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131A_1_midterm_solution

131A_1_midterm_solution - EE 131A Midterm Solution Spring...

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EE 131A Midterm Solution Spring 2008 K. Yao 1. a. S = { HHH, HHT, HTH, THH, TTH, THT, HTT, TTT } A = { HTT, HHT, HTH, HHH } B = { HHT, HTH, THH } b. A B = { HHT, HTH } . P ( A | B ) = P ( A B ) P ( B ) = 2 / 8 3 / 8 = 2 3 c. P ( B | A ) = P ( A B ) P ( A ) = 2 / 8 4 / 8 = 1 2 2. a. (Method 1) The plane fails if all 4-engines fail with a probability of q 4 or any 3 out of the 4 engines fail with a probability of C 4 3 q 3 (1 - q ) = 4 q 3 (1 - q ). Thus, p (4) = 1 - q 4 - 4 q 3 (1 - q ) = 1 - 10 - 4 - 4 × 10 - 3 × 0 . 9 = 0 . 9963 (Method 2) The plane is safe if all 4-engines are working with probability p 4 , or any 3 out of 4 engines are working with probability C 4 3 p 3 q , or any 2 out of 4 engines are working with probability C 4 2 p 2 q 2 . Thus, P (4) = p 4 + C 4 3 p 3 q + C 4 2 p 2 q 2 = (0 . 9) 4 + 4 × (0 . 9) 3 (0 . 1) + 6 × (0 . 9) 2 (0 . 1) 2 = 0 . 9963 b. (Method 1) Similarly, the plane fails if all 3 engines fail with a probability of q 3 or any 2 out of the 3 engines fail with a probability of C 3 2 q 2 (1 - q ) = 3 q 2 (1 - q ). Thus, p (3) = 1 - q 3 - 3 q 2 (1 - q ) = 1 - 10 - 3 - 3 × 10 - 2 × 0 . 9 = 0 . 972 (Method 2) The plane is safe if all 3 engines are working with probability p 3 , or any 2 out of 3 engines are working with probability C 3 2 p 2 q , or any 2 out of 4 engines are working with probability C 4 2 p 2 q 2 . Thus, P (3) = p 3 + C 3 2 p 2 q = (0 . 9) 3
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