Let P(A ∩B) = 0.3, and P(A ∩ B^c) = 0.15. Compute P(B|A). A) 0.33 B) 0.45 C) 0.5 D) 0.67

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Quizplus · Accepted Answer

Using Bayes' theorem, we have P(B|A) = P(A ∩ B)/P(A) = P(A|B) * P(B) / P(A). We are given P(A ∩ B) = 0.3 and we can find P(A) using the law of total probability: P(A) = P(A ∩ B^c) + P(A ∩ BS0U0P0cS0S0P1) = 0.15 + (P(A ∩ BS0U0P0S0S0P0) * P(BS0U0P0S0S0P0) + P(A ∩ BS0U0P0S0S0P1) * P(BS0U0P0S0S0P1) + P(A ∩ BS0U0P1S0S0P0) * P(BS0U0P1S0S0P0) + P(A ∩ BS0U0P1S0S0P1) * P(BS0U0P1S0S0P1)) = 0.15 + P(BS0U0P0S0S0P0) * P(A ∩ BS0U0P0S0S0P0 | BS0U0P0S0S0P0) + P(BS0U0P0S0S0P1) * P(A ∩ BS0U0P0S0S0P1 | BS0U0P0S0S0P1) + P(BS0U0P1S0S0P0) * P(A ∩ BS0U0P1S0S0P0 | BS0U0P1S0S0P0) + P(BS0U0P1S0S0P1) * P(A ∩ BS0U0P1S0S0P1 | BS0U0P1S0S0P1), where we used the fact that BS0U0P0cS0S0P1 and B^c are complements and therefore the probability of A given one is the probability of its complement given the other. We are not given the probabilities P(BS0U0P0S0S0P0), P(BS0U0P0S0S0P1), P(BS0U0P1S0S0P0), and P(BS0U0P1S0S0P1), which prevent us from finding P(A) and therefore P(B|A). However, we do know that P(A) ≤ P(A ∩ B) / P(B) = 0.3 / P(B). Since P(B|A) is the conditional probability that B occurs given A occurs, it is intuitively reasonable that P(B|A) increases as the probability of B increases or as the probability of A decreases. Therefore, we can say that P(B|A) ≥ 0.3 / (P(BS0U0P0S0S0P0) + P(BS0U0P0S0S0P1) + P(BS0U0P1S0S0P0) + P(BS0U0P1S0S0P1)). We can see from the answer choices that the only one that satisfies this inequality is D: P(B|A) ≥ 0.3 / 0.45 = 0.67. Note that we can't determine whether P(B|A) is greater than 0.67 or not, but we know that it has to be at least 0.67 based on the given information.