The probability of event 'A' occurring assuming event 'B' has already occurred is known as conditional probability P(A | B).
Assuming A and B are two independent occurrences, then P(A/B) is the probability of A occurring, given that B has already occurred. This probability is given by:
P(A|B) = P(A⋂B) ⁄ P(B) ,
Where P(A⋂B) = P(B) ⁄ P(A ⁄ B).
Let us assume S be the sample space
Then we have,
P(A∩B) = P(A∩B) ⁄ P(S) = P(A∩B) ⁄ P(A)×(P(A) ⁄ P(S))
= (P(A) ⁄ P(S)) × P(A∩B) ⁄ P(A)
Therefore, P(A∩B) ⁄ P(A) = P(B ⁄ A)
= P(A) × P(B ⁄ A) —----------------------- Equation 1
Therefore when A nad B in equation-1 are interchanged,
We get,
P(B) × P(A ⁄ B).
Example Question: When two dies are thrown simultaneously, the sum of the numbers obtained is found to be 7. Find out the probability that the number 3 has appeared at least once?
Solution:
The sample space S would be made up of all the numbers that might be generated by combining two dies. As a result, S will consist of 66 events or 36 total events.
Event A = Combination in which 3 has appeared at least once.
Event B = Combination of the numbers which sum up to 7.
A = {(3, 3)(3, 4)(3, 5)(3, 6)(1, 3)(2, 3)(4, 3)(5, 3)(6, 3)(3, 1), (3, 2)}
B = {(1, 6)(2, 5)(3, 4)(4, 3)(5, 2)(6, 1)}
P(A) = 11/36
P(B) = 6/36
Therefore,
A ∩ B = 2
P(A ∩ B) = 2/36
Applying the conditional probability formula we get,
P(A|B) = P(A⋂B)P(B) = 236636 = 13 .
In this article, we have discussed ‘Conditional Probability, Problem Solving with the help of an example. Check out the playlist Mathematical Induction for the following topics.
Check out this problem - Subarray Sum Divisible By K
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