The probability of A, given B has occurred is denoted by P(AB). Conditional probability of an event A is calculated when the event B has already occurred by the given formula.
$P(AB)$ = $\frac{P(A and B)}{P(B)}$
Example:
In an exam, two reasoning problems, 1 and 2, are asked. 35% students solved problem 1 and 15% students solved both the problems. How many students who solved the first problem will also solve the second one?
Solution:
Probability of student solving problem $1, P(1) = 0.35$
Probability of student solving both problem, $P(1\ and\ 2) = 0.15$
Probability of solving 2 if 1 is solved, $P(21)$ = $\frac{P(1 and 2)}{P(1)}$ = $\frac{0.15}{0.35}$ = $0.428$
Conditional Probability Word Problems
Problem 1:
Out of 50 people surveyed in a study, 35 smoke in which there are 20 males. What is the probability the if the person surveyed is a smoker then he is a male?
Solution:
Probability of the person being male and a smoker, $P(A\ and\ B)$ = $\frac{20}{50}$
Probability of person being smoker, $P(A)$ = $\frac{35}{50}$
Probability of a person being male if he is smoker, $P(BA)$ = $\frac{P(A and B)}{P(A}$ = $\frac{20}{35}$ = $\frac{1}{7}$
Hence, the compound probability that if the person surveyed is a smoker then he will be a male = $\frac{1}{7}$Problem 2:
The probability of raining on Sunday is $0.07$. If today is Sunday then find the probability of rain today.
Solution:
Probability that it is raining and the day is Sunday, $P(A\ and\ B) = 0.07$
Probability that is is Sunday, $P(B)$ = $\frac{1}{7}$
Probability that it will rain if today is Sunday, $P(AB)$ = $\frac{0.07}{\frac{1}{7}}$ = $0.49$
Hence, the compound probability of raining if it is Sunday is $0.49$.Problem 3:
In a school the third language has to be chosen between Hindi and French. If a student has taken French then what is the probability that he will take Hindi, if the probability of taking Hindi is 0.34?
Solution:
Probability of taking French and Hindi, $P(A\ and\ B) = 0$ as they are mutually exclusive events.
Probability of taking French, $P(B) = 0.34$
Probability of taking Hindi if French has been opted, $P(AB)$ = $\frac{P(A and B)}{P(B)}$ = $\frac{0}{0.34}$ = $0$
Compound probability of mutually exclusive events is 0.Problem 4:
Using a Venn diagram show the probability of occurrence of B if A has already occurred.
Solution:
Here is the venn diagram showing probability of A and B.
The shaded area shows the probability of B if A has already occurred.
Problem 5:
The given table shows the data of 10 point holders in a given class of 30. Find the probability that the student getting 10 point is a girl.

10 pointer 
Not 10 pointer

Girl 
3 
8 
Boy 
6 
13 
Solution:
From the table we can retrieve the given information,
Total boys = 19
Total girls = 11
Girl getting 10 point = 3
Probability that the person getting 10 point is a girl = $\frac{3}{11}$
It can be explained using the formula $P(AB)$ = $\frac{P(A and B)}{P(B)}$ as given here,
Probability of the student being a girl, $P(B)$ = $\frac{11}{30}$
Probability of the student getting 10 point and being a girl, $P(A\ and\ B)$ = $\frac{3}{30}$
Probability that the 10 pointer being selected is a girl, $P(AB)$ = $\frac{P(A and B)}{P(B)}$ = $\frac{3}{11}$