 # Conditional Probability

Conditional probability is one of the most important concepts in the topic of probability. In CAT exam, a few of the questions based on probability are always included in the CAT quantitative aptitude section every year. These questions based on conditional probability are easily solvable and scoring which can help the CAT aspirants to score more in the CAT exam.

## Theory of Conditional Probability:

To understand the concept of conditional probability, it is important to know about dependent and independent events. In independent events, each event is never affected by the other events while in dependent events, each event can be affected by other events.

The concept of conditional probability is based on the concept of dependent events, where the probability that an event B takes place will depend on whether another event A has or has not taken place.

The conditional probability of an event B, given the event A; denoted by P(B/A) is defined as

P(B/A) = P(A ∩ B)/P(A) where P(A) ≠ 0

Here, P(A ∩ B) is the joint probability of A and B, or the probability of both A and B together.

Let us understand this formula with an example:

A teacher took two tests in a class. In the tests, 42% of the total students in that class passed the first test while only 25% of the students passed both the tests. Find out the percent of those students who passed the first test has also passed the second test.

Solution:

This problem describes a conditional probability since it asks us to find the probability that the second test was passed given that the first test was passed.

P(Second/First) = P(First ∩ Second)/P(first) = 0.25/0.42 = 0.60 = 60%

## Illustrations on Conditional Probability for CAT

Example 1:

In a certain factory, there are 5 working days in a week. If the probability that it is Thursday and that an worker is absent is 0.03. Find the probability that an worker is absent, given that today is Thursday.

Solution:

Since there are 5 working days in a week, the probability that it is Thursday is 0.2.

P(Absent/Thursday) = P(Absent ∩ Thursday)/P(Thursday) = 0.03/0.2 = 0.15 = 15% .

Example 2:

At a CBSE school, 18% of all students play football and basketball and 32% of all students play football. What is the probability that a student plays basketball given that the student plays football?

B → Basketball F → Football

P(B/F) = P(B ∩ F)/P(F) = 18/32 = 0.56 = 56%