SolutionA - 2 ,b = 4 . The pdf is shown below: P ( X <...

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IE 300 / GE 331 Spring 2009 Midterm Exam (A) Solution Problem 1. Let Z be the event of having cancer; P be the event of positive result. Therefore, P ( Z ) = 0 . 005 , P ( P | Z ) = 0 . 98 , P ( P C | Z C ) = 0 . 98. (a) P ( Z C | P ) = P ( P | Z C ) P ( Z C ) P ( P ) = P ( P | Z C ) P ( Z C ) P ( P | Z ) P ( Z ) + P ( P | Z C ) P ( Z C ) = (1 - P ( P C | Z C )(1 - P ( Z )) P ( P | Z ) P ( Z ) + (1 - P ( P C | Z C ))(1 - P ( Z )) = (0 . 02)(0 . 995) (0 . 98)(0 . 005) + (0 . 02)(0 . 995) = 0 . 802 (b) Expected # of positive results= 10000 × P ( P ) = 10000[(0 . 98)(0 . 005) + (0 . 02)(0 . 995)] = 248 . Problem 2. (a) There are ± 3 2 ² = 3 ways for him to pick the two coins, only one of these gives him two fair coins. Hence P (did not pick the biased coin) = 1 / 3 . (b) Let A denote the event that Dilbert did not pick the biased coin, i.e., picked the two fair coins, and B the event that the two tosses resulted in one Head and one Tail. We know from (a) that P ( A ) = 1 / 3. It is easy to see that P ( B | A ) = 1 / 2. P ( B | A C ) = P (Heads on biased, Tails on fair) + P (Heads on fair, Tails on biased) 1
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IE 300 / GE 331: Midterm Exam (A) Solution 2 = 3 / 4 × 1 / 2 + 1 / 4 × 1 / 2 = 1 / 2 . P ( A | B ) = P ( B | A ) P ( A ) P ( B | A ) P ( A ) + P ( B | A C ) P ( A C ) = 1 / 2 × 1 / 3 1 / 2 × 1 / 3 + 1 / 2 × 2 / 3 = 1 / 3 . Problem 3. (a) For a uniform r.v., E [ X ] = a + b 2 and Var [ X ] = ( b - a ) 2 12 a = -
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Unformatted text preview: 2 ,b = 4 . The pdf is shown below: P ( X < 0) = Z-2 1 / 6 d x = x 6 -2 = 1 / 3 . (b) We use LOTUS E h | X | i = Z | X | f X ( x ) d x = 1 / 6 h Z-2 (-x ) d x + Z 4 x d x i = 1 / 6 h-x 2 2 -2 + x 2 2 4 i = 5 / 3 . (c) Y = | X | takes values in [0 , 4], hence F Y ( ) = 0 for < 0 and F Y ( ) = 1 for > 4. For 0 2, F Y ( ) = P { Y } = P {- X } = / 3. For 2 4, F Y ( ) = P { Y } = P { X } = ( + 2) / 6. Hence, f Y ( ) = d F Y ( ) d = 1 / 3 , 2 , 1 / 6 , 2 < 4 , , else . Problem 4. (a) P ( X k ) = + X i = k i e- i ! (b) Let P ( X k ) = P r , then P (all bikes are in use in on exactly 3 days) = 6 3 ! P 3 r 1-P r 3...
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This note was uploaded on 09/08/2009 for the course GE 331 taught by Professor Negarkayavash during the Spring '09 term at University of Illinois at Urbana–Champaign.

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SolutionA - 2 ,b = 4 . The pdf is shown below: P ( X <...

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