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Unformatted text preview: Introduction to Probability Theory Homework Set III Solutions 1. There are 6 red and 4 black balls in a bag. (a) I sample from the bag n times without replacement ( 1 ≤ n ≤ 10 ). Let X n denote the number of red balls in the sample. Compute the density of X n (i.e, compute f X n ( x ) = P ( X n = x ) for each x ∈ R ). Solution: I would like to simply say that the density function is f X n ( x ) = ( ( 6 x )( 4 n x ) ( 10 n ) if x = 0 , 1 ,...,n otherwise. I’m a little bit unhappy about this solution, since for n ≥ 5 it requires computing such quantities as ( 4 5 ) . This is formally defined ( ( 4 5 ) = (4) 5 5! = 4 . 3 . 2 . 1 . 5! = 0 ) but it’s a little bit odd since we haven’t really though about binomial coefficients ( n k ) in which k > n . I would prefer to look at the ranges of X 1 ,...,X 10 separately and write down a different density function for each one. For n = 1 , the possible values that X n can take on are and 1 (these are the different numbers of red balls that we could take out), and we have f X 1 ( x ) = ( ( 6 x )( 4 1 x ) ( 10 1 ) if x = 0 , 1 otherwise. For n = 2 , the possible values that X n can take on are , 1 and 1 and we have f X 2 ( x ) = ( ( 6 x )( 4 2 x ) ( 10 2 ) if x = 0 , 1 , 2 otherwise. For n = 3 , the possible values that X n can take on are , 1 , 2 and 3 and we have f X 3 ( x ) = ( ( 6 x )( 4 3 x ) ( 10 3 ) if x = 0 , 1 , 2 , 3 otherwise. For n = 4 , the possible values that X n can take on are , 1 , 2 , 3 and 4 and we have f X 4 ( x ) = ( ( 6 x )( 4 4 x ) ( 10 4 ) if x = 0 , 1 , 2 , 3 , 4 otherwise. 1 For n = 5 , the possible values that X n can take on are 1 , 2 , 3 , 4 and 5 (note that if we take out 5 balls we must take out at least one red one) and we have f X 5 ( x ) = ( ( 6 x )( 4 5 x ) ( 10 5 ) if x = 1 , 2 , 3 , 4 , 5 otherwise. For n = 6 , the possible values that X n can take on are 2 , 3 , 4 , 5 and 6 (note that if we take out 6 balls we must take out at least two red ones) and we have f X 6 ( x ) = ( ( 6 x )( 4 6 x ) ( 10 6 ) if x = 2 , 3 , 4 , 5 , 6 otherwise. For n = 7 , the possible values that X n can take on are 3 , 4 , 5 and 6 (note that if we take out 7 balls we must take out at least three red ones, but we can’t take out more than 6 ) and we have f X 7 ( x ) = ( ( 6 x )( 4 7 x ) ( 10 7 ) if x = 3 , 4 , 5 , 6 otherwise. For n = 8 , the possible values that X n can take on are 4 , 5 and 6 and we have f X 8 ( x ) = ( ( 6 x )( 4 8 x ) ( 10 8 ) if x = 4 , 5 , 6 otherwise. For n = 9 , the possible values that X n can take on are 5 and 6 and we have f X 9 ( x ) = ( ( 6 x )( 4 9 x ) ( 10 9 ) if x = 5 , 6 otherwise. Finally, for n = 10 , the only possible value that X n can take on is 6 and we have f X 10 ( x ) = ( ( 6 x )( 4 10 x ) ( 10 10 ) = 1 if x = 6 otherwise....
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 Spring '09
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