Conditional Probability

Proof

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 .

FAQs

1. Explain conditional probability with the help of a diagram.
   From the figure shown below, S specifies the sample space, while A and B are two events. In the case when event B has already occurred, our sample space S is limited to B since the probability of an event occurring is now contained within B.

Key Takeaways

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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