hw2sol - 3-110 a) Let X denote the number of calls in one...

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3-88 Let X = # of opponents until the player is defeated (i.e. # of trials until first success). Success = player defeated p=0.2 (a) Find f(x) for number of successes until first failure. X~geometric with p = 0.2 f(x) = (1 – p) x – 1 p = 0.8 (x – 1) *0.2 (b) Find P(X 3) = 1- P(X<3) = 1 – [P(X=1) + P(X=2)] = 0.64 Remember X is counting the # of opponents played including the last opponent who defeats the player. So when X=n that means the player defeated n-1 opponents and lost to the n th opponent (c) μ = E(X) = 1/p = 5 (d) P(X 4) = 1- P(X<4) = 1- [P(X=1)+P(X=2)+P(X=3)] = 0.512 (e) Let Y = number of games played until the player contests four or more opponents (count # of trials until first success) Success = contesting 4 or more opponents p=0.512 Y ~ geometric with p=0.512 E(Y) = 1/p o = 1.95 games
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Unformatted text preview: 3-110 a) Let X denote the number of calls in one hour. Then, X is a Poisson random variable with = 10. P X e ( ) ! . = = =-5 10 5 0 0378 10 5 . b) P X e e e e ( ) ! ! ! . = + + + =----3 10 1 10 2 10 3 0 0103 10 10 10 2 10 3 c) Let Y denote the number of calls in two hours. Then, Y is a Poisson random variable with = 20. P Y e ( ) ! . = = =-15 20 15 0 0516 20 15 d) Let W denote the number of calls in 30 minutes. Then W is a Poisson random variable with = 5. P W e ( ) ! . = = =-5 5 5 01755 5 5 3-118 a) Let X denote the failures in 8 hours. Then, X has a Poisson distribution with = 0.16. 8521 . ) ( 16 . = = =-e X P b) Let Y denote the number of failure in 24 hours. Then, Y has a Poisson distribution with = 0.48. 3812 . 1 ) ( 1 ) 1 ( 48 =-= =-= -e Y P Y P...
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This note was uploaded on 05/01/2008 for the course IE 360 taught by Professor Chen during the Spring '08 term at University of Louisville.

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