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
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This note was uploaded on 04/15/2010 for the course ECSE 4500 taught by Professor Woods during the Spring '08 term at Rensselaer Polytechnic Institute.

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ProbSetFour - Rensselaer Electrical, Computer, and Systems...

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