MidtermSolution

# MidtermSolution - Midterm Solutions August 1 2007 1...

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Midterm Solutions August 1, 2007 1. ( 10 2 ) + ( 8 2 ) + ( 6 2 ) ( 24 2 ) 2. ( 10 6 ) 6! 10 6 = 0 . 1512 3. (a) P ( X 11) = R 11 3 e - 3 x dx = e - 3 . 11 (b) For x 0, P ( X 2 x ) = P ( X x ) = 1 - P ( X > x ) = 1 - e - 3 x . ( d/dx )(1 - e - 3 x ) = 3 2 x e - 3 x . f X 2 ( x ) = 3 2 x e - 3 x for x > 0 and 0 elsewhere. 4. A is the event the 1st airline goofed, B for the 2nd airline, and C for the 3rd airline. P ( A ) = 4 / 10, P ( B ) = 2 / 10, P ( C ) = 1 / 10. P ( A | A B C ) = P ( A ( A B C )) P ( A B C ) = P ( A ) P ( A B C ) . P ( A B C ) = P ( A ) + P ( B ) + P ( C ) - P ( AB ) - P ( AC ) - P ( BC ) + P ( ABC ) = 4 / 10+2 / 10+1 / 10 - (4 / 10)(2 / 10) - (4 / 10)(1 / 10) - (2 / 10)(1 / 10)+ (4 / 10)(2 / 10)(1 / 10). P ( A ) = 4 / 10. The resulting probability is given by 400
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