255_Ch_5_PracticeProbs_Solns.pdf

255_Ch_5_PracticeProbs_Solns.pdf - 255 Chapter 5 Solutions...

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255 Chapter 5 Solutions Section 5.1 1 . a. 0.2 b. 0.42 c. 0.7 d. 0.5 e. P (0 , 0) = 0 . 1 , P X (0) P Y (0) = (0 . 16)(0 . 24) = 0 . 1 so X and Y are not independent. 3 . a. p(1,1)=0.15 b. 0.4 c. 0.22 d. P (exactly 4)=0.17 and P (atleast 4)-0.46 e. P 1 (0) = P ( X 1 = 0) = p (0 , 0) + p (0 , 1) + p (0 , 2) + p (0 , 3) = 0 . 19 , P 1 (1) = 0 . 3 etc. Thus E ( X 1 ) = 1 . 7 f. P 2 (0) = P ( X 2 = 0) = p (0 , 0) + p (1 , 0) + p (2 , 0) + p (3 , 0) + p (4 , 0) = 0 . 19 etc g. p (4 , 0) = 0 yet p 1 (4) = 0 . 12 > 0 and p 2 (0) = 0 . 19 > 0 so p ( x 1 , x 2 ) = p 1 ( x 1 ) p 2 ( x 2 ) and thus the two variables are not independent. 4 . a. 0.0025 b. 0.077 c. 0.879 d. The sum of probabilities for blue, orange and green is 0.6. The probability of at least 7 is 0.382. 6 . a. 0.0518 b. 0.4014 c. p ( x, y ) = x ! y !( x - y )! 0 . 6 y 0 . 4 ( x - y ) p x ( x ) p y (4) = 0 . 0194 p y (3) = 0 . 1058 p y (2) = 0 . 2678 p y (1) = 0 . 359 p y (0) = 0 . 248 9 . a. K=3/380000 b. 0.3024 1
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c. 0.3593 d. 10 Kx 2 + 0 . 05 for 20 x 30 e. f y ( y ) is obtained by substituting y for x in (d). X and Y are not independent. 12 . a. 0.05 b. The marginal pdf of X is e - x , x 0 . The marginal pdf of Y is 1 (1+ y ) 2 , 0 y . The two random variables are not independent. c. 0.3 14 . a. (1 - e - λt ) 10 b. P(sucess)= 1 - e - λt , P(k successes among 10 trials)= 10! k !(10 - k )! (1 - e - λt ) k ( e λt ) 10 - k c. P(exactly 5 fail)= ( 9 5 ) (1 - e - λt ) 5 ( e - λt ) 4 ( e - μt ) + ( 9 4 ) (1 - e - λt ) 4 (1 - e - μt )( e - λt ) 5 Section 5.2 18 . a. 14.1 b. 9.6 21 . E ( XY ) = L 2 22 . Revenue = 15 . 4 23 . E ( h ( X, Y )) = 1 4 24 . M Z ( t ) = E ( exp ( tX + tY )) = E ( exp ( tX )) E ( exp ( tY
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