131A_1_midterm_sol_06

131A_1_midterm_sol_06 - 1 2 3 4 5 6 1 1/2 1/4 1/8 F X(x x c...

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EE 131A Midterm Exam Solution Winter 2006 K. Yao 1. a. 2 x 4 x 4 x 3 x 3 x 2 x 2 x 1 x 1 = 1152 . b. 4 x 7! = 2880 . 2. a. P ( B 1 ) = 1 / 4 , P ( B 2 ) = 1 / 4 , P ( B 3 ) = 2 / 4 = 1 / 2 . b. P ( A | B 1 ) = 5 / 10 = 1 / 2 , P ( A | B 2 ) = 10 / 15 = 2 / 3 , P ( A | B 3 ) = 5 / 15 = 1 / 3 . c. Use the Theory of Total Probability. P ( A ) = P ( A B 1 )+ P ( A B 2 )+ P ( A B 3 ) = P ( A | B 1 ) P ( B 1 )+ P ( A | B 2 ) P ( B 2 ) + P ( A | B 3 ) P ( B 3 ) = (1 / 2) x (1 / 4) + (2 / 3) x (1 / 4) + (1 / 3) x (1 / 2) = 11 / 24 . d. Use Bayes Rule. P ( B 1 | A ) = P ( A | B 1 ) P ( B 1 ) P ( A | B 1 ) P ( B 1 ) + P ( A | B 2 ) P ( B 2 ) + P ( A | B 3 ) P ( B 1 ) = (1 / 2) x (1 / 4) (11 / 24) = 3 / 11 . 3. a. Since there are 6 possible number from 1 to 18 divisible by 3, and each one is equally likely, then the probability of winning at each attempt is p = 6 / 18 = 1 / 3 . The probability of losing at each attempt is q = 2 / 3 . Since this is a Bernoulli trail of length n = 8 , then p (3) = C 8 3 (1 / 3) 3 (2 / 3) 5 = 1792 / 6561 = 0 . 273. b. 8 X k =3 p ( k ) = 1 - 2 X k =0 p ( k ) = 1 - C 8 0 p 0 q 8 - C 8 1 p 1 q 7 - C 8 2 p 2 q 8 = 3489 6561 = 0 . 532 4. a. F X ( x ) = n =1 (1 / 2 n ) U ( x - n ) , < x < . b. The cdf F X ( x ) is given below.
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Unformatted text preview: 1 2 3 4 5 6 . .. 1 1/2 1/4 1/8 F X (x) x c. f X ( x ) = ∑ ∞ n =1 (1 / 2 n ) δ ( x-n ) , ∞ < x < ∞ . d. The pdf f X ( x ) is given below. (½) (x-1) 6 . .. 1 2 3 4 5 (¼) (x-2) (1/8) (x-3) f X (x) x . . . e. Since (1 / 2) + (1 / 4) + (1 / 8) + (1 / 16) = (15 / 16) , the smallest x satisfying P ( X ≤ x ) = (15 / 16) is 4 , since P ( X ≤ 3 . 99999 ... ) = (14 / 16) . f. Since P ( X ≤ 4 . 99999 ... ) = (15 / 16) but P ( X ≤ 5) = (31 / 32) , thus the largest x satisfying P ( X ≤ x ) = (15 / 16) is 4 . 99999 ... = 5-....
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This note was uploaded on 06/09/2010 for the course EE 131A taught by Professor Lorenzelli during the Spring '08 term at UCLA.

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