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 125 4 p 2 is
A. A
Minimum of the chosen numbers is at most 40 It means we have to choose all the numbers from 1 to 40 only from 41 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 41 to 100   = 60 × 60 × 60 = 60 3 Total number of favourable cases = 100 3 - 60 3 So, p 2 = 100 3 - 60 3 100 3 = 20 3 5 3 - 3 3 20 3 × 5 3 = 98 125 Hence, 125 4 p 2 = 125 4 × 98 125 = 24 . 50
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