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hw5_sol

# hw5_sol - IEOR 172 Probability and Risk Analysis for...

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IEOR 172: Probability and Risk Analysis for Engineers, Fall 2007 Homework 5 Solution Chapter 4 Question 28 3 · 4 20 = 3 5 . Question 41 10 X i =7 10 i ! 1 2 10 Question 42 5 3 ! p 3 (1 - p ) 2 + 5 4 ! p 4 (1 - p ) + p 5 3 2 ! p 2 (1 - p ) + p 2 ⇐⇒ 6 p 3 - 15 p 2 + 12 p - 3 0 ⇐⇒ 6( p - . 5)( p - 1) 2 0 ⇐⇒ p . 5 Question 44 α n X i = k n i ! p i 1 (1 - p 1 ) n - i + (1 - α ) n X i = k n i ! p i 2 (1 - p 2 ) n - i Question 46 Denote p as the probability of making correct decision: p = . 8 · . 65 + . 9 · . 35 = . 835 So the probability of the jury did the correct decision: P { correct decision } = . 65 12 X i =9 12 i ! p i (1 - p ) 12 - i + . 35 12 X i =4 12 i ! = . 92 . The probability of defendants is convicted: P { correct decision } = . 65 12 X i =9 12 i ! p i (1 - p ) 12 - i + . 35 3 X i =1 12 i ! p i (1 - p ) 12 - i = . 57 1

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Question 50 (a) P( h, t, t | 6 heads) = P( h, t, t, 6 heads) P(6 heard) = P( h, t, t )P(6 heads | h, t, t ) P(6 heads) = pq 2 7 5 ! p 5 q 2 10 6 ! p 6 q 4 = 1 10 (b) P( t, h, t | 6 heads) = P( t, h, t, 6 heads) P(6 heard) = P( t, h, t )P(6 heads | h, t, t ) P(6 heads) = pq 2 7 5 ! p 5 q 2 10 6 ! p 6 q 4 = 1 10 Problem 1 Suppose we order n newspapers, then the expected profit E[ X ] = n X k =0 1 25 ( k + . 1( n - k )) + 24 X k = n +1 1 25 n - . 3 n = 1 25 . 9 n ( n + 1) 2 + . 1 n ( n + 1) + n (24 - n ) ! - . 3 n = 1 25 24 . 55 n - . 45 n 2 - . 3 n. The first order optimal condition shows that 24 . 55 - . 9 n - 7 . 5 = 0 ⇐⇒ n = 18 . 94 . Since n could only be integer, we round up and down, and the optimal order is 19.
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