability and its significance in probability theory.
Understanding Conditional Probability \( P(A|B) \)
Conditional probability is essentially the likelihood of one event happening based on the occurrence of another event. It helps us understand how events are connected.
If the probability of one event happening doesn’t affect the probability of another event, they are independent events. For instance, if we toss two coins, the probability of getting heads on the second toss does not depend on whether we got heads or tails on the first toss. On the other hand, if the occurrence of one event affects the probability of another event, they are dependent events.
Conditional probability is represented as `P(A | B)`, meaning the probability of `A` given `B` has occurred. In simpler terms, it calculates the chance of one event happening when a certain condition is met.
Understanding the Probability Formula \( P(A|B) \)
What is the probability of a given `b` formula? Probability of `A` given `B`, denoted as `P(A|B)`, comes into play when events are dependent on each other. It tells us the likelihood of event `A` occurring given that event `B` has already happened. To compute this, we use the formula:
`P(A|B) = \frac{P(A \cap B)}{P(B)}`
where,
`P(A ∩ B)` indicates the probability of events `A` and `B` happening at the same time
`P(B)` signifies the probability of event `B` occurring independently.
Similarly, `P(B|A) = \frac{P(A \cap B)}{P(A)}`
These formulas lead to the product rules of probability:
\( P(A \cap B) = P(A|B) \times P(B) \)
\( P(A \cap B) = P(B|A) \times P(A) \)
If events `A` and `B` are independent, then \( P(A|B) = P(A) \) and \( P(B|A) = P(B) \). In this case the formulas mentioned above turn into:
\( P(A \cap B) = P(A) \times P(B) \)
It's crucial to note that \( P(A|B) \) is undefined if \( P(B) = 0 \).
Special Cases of Conditional Probability \( P(A|B) \)
Case `1`: When `A` and `B` are mutually exclusive (meaning they can't both happen at once), then (\( A \cap B = \emptyset \)), thus making \( P(A|B) = 0 \).
Case `2`: If `B` is a subset of `A`, then \( A \cap B = B \), leading to `P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{P(B)}{P(B)} = 1`.
Case `3`: When `A` is a subset of `B`, then \( A \cap B = A \), resulting in `P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{P(A)}{P(B)}`.
Bayes Theorem
Bayes' theorem, named after Reverend Thomas Bayes, is a fundamental concept in probability theory and statistics. It provides a way to revise or update the probability of an event based on new evidence or information.
Bayes' theorem is stated as follows:
`P(A|B) = \frac{P(B|A) \times P(A)}{P(B)}`
where:
- \( P(A|B) \) is the updated probability of event `A` occurring given that event `B` has occurred. This is called the posterior probability.
- \( P(B|A) \) is the probability of event `B` occurring given that event `A` has occurred. This is called the likelihood.
- \( P(A) \) and \( P(B) \) are the probabilities of events `A` and `B` occurring independently. These are called the prior probabilities.
In simpler terms, Bayes' theorem allows us to update our beliefs about the likelihood of an event `A` occurring, given the occurrence of another event `B`. It does so by combining our prior knowledge or beliefs about the probabilities of `A` and `B`, along with new evidence provided by the conditional probability of `B` given `A`.
Bayes' theorem has widespread applications in various fields, including machine learning, medical diagnosis, spam filtering, and more. It enables us to make better-informed decisions by incorporating new information into our existing knowledge or beliefs about uncertain events.
Applications of Conditional Probability \( P(A|B) \)
`1`. Medical Diagnosis: Conditional probability is extensively used in medical diagnosis. Doctors often rely on various tests to diagnose diseases. The accuracy of these tests and the prevalence of the disease in the population play crucial roles in determining the probability that an individual has the disease given a positive or negative test result.
`2`. Weather Forecasting: Weather forecasting involves predicting the likelihood of different weather conditions based on current weather patterns, historical data, and atmospheric conditions. Conditional probability helps meteorologists estimate the probability of specific weather events occurring given certain conditions, such as the probability of rain given cloudy skies.
