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# hw4 - MATH 471 HW 4 Solutions Pengsheng Ji Oce Hours...

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MATH 471 HW 4 Solutions Pengsheng Ji Office Hours: 4:05-5:05 PM Tuesday 5:20-6:20 PM Thursday 218 Mallot Hall 2.4.6 Let A=”1 was received”, B=”1 was sent”. Since they are received as sent with probability 0.9 but errors occur with probability 0.1, we have P ( A | B ) = 0 . 9 , P ( A c | B ) = 0 . 1 , P ( A c | B c ) = 0 . 9 , P ( A | B c ) = 0 . 1 . Then P ( A B ) = P ( A | B ) P ( B ) = 0 . 9 * 0 . 5 = 0 . 45 , P ( A B c ) = P ( A | B c ) P ( B c ) = 0 . 1 * 0 . 5 = 0 . 05 . Hence, the probability that a 1 was sent given that we received a 1 is given by P ( B | A ) = P ( A B ) P ( A ) = 0 . 45 0 . 45 + 0 . 05 = 0 . 9 . 2.4.17 Let A=”the person voted”, B 1 =”a Conservative”, B 2 =”a Liberal”, B 3 =”an Independent.” By the Bayes Formula, we have the probability she is Liberal given that she is a voter is given by P ( B 2 | A ) = P ( B 2 ) P ( A | B 2 ) 3 i =1 P ( B i ) P ( A | B i ) = 0 . 5 * 0 . 8 0 . 3 * 2 / 3 + 0 . 5 * 0 . 8 + 0 . 2 * 0 . 5 = 4 / 7 . 2.6.11 Denote these three events by A,B and C, respectively. Then P(A)=1/4, P(B)=1/3, P(C)=1/2. Since they are independent, the probability exactly one will occur is P ( A B c C c ) + P ( A c B C c ) + P ( A c B c C ) = 1 4 * (1 - 1 3 ) * (1 - 1 2 ) + (1 - 1 4 ) * 1 3 * (1 - 1 2 ) + (1 - 1 4 )(1 - 1 3 ) * 1 2 = 11 24 .

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2.6.23 (a) Let A=”a driver will have an accident”, B i , i = 1 , 2 , 3 , 4, be the four groups from the youngest to the oldest. By the Law of Total Probability, the probability
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