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151sol2

# First Course in Probability, A (7th Edition)

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Math 151 HW 2 Solutions Problems 4. Let A be the probability that at least on of a pair of dice lands on 6. Easy computaion gives: P ( A | i = 2) = 0, P ( A | i = 3) = 0, P ( A | i = 4) = 0, P ( A | i = 5) = 0, P ( A | i = 6) = 0, P ( A | i = 7) = 2 / 6, P ( A | i = 8) = 2 / 5, P ( A | i = 9) = 2 / 4, P ( A | i = 10) = 2 / 3, P ( A | i = 11) = 2 / 2, P ( A | i = 2) = 1 / 1. 18. Let E, I, L, C be the event that a person voted in the local election, is an independent, is a liberal, is a conservative, respectively. (a) Using Bayes Formula, P ( I | E ) = P ( IE ) P ( E ) = P ( I ) P ( E | I ) P ( I ) P ( E | I )+ P ( L ) P ( E | L )+ P ( C ) P ( E | C ) = (0 . 46)(0 . 35) (0 . 46)(0 . 35)+(0 . 30)(0 . 62)+(0 . 24)(0 . 58) = 0 . 331 Similar computation gives (b) 0.383 (c) 0.286 (d) (0 . 46)(0 . 35) + (0 . 30)(0 . 62) + (0 . 24)(0 . 58) = 0 . 4862, which means that 48.62 percent of voters participated. 26. Let C, M, W be event of being colorblind, man, woman, respectively. We want to calculate: P ( M | C ) = P ( MC ) P ( C ) = P ( M ) P ( C | M ) P ( M ) P ( C | M )+ P ( W ) P ( C | W ) Case1: (0 . 5)(0 . 05) (0 . 5)(0 . 05)+(0 . 5)(0 . 0025) = 0 . 95 Case2: (2 / 3)(0 . 05) (2 / 3)(0 . 05)+(1 / 3)(0 . 0025) = 0 . 98 32. Let C j j = 1 , 2 , 3 , 4 be the event that family has 1,2,3,4 children respectively. Let E be the event that child is eldest child. Using Bayes’s formula: (a) P ( C 1 | E ) = P ( C 1 E ) P ( E ) = P ( C 1 ) P ( E | C 1 ) P ( C 1 ) P ( E | C 1 )+ P ( C 2 ) P ( E | C 2 )+ P ( C 3 ) P ( E | C 3 )+ P ( C 4 ) P ( E | C 4 ) = 1(0 . 1) 1(0 . 1)+(1 / 2)0 . 25+(1 / 3)0 . 35+(1 / 4)0 . 3 = 0 . 24 (b) Similarly, we get 0.18

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151sol2 - Math 151 HW 2 Solutions Problems 4 Let A be the...

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