If for two events [math: A] and [math: B] , [math: Pr(A)=920] Pr(A)= \frac{9}{20} , [math: Pr(B)=12] Pr(B)= \frac{1}{2} , and [math: Pr(A∩B)=15] Pr(A\cap B)= \frac{1}{5} , then [math: P(B|A)] is equal to :

Question

If for two events [math: A] and [math: B] , [math: Pr(A)=920] Pr(A)= \frac{9}{20} , [math: Pr(B)=12] Pr(B)= \frac{1}{2} , and [math: Pr(A∩B)=15] Pr(A\cap B)= \frac{1}{5} , then [math: P(B|A)] is equal to :

BlackTom AI · Accepted Answer

Let’s break down the given information and compute step by step.
First, identify what we need: P(B|A) = P(A ∩ B) / P(A).
Option A is 9/40. This would imply a ratio of 0.225, which would correspond to P(A ∩ B) being 0.225 times P(A). But the actual values must come from the provided fractions, so we test against the correct formula rather than guess.
Option B is 9/20. This equals 0.45. If P(B|A) were 9/20, then P(A ∩ B) would need to be (9/20) × P(A) = (9/20) × (9/20) = 81/400, which is not equal to the given P(A ∩ B) = 1/5. Hence this option is inconsistent with the data.
Option C is 4/9. This is approximately 0.444..., which will be obtained by the correct computation P(A ∩ B) / P(A) = (1/5) / (9/20). Perform the division: (1/5) ÷ (9/20) = (1/5) × (20/9) = 20/45 = 4/9. This matches the provided intersection and A probabilities exactly.
Option D is 2/5. This equals 0.4. If P(B|A) were 2/5, then P(A ∩ B) would be (2/5) × P(A) = (2/5) × (9/20) = 18/100 = 9/50, which does not equal the given P(A ∩ B) = 1/5. Therefore this option is not correct.
To summarize the correct computation: P(B|A) = P(A ∩ B) / P(A) = (1/5) / (9/20) = (1/5) × (20/9) = 4/9, which corresponds to option C.

MUF0142 Fund. Mathematics Unit 2 - Semester 2, 2025

If for two events [math: A] and [math: B] , [math: Pr(A)=920] Pr(A)= \frac{9}{20} , [math: Pr(B)=12] Pr(B)= \frac{1}{2} , and [math: Pr(A∩B)=15] Pr(A\cap B)= \frac{1}{5} , then [math: P(B|A)] is equal to :

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