Let P(A) = 0.4, P(B|A) = 0.5, and P(B|A^c) = 0.25. Compute P(A|B). A) 0.20 B) 0.35 C) 0.57 D) 0.80

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We can use Bayes' theorem to solve for P(A|B): P(A|B) = P(B|A) * P(A) / P(B) To solve for P(B), we can use the law of total probability: P(B) = P(B|A) * P(A) + P(B|A^c) * P(A^c|A') * P(A') where A' represents the complement of A, i.e. not A. To solve for P(A^c|A'), we can use the law of total probability again: P(A^c|A') = P(A^c|B'A') * P(B'A') + P(A^c|BA') * P(BA') where BA' represents the event where B occurs but not A and B'A' represents the complement of both A and B. We are not given direct information about P(A^c|B'A') or P(A^c|BA'), so we cannot solve for P(B) or P(A|B) exactly. However, we can use the given information to determine that P(B'A') > P(BA'). To see why this is true, note that P(B|A) > P(B|A') since B is more likely to occur if A occurs. And since P(A^c|A) and P(A^c|A') are positive, it follows from the law of total probability that P(A^c|B'A') > P(A^c|BA'). Therefore, P(B'A') is larger than P(BA') since P(B|A') and P(B|A) are positive, and P(A^c|B'A') > P(A^c|BA'). Since P(BA') < P(B) and P(B|A) < 1, it follows that P(A|B) > P(A). Intuitively, this makes sense because given that we know B occurred, it is more likely that A also occurred since P(B|A) is positive. Therefore, the most likely value for P(A|B) is 0.57, corresponding to choice (C).