麻省理工学院概率论与数理统计考试题.pdf

18.05. Test 1. (1) Consider events A = {HHH at least once} and B = {TTT at least once}. We want to find the probability P (A � B). The complement of A � B will be c c A � B , i.e. no TTT or no HHH, and c c P (A � B) = 1 − P (A � B ). To find the last one we can use the probability of a union formula c c c c c c P (A � B ) = P (A ) + P (B ) − P (A � B ). c Probability of A , i.e. no HHH, means that on each toss we don’t get HHH. The probability not to get HHH on one toss is 7/8 and therefore, c 7�10 P (A ) = . 8 c c c The same for P (B ). Probability of A �B , i.e. no HHH and no TTT, means that on each toss we don’t get HHH and TTT. The probability not to get HHH and TTT on one toss is 6/8 and, therefore, c c 6 �10 P (A � B ) = . 8 Finally, we get, P (A � B) = 1 − 7�10 + 7�10 − 6 �10�. 8 8 8 (2) We have P (F ) = P (M ) = 0.5, P (CB M ) = 0.05 and P (CB F ) = 0.0025. | | Using Bayes’ formula, P (CB |M )P (M ) 0.05 × 0.5 P (M |CB) = = P (CB |M )P (M ) + P (CB |F )P (F ) 0.05 × 0.5 + 0.0025 × 0.5 (3) We want to