Three numbers are chosen at random, one after another with replacement, from the set S=\ 1,2,3, , 100\ . Let p_ 1 be the probability that the maximum of chosen numbers is at least 81 and p_ 2 be the probability that the minimum of chosen numbers is at most 40 . The value of 625 4 p 1 is
A. A
Maximum of the chosen numbers is at least 81 It means we have to choose at least one number from 81 to 100 Total number of possible selections = 100 × 100 × 100 = 100 3 Favourable cases = Total - unfavourable cases Unfavourable cases are those in which we have selected all the three numbers form 1 to 80   = 80 × 80 × 80 = 80 3 Total number of favourable cases = 100 3 - 80 3 So, p 1 = 100 3 - 80 3 100 3 = 20 3 5 3 - 4 3 20 3 × 5 3 = 125 - 64 125 = 61 125 Hence, 625 4 p 1 = 625 4 × 61 125 = 76 . 25
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