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Hmwk9Sol

# Hmwk9Sol - ISyE 2027 R D Foley Probability with...

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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 - λt e - μt = 1 - e -( λ + μ ) t where t > 0. (b)The p.d.f of Z is ( λ + μ ) e -( λ + μ ) t . (c) Z has exponential distribution with parameter λ + μ . (d)The expected value of Z is 1 λ + μ = 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 0 Pr { X < V | V = t } 7 e - 7 t dt = Z 0 Pr { X < t } 7 e - 7 t dt = Z 0 ( 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

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Hmwk9Sol - ISyE 2027 R D Foley Probability with...

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