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 β  bbcbd Section β‘ 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) = β = 4.375 ) = 23.25 E X )=β D X) = E X ) β 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 ) β 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 β 2) Γ 1 4 1βͺ3 5] = 1 β 1 Γ Γ β 5 β 3 ) Γ = 2=1 Then c = 5. b) P[1 X 3] = 1 β 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 β1 + ), 2 4 1, 5 0 2 5 d) Expectation and Variance E X) = ) = β E X )= ) D X) = E ) β E X) = e) P[X 1] = 1 β P[X = + β β + β 1] = 1 β = β =+ = + = 1.4983 = 2 ,1 15 4 f x|A) = ,2 15 0, Q3. a) 0.9 Γ 0.7 Γ 0.5 = 0.315 + 2 5 = =+ = b) 0.9 Γ 0.7 = 0.63 c) Let us define the event βget the \$100β as A and the event βanswer the question 2 incorrectlyβ as B, then P[A ] = 1 β P[A] = 1 β 0.315 = 0.685 P[B] = 0.9 Γ 1 β 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[\$] β\$ \$ 0 0.1 0.27 0.315 0.315 \$ [\$] = 2.7 + 6.3 + 31.5 = 40.5 Q4. a) Γ = b) Γ + Γ = 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 Γ + 0.9 Γ = d) According to Bayesβ Theorem P[in Group B|vote for C1] = e) P[C1] = β 0.8 0.2 [ & [ Γ 0.1 0.9 = Γ. [ = / / = ...
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## This note was uploaded on 10/16/2011 for the course ELEC 308,315,10 taught by Professor Prof.shenghuisong during the Spring '11 term at CUHK.

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