# ProbSetFour - Rensselaer Electrical Computer and Systems...

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Rensselaer Electrical, Computer, and Systems Engineering Department ECSE 4500 Probability for Engineering Applications PS#4 Solutions March 16, 2010 1. (2.09) First we need to calculate the probability that X is less than 1 and that it is greater than 2. Now P [ X < 1] = P [ X 1] = F X (1) = 1 e 1 . P [ X > 2] = 1 P [ X 2] = 1 F X (2) = 1 (1 e 2 ) = e 2 . Since the events are disjoint, the probability that X < 1 or X > 2 is P [ { X < 1 } { X > 2 } ] = P [ X < 1] + P [ X > 2] = 1 e 1 + e 2 = 0 . 767 . 2. (2.11) Let the number of bulbs produced by A and B be n A and n B respectively. We have n A + n B = n , and n is the total number of the bulbs. So P [ A ] = n A n = 1 4 and P [ B ] = n B n = 3 4 . Since we have F X ( x | A ) = (1 e 0 . 2 x ) u ( x ) , F X ( x | B ) = (1 e 0 . 5 x ) u ( x ) , then F X ( x ) = F X ( x | A ) P ( A ) + F X ( x | B ) P ( B ) = 1 4 (1 e 0 . 2 x ) u ( x ) + 3 4 (1 e 0 . 5 x ) u ( x ) . So F (2) = 1 4 (1 e 0 . 2 × 2 ) + 3 4 (1 e 0 . 5 × 2 ) = 0 . 56 , F (5) = 1 4 (1 e 0 . 2 × 5 ) + 3 4 (1 e 0 . 5 × 5 ) = 0 . 85 , F (7) = 1 4 (1 e 0 . 2 × 7 ) + 3 4 (1 e 0 . 5 × 7 ) = 0 . 92 .

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