Spring11_midterm_sol

Spring11_midterm_sol - ELEC210 Spring 2011 Midterm Answer...

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Unformatted text preview: ELEC210 Spring 2011 Midterm Answer Sheet Section Ⅰ bbcbd Section Ⅱ Q1. a) P[3 X 6] = P[X = 4 or 5 or 6] = P[1 X 6|X = 2] = P[X = 3 or 4 or 5|X 2] = [ [ = = b) The cdf of X kp k) = c) E X) = ∑ = 4.375 ) = 23.25 E X )=∑ D X) = E X ) − E X) = 4.1094 d) P[3 7] = X = , therefore th e conditional pmf of X is listed as foll ows, K P 3 5 6 7 1 3 |) 4 1 6 1 12 1 4 1 6 e) The probability of Y Y P E Y) = , E Y )= 1 ) 2 3 4 3 16 6 16 4 16 3 16 , D Y) = E Y ) − E Y) = = 0.9961 Q2. a) Total probability is equal to 1, there fore we have the area under the fun ction f x) is 1. c + c − 2) × 1 4 1∪3 5] = 1 − 1 × × − 5 − 3 ) × = 2=1 Then c = 5. b) P[1 X 3] = 1 − P[0 X X x, 0 x 2 ,2 x 5, c) The function f x) = 0, therefore otherwis e 0, 1 , 0 F X) = 1 6 1 −1 + ), 2 4 1, 5 0 2 5 d) Expectation and Variance E X) = ) = ∙ E X )= ) D X) = E ) − E X) = e) P[X 1] = 1 − P[X = + ∙ ∙ + − 1] = 1 − = ∙ =+ = + = 1.4983 = 2 ,1 15 4 f x|A) = ,2 15 0, Q3. a) 0.9 × 0.7 × 0.5 = 0.315 + 2 5 = =+ = b) 0.9 × 0.7 = 0.63 c) Let us define the event “get the $100” as A and the event “answer the question 2 incorrectly” as B, then P[A ] = 1 − P[A] = 1 − 0.315 = 0.685 P[B] = 0.9 × 1 − 0.7) = 0.27 0.27 P[A ] P[B|A ] = = = = 0.3942 [A ] 0.685 P[A ] Since B is a subcase of A , therefore P[A = P[ ]. d) The probability of $ $ E[$] = 10 20 100 P[$] ∑$ $ 0 0.1 0.27 0.315 0.315 $ [$] = 2.7 + 6.3 + 31.5 = 40.5 Q4. a) × = b) × + × = c) P[Vote for C2] = P[Vote for C2|A group]P[A group] + P[Vote for C2|B group]P[B group] = 0.2 × + 0.9 × = d) According to Bayes’ Theorem P[in Group B|vote for C1] = e) P[C1] = ∑ 0.8 0.2 [ & [ × 0.1 0.9 = ×. [ = / / = ...
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