`3`. Machine Learning and Artificial Intelligence: In machine learning and AI applications, conditional probability is used in algorithms for classification, prediction, and recommendation systems. For example, in spam filtering, algorithms use conditional probability to classify emails as spam or not spam based on features such as the presence of certain words or phrases given the email's content.
`4`. Financial Modeling: Conditional probability is employed in financial modeling and risk assessment to estimate the likelihood of various financial events occurring under different economic conditions. Banks and financial institutions use conditional probability to evaluate the risk associated with investments, loans, and insurance policies.
`5`. Quality Control and Manufacturing: In manufacturing processes, conditional probability is used to monitor and improve product quality. By analyzing the relationship between input variables (such as raw materials and production methods) and output variables (such as product defects), companies can identify factors that affect product quality and take corrective actions to minimize defects and improve overall product reliability.
Solved Examples
Example `1`: Suppose we have a fair coin, and we want to find the probability of getting a head given that the coin landed on heads in the previous toss.
Solution:
Let:
Event `A` be getting a head in the current toss.
Event `B` be getting a head in the previous toss.
We want to find \(P(A|B)\), the probability of getting a head in the current toss given that the previous toss resulted in a head.
Since the coin is fair, `P(A) = P(B) = \frac{1}{2}`.
The probability of getting two consecutive heads:
`P(A \cap B) = P(A) \times P(B)`
`= \frac{1}{2} \times \frac{1}{2}`
`= \frac{1}{4}`
Using the conditional probability formula:
`P(A|B) = \frac{P(A \cap B)}{P(B)}`
`= \frac{\frac{1}{4}}{\frac{1}{2}}`
`= \frac{1}{2}`
So, the probability of getting a head in the current toss given that the previous toss resulted in a head is `\frac{1}{2}`.
Example `2`: Suppose we have a standard deck of `52` cards, and we draw one card at random. We want to find the probability of drawing a queen given that the card drawn is a face card (jack, queen, or king).
Solution:
Let:
Event `A` be drawing a queen.
Event `B` be drawing a face card.
There are `12` face cards in the deck (`4` jacks, `4` queens, and `4` kings).
So, `P(B) = \frac{12}{52} = \frac{3}{13}`
Of these face cards, there are `4` queens.
So, `P(A \cap B) = \frac{4}{52} = \frac{1}{13}`
Using the probability of `A` given `B` formula:
`P(A|B) = \frac{P(A \cap B)}{P(B)}`
`= \frac{\frac{1}{13}}{\frac{3}{13}}`
`= \frac{1}{3}`
So, the probability of drawing a queen given that the card drawn is a face card is `\frac{1}{3}`.
Example `3`: In a survey among a group of students, `80%` play cricket, `60%` play football, and `50%` play both sports. If a student is chosen at random and it is known that the student plays football, what is the probability that the student also plays cricket?
Solution:
Let:
Event `A` be the student plays cricket.
Event `B` be the student plays football.
We want to find \(P(A|B)\), the probability that the student plays cricket given that he/she plays football.
Now,
- \( P(A) \): Probability of a student playing cricket `= 0.80`
- \( P(B) \): Probability of a student playing football `= 0.60`
- \( P(A \cap B) \): Probability of a student playing both cricket and football `= 0.50`
Using the conditional probability formula:
`P(A|B) = \frac{P(A \cap B)}{P(B)}`
`= \frac{0.50}{0.60}`
`= \frac{5}{6}`
So, the probability that a student plays cricket given that the student plays football is `\frac{5}{6}`.
Example `4`: Suppose a couple plans to have two children. We want to find the probability that they will have two girls given that they already have a girl.
Solution:
Let:
Event `A` be having a girl as the first child.
Event `B` be having a girl as the second child.
We want to find \(P(B|A)\), the probability of having a girl as the second child given that the first child is a girl.
Since the gender of each child is independent and equally likely, so, `P(A) = P(B) = \frac{1}{2}`.
