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Unformatted text preview: 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!...
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 Spring '08
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