# Hw9sol - dx = y 6 + 1 4 f Y | X ( y | x ) = f X,Y ( x,y ) f...

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ISyE 2027B Probability with Applications Fall 2010 Homework 9 Solutions 1. P ( V > 40 , 000 | V > 10 , 000) = P ( { V > 40 , 000 } ∩ { V > 10 , 000 } ) P ( V > 10 , 000) = P ( V > 40 , 000) P ( V > 10 , 000) = P ( Y > 0 . 4) P ( Y > 0 . 1) = R 1 0 . 4 5(1 - y ) 4 dy R 1 0 . 1 5(1 - y ) 4 dy = 0 . 6 5 0 . 9 5 = 0 . 1317 2. (a) P ( N = 3 ,X = 2) = P ( X = 2 | N = 3) P ( N = 3) = ± 3 1 ² × ( 1 2 ) 2 × (1 - 1 2 ) × 1 6 = 1 16 (b) P ( X = 5) = P ( X = 5 ,N = 5) + P ( X = 5 ,N = 6) = P ( X = 5 | N = 5) P ( N = 5) + P ( X = 5 | N = 6) P ( N = 6) = ± 5 5 ² × ( 1 2 ) 5 × 1 6 + ± 6 5 ² × ( 1 2 ) 5 × (1 - 1 2 ) × 1 6 = 1 48 (c) Condition on N , X Bin ( N, 1 / 2), so E [ X | N ] = N/ 2. Thus E [ X ] = E [ E [ X | N ]] = E [ N/ 2] = (1 + 2 + 3 + 4 + 5 + 6) / 6 × 1 / 2 = 7 / 4. 3. P U (0 , 1], f p ( p ) = 1 on (0 , 1]. P (Exactly 2 correct) = Z 1 0 P (Exactly 2 | P = p ) f P ( p ) dp = Z 1 0 ± 3 2 ² p 2 (1 - p ) dp = 1 4 4. When T 1 / 2, U never exceeds 1 / 2. We focus on T > 1 / 2. T U (0 , 1], f T ( t ) = 1 on (0 , 1]. P ( U > 1 2 ) = Z 1 1 / 2 P ( U > 1 2 | T = t ) f T ( t ) dt = Z 1 1 / 2 Z t 1 / 2 1 t du ! dt = Z 1 1 / 2 t - 1 / 2 t dt = Z 1 1 / 2 (1 - 1 2 t ) dt = 1 2 + 1 2 ln 1 2 = 1 2 (1 - ln 2) 5. f X ( x ) = Z 5 4 x + y 6 dy = x 6 + 3 4 f Y ( y ) = Z 2 1 x + y 6

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Unformatted text preview: dx = y 6 + 1 4 f Y | X ( y | x ) = f X,Y ( x,y ) f X ( x ) = 1 6 ( x + y ) x 6 + 3 4 E [ Y | X = 3 / 2] = Z 5 4 y 1 6 ( 3 2 + y ) 3 2 6 + 3 4 dy = Z 5 4 ( y 4 + y 2 6 ) dy = 325 72 1 ISyE 2027B Probability with Applications Fall 2010 f X | Y ( x | y ) = f X,Y ( x,y ) f Y ( y ) = 1 6 ( x + y ) y 6 + 1 4 E [ X | Y = 4] = Z 2 1 x 1 6 ( x + 4) 4 6 + 1 4 dx = Z 2 1 2 11 ( x 2 + 4 x ) dx = 50 33 2...
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## This note was uploaded on 02/21/2011 for the course ISYE 2027 taught by Professor Zahrn during the Fall '08 term at Georgia Tech.

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Hw9sol - dx = y 6 + 1 4 f Y | X ( y | x ) = f X,Y ( x,y ) f...

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