mm2_solutions_complete

# mm2_solutions_complete - Problem 4.1 Let x1 x10 be a...

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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! 10! = 0 . 0595 P ( X = 5) = 5 · 4 · 3 · 2 · 5 · 5! 10! = 0 . 0198 P ( X = 6) = 5! · 5 · 4! 10! = 0 . 004 One can note that since there are only 5 male persons, P ( X = 7) = ... = P ( X = 10) = 0. 1

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Problem 4.2 Let us consider a r.v. X head representing a number of heads obtained in the measurements. X head has a the following sample space: X head ∈ { 0 , 1
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## This note was uploaded on 02/03/2008 for the course STAT none taught by Professor None during the Spring '08 term at University of Iowa.

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mm2_solutions_complete - Problem 4.1 Let x1 x10 be a...

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