The probability of having two girls:
`P(A \cap B) = P(A) \times P(B)`
`= \frac{1}{2} \times \frac{1}{2}`
`= \frac{1}{4}`
Using the conditional probability formula:
`P(B|A) = \frac{P(A \cap B)}{P(A)}`
`= \frac{\frac{1}{4}}{\frac{1}{2}}`
`= \frac{1}{2}`
So, the probability of having a girl as the second child given that the first child is a girl is `\frac{1}{2}`.
Example `5`: When a fair die is rolled, what is the probability of rolling an even number, given that the number rolled is greater than four?
Solution:
Let:
Event `A` be getting an even number.
Event `B` be getting a number greater than four.
We want to find \(P(A|B)\), the probability of getting an even number given that the number rolled is greater than four.
`A` is the event of getting an even number. So, \(A = \{2, 4, 6\} \)
Thus, `P(A) = \frac{3}{6} = \frac{1}{2}`
`B` is the event of getting a number greater than four. So, \(B = \{5, 6\} \)
Thus, `P(B) = \frac{2}{6} = \frac{1}{3}`
Since, \(A \cap B = \{ 6\} \), so, `P(A \cap B) = \frac{1}{6}`
Using the conditional probability formula:
`P(A|B) = \frac{P(A \cap B)}{P(B)}`
`= \frac{\frac{1}{6}}{\frac{1}{3}}`
`= \frac{1}{2}`
So, the probability of getting an even number, given that the number rolled is greater than four, is `\frac{1}{2}`.
Practice Problems
Q`1`. The probability of a student passing in English is `\frac{5}{8}` and of the student passing in both English and Hindi is `\frac{1}{2}` . What is the probability of that student passing in Hindi knowing that he passed in English?
- `\frac{1}{2}`
- `\frac{9}{8}`
- `\frac{4}{5}`
- `\frac{5}{4}`
Answer: c
Q`2`. A fair six-sided die is rolled. If it is known that the number rolled is odd, what is the probability that it is also prime?
- `\frac{1}{2}`
- `\frac{1}{3}`
- `\frac{4}{5}`
- `\frac{2}{3}`
Answer: d
Q`3`. In a survey among few people, `60%` read Hindi newspaper, `40%` read English newspaper and `20%` read both. If a person is chosen at random and if he already reads Hindi newspaper find the probability that he also reads English newspaper.
- `\frac{1}{2}`
- `\frac{1}{3}`
- `\frac{1}{4}`
- `\frac{1}{5}`
Answer: b
Q`4`. Two cards are drawn from a deck of `52` cards where the first card is NOT replaced before drawing the second card. What is the probability that both cards are jack?
- `\frac{9}{24}`
- `\frac{1}{221}`
- `\frac{1}{157}`
- `\frac{2}{3}`
Answer: b
Frequently Asked Questions
Q`1`. What does \( P(A|B) \) represent?
Answer: \( P(A|B) \) represents the probability of event `A` occurring given that event `B` has already occurred. It denotes the likelihood of event `A` given the occurrence of event `B`.
Q`2`. How is \( P(A|B) \) calculated?
Answer: \( P(A|B) \) is calculated using the formula:
`P(A|B) = \frac{P(A \cap B)}{P(B)}`
where \( P(A \cap B) \) is the probability of both events `A` and `B` occurring simultaneously, and \( P(B) \) is the probability of event `B` occurring.
Q`3`. What does \( P(A \cap B) \) represent?
Answer: \( P(A \cap B) \) represents the probability of both events `A` and `B` occurring simultaneously. It denotes the intersection of events `A` and `B`.
Q`4`. What if \( P(B) = 0 \)?
Answer: If \( P(B) = 0 \), the conditional probability \( P(A|B) \) is undefined. This is because division by zero is undefined in mathematics.
Q`5`. How does independence affect \( P(A|B) \)?
Answer: If events `A` and `B` are independent, then \( P(A|B) = P(A) \). In other words, the occurrence of event `B` does not affect the probability of event `A` occurring.