For married couples living in Springfield MA, the probability that the husband w
ID: 3236759 • Letter: F
Question
For married couples living in Springfield MA, the probability that the husband will vote on a bond referendum is 0.21, the probability that his wife will vote in the referendum is 0.28, and the probability that both husband and wife will vote is 0.15. What is the probability that. a) At least one member of a married couple will vote? (Husband, wife or both) b) A wife will vote, given that her husband will vote? c) A husband will vote, given that his wife will not vote? d) Can we assume that the decision of the wife to vote is independent of her husband decisions and why?Explanation / Answer
Let P(A) = Probability that the husband will vote = 0.21 P(B) = Probability that the wife will vote = 0.28 P(A B) = Probability both will vote = 0.15 a) To find P(atleast 1 member of the married couple will vote) = P(1 of husband wife will vote) + P(both will vote) = P(husband will vote and wife will not) + P(wife will vote and husband will not) + P(both will vote) = [P(A) - P(AB) ] + [P(B) - P(AB) ] + P(A B) = (0.21 - 0.15) + (0.28 - 0.15) + 0.15 = 0.21 + 0.28 - 0.15 = 0.34 P(atleast 1 member of the married couple will vote) = 0.34 b) To find P(a wife will vote given her husband will vote) = P(B / A) = P(A B) / P(A) = 0.15 / 0.21 = 0.7143 P(a wife will vote given her husband will vote) = 0.7143 c) To find P(a husband will vote given his wife will not vote) = P(A/B') P(B') = P(wife will not vote) = 1 - P(B) = 1 - 0.28 = 0.72 P(A/B') = P(AB')/P(B') = [P(A) - P(AB)] / P(B') = (0.21 - 0.15)/0.72 = 0.0833 P(a husband will vote given his wife will not vote) = 0.0833 d) P(A) x P(B) = 0.21 x 0.28 = 0.0588 P(AB) which is 0.15 Since P(AB) P(A) x P(B) the events A and B are not independent that is The decision of the wife to vote is not independent on her husband's decision
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