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hw2_sol

# P w p w a p a p w b p b p 2 q 2 by total probability

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P ( W ) = P ( W | A ) P ( A ) + P ( W | B ) P ( B ) = p 2 + q 2 by total probability. Since P ( W A ) = P ( W | A ) P ( A ) = p 2 , we want to conditions on p and q so that 1 2 ( p 2 + q 2 ) = p 2 . This results in p = q . 8. Let A = Top route is working, B = Bottom route is working and C = Bottom left link is working. Notice that C B because the bottom route works if and only if both of the bottom links work. (a) The event that Acahti is connected to the Big City is A B P ( A B ) = P ( A )+ P ( B ) - P ( A B ) = (1 - p ) 2 +(1 - p ) 2 - (1 - p ) 4 = 2(1 - p ) 2 - (1 - p ) 4 Because P ( A ) is equal to the probability that the two top links are working ( (1 - p ) × (1 - p ) ) and similarly for B and also P ( A B ) = P ( A ) P ( B ) because the events A and B are independent. (b) The probability that Acahti is connected to the Big City given that the bottom left link is working is P ( A B | C ) = P (( A B ) C ) P ( C ) = P (( A C ) ( B C )) 1 - p = P (( A C ) B ) 1 - p . However P (( A C ) B ) = P ( A C ) + P ( B ) - P ( A C B ) = P ( A ) P ( C ) + P ( B ) - P ( A ) P ( B ) = (1 - p ) 3 + (1 - p ) 2 - (1 - p ) 4 . Consequently, after simpliﬁcations we ﬁnd P ( A B | C ) = 1 - 2 p 2 + p 3 . 3

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(c) We want to compute P ( B | ( A B ) C ) , this is done as follows P ( B | ( A B ) C ) = P ( B ( A B ) C ) P (( A B ) C ) = P ( B ) P (( A B ) C ) , where the last equality follows from the fact that B C and B ( A B ) . Consequently P ( B | ( A B ) C ) = P ( B ) P ( C ) P (( A B ) | C ) = (1 - p ) 2 (1 - p )(1 - 2 p 2 + p 3 ) = 1 1 + p - p 2 . (d) We want P ( A B ) = 2(1 - p ) 2 - (1 - p ) 4 > 0 . 99999 . Letting x = (1 - p ) 2 , the last inequality becomes 2 x - x 2 - 0 . 99999 > 0 . So that x > 0 . 9968 or equivalently p < 0 . 00158 . 9. Note that for any k 1 , we have p X ( k ) p X ( k - 1) = e - λ λ k k ! e - λ λ k - 1 ( k - 1)! = λ k . Now if k k * , then we have p X ( k ) p X ( k - 1) = λ k λ k * λ λ = 1 , so p X ( k ) is nondecreasing from 0 up to k * . If k > k * , then we must have k > λ , so p X ( k ) p X ( k - 1) = λ k < 1 . Thus p X ( k ) is strictly decreasing from k * onward. 10. (a) The hashed region corresponds to the event { Z 1 / 3 } . - 6 x y 1 1 ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± 1 / 3 1 / 3 (b) Note that the probability of any event equals its area. From the picture, we have F Z ( z ) = Pr( Z z ) = z + (1 - z ) z = 2 z - z 2 , ( z [0 , 1]) and F Z ( z ) = 0 for z < 0 , and F Z ( z ) = 1 for z > 1 . 4
-0.5 0 0.5 1 1.5 -1 -0.5 0 0.5 1 1.5 2 F Z (z) z (c) X takes a continuum of values, so it is not discrete. Its CDF is differentiable, so it is continuous. Its PDF is f Z ( z ) = 0 for z ( -∞ , 0) (1 , ) , because F Z is constant there. For z [0 , 1] f Z ( z ) = d dz F Z ( z ) = 2 - 2 z ( z [0 , 1]) . (d) We have Pr( Z > 3 / 4) = 1 - Pr( Z 3 / 4) = 1 - F Z (3 / 4) = 1 / 16 . 5
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P W P W A P A P W B P B p 2 q 2 by total probability Since...

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