HW 2 Solutions

# HW 2 Solutions - 2.42 Given P(H=1 = 0.2 P(H=2 = 0.05 P(H=0...

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2.42 Given: P(H=1) = 0.2, P(H=2) = 0.05, P(H=0) = 0.75 Where H = number of hurricanes in a year P(J=H=1) = P(D H=1) = 0.99 P(J=H=2) = P(D H=2) = 0.80 Where J and D denote survival of jacket and deck substructure respectively Assume J and D are statistically independent (a) For the case of one hurricane, i.e. H=1 P(damage) = 1-P(JD H=1) = 1- P(J H=1) P(D H=1) = 1 - 0.99x0.99 = 0.0199 (b) For next year where the number of hurricanes is not known. P(no damage) = P(JD H=0)P(H=0)+ P(JD H=1)P(H=1)+ P(JD H=2)P(H=2) = 1x0.75 + 0.99x0.99x0.2 + 0.8x0.8x0.05 = 0.75 + 0.196 + 0.032 = 0.978 (c) P(H=0 JD) = ( 0) ( 0) 1 0.75 0.767 ( ) 0.978 P JD H P H P JH = = × = =

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2.54 (a) T = wait time in queue (in min.) N = number of trucks in queue P(T<5 N=2) = P(loading time for both trucks ahead will take only 2 min. each) = 0.5x0.5 = 0.25 (b) P(T<5) = P(T<5 N=0)P(N=0)+ P(T<5 N=1)P(N=1)+ P(T<5 N=2)P(N=2) Note that if the number of trucks in queue is 3 or more, the waiting time will definitely exceed 5. Hence those items will not contribute any probabilities. Hence P(T<5) = 1x0.175 + 1x0.125 + 0.25x0.3 = 0.375