mm2_solutions - Problem 4.1 Let x1 x10 be a sequence of...

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Problem 4.1 Let x 1 , . . . , x 10 be a sequence of ranked persons where x i can be either a male or a female. If X = 1, then there is a female person on the first place of the sequence. The number of ways we can create a sequence of 10 persons that starts with a female is equal to 5 · 9! (There are 5 different persons that can be put on the first place; the other 9 persons can be placed arbitrary). The total number of permutations of 10 persons is equal to 10!. Thus, P ( X = 1) = 5 · 9! 10! = 0 . 5 If X = 2, then the sequence starts with M, F, . . . . The first male can be chosen in 5 different ways; there are 5 ways to choose the second female; the other 8 persons can be placed arbitrary. There- fore, P ( X = 2) = 5 · 5 · 8! 10! = 0 . 2778 If X = 3, then the sequence starts with M, M, F, . . . . The first male can be chosen in 5 different ways; the second male can be chosen in 4 ways. Thus, P ( X = 3) = 5 · 4 · 5 · 7! 10! = 0 . 1389 In the similar fashion we can find the rest of probabilities: P ( X = 4) = 5 · 4 · 3 · 5 · 6!
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