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Unformatted text preview: ISyE 2027 Probability with Applications Fall 2011 R. D. Foley Solution to Homework 9 December 12, 2011 1. Solution: (a)The c.d.f of Z is F Z ( t ) = Pr { Z 6 t } = 1 Pr { Z > t } = 1 Pr { X > t , Y > t } = 1 Pr { X > t } Pr { Y > t } = 1 e t e t = 1 e( + ) t where t > 0. (b)The p.d.f of Z is ( + ) e( + ) t . (c) Z has exponential distribution with parameter + . (d)The expected value of Z is 1 + = 1 3 + 7 = 1 10 , so the expected value of Z is 6 minutes. 2. Solution: Suppose X denotes the length of time until A finishes the race. Y denotes the length of time until B finishes the race. Z denotes the length of time until C finishes the race. Then X , Y and Z are independent, exponentially distributed random variables with parameters 2 per hour, 3 per hour and 4 per hour. (a)The probability that A takes longer than 1 hour is Pr { X > 1 } = e 2 . (b)Let V = min { Y , Z } , so V has exponential distribution with parameter 3 + 4 = 7 per hour. Thus, the probability that A wins is Pr { X < V } = Z Pr { X < V  V = t } 7 e 7 t dt = Z Pr { X < t } 7 e 7 t dt = Z ( 1 e 2 t ) 7 e 7 t dt = 2 9 (c)Let W denote the length of time until the winning turtle finishes the race. Then W = min { X , Y , Z } , so the c.d.f of W is 1 e 9 t . Thus, the expected time of the winning turtle is 1 9 hour.hour....
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This note was uploaded on 01/22/2012 for the course ISYE 2027 taught by Professor Zahrn during the Fall '08 term at Georgia Institute of Technology.
 Fall '08
 Zahrn

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