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hw1 - solutions - f08

# hw1 - solutions - f08 - IE 336 Operations Research...

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IE 336: Operations Research - Stochastic Models Homework Set #1 Solutions Fall 2008 1. (Hint: Try considering A ( B 1 B 2 ∪ · · · ∪ B n ) = ( A B 1 ) ( A B 2 ) ∪ · · · ∪ ( A B n ) . ) Figure 1: Venn Diagram for 4 B events Based on the Venn Diagram we can say that A = ( A B 1 ) ( A B 2 ) ∪ · · · ∪ ( A B n ) . (1) Taking the probability we have, P ( A ) = P ( A B 1 ) P ( A B 2 ) ∪ · · · ∪ P ( A B n ); (2) by the third probability axiom, P ( A ) = P ( A B 1 ) + P ( A B 2 ) + · · · + P ( A B n ) . (3) Using the definition of conditional probability, P ( A B i ) = P ( B i ) P ( A | B i ) . (4) So, subsituting (4) into (3) we obtain, P ( A ) = P ( B 1 ) P ( A | B 1 ) + P ( B 2 ) P ( A | B 2 ) + · · · + P ( B n ) P ( A | B n ) (5) P ( A ) = n X i =1 P ( B i ) P ( A | B i ) . ( Done ) (6) 1

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2. (a) B = { 1,3 } , C = { 3,4 } (b) If two events are independent then P ( A B ) = P ( A ) P ( B ) . Event ( A B ) = { 1 } ; so, P ( A B ) = 1 4 . We know that P ( A ) = 1 2 and P ( B ) = 1 2 . Thus, P ( A ) P ( B ) = 1 2 · 1 2 = 1 4 = P ( A B ); therefore, A and B are independent. ( Done ) (c) Similar to part (B) except we need event ( A C ). Event ( A C ) = so, P ( A C ) = 0 . We know that P ( A ) = 1 2 and P ( C ) = 1 2 . Thus, P ( A ) P ( C ) = 1 2 · 1 2 = 1 4 6 = P ( A C ); therefore, A and C are not independent. ( Done ) (Note: A and C form a partition of the sample space) 3. We know that R -∞ f ( x ) dx = 1, so R 0 ce - 5 x dx = 1. c = 5. E [ X ] = R 0 x · 5 e - 5 x dx = 5 5 . Note: (a) Some integration tables will have the appropriate form of the integral. However, should you not
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hw1 - solutions - f08 - IE 336 Operations Research...

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