MATH
Math121 Discussion - 10th week solutions

Math121 Discussion - 10th week solutions - MATH 121...

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MATH 121 DISCUSSION WEEK 10 SOLUTIONS 1) (a) Let A = draw even # from first pile, B = draw odd # from second pile, C = draw # less than 3 from third pile. A = { 2, 4, 6, 8 } , B = { 1, 3, 5, 7 } , C = { 1, 2 } , so P(A) = 4 8 , P(B) = 4 8 , P(C) = 2 8 . P(A B C) = P(A) × P(B) × P(C) = ( 4 8 )( 4 8 )( 2 8 ) = ( 1 2 )( 1 2 )( 1 4 ) = 1 16 . (b) There are 8 × 8 × 8 = 512 possible combinations of three cards. There are 8 ways you can draw the same # from each pile: { 1, 1, 1 } , { 2, 2, 2 } , { 3, 3, 3 } , { 4, 4, 4 } , { 5, 5, 5 } , { 6, 6, 6 } , { 7, 7, 7 } , { 8, 8, 8 } . Thus, P(draw all the same #s) = 8 512 = 1 64 . (c) P(not all the same #s) = 1 - P(all the same #s) = 1 - 8 512 = 512 512 - 8 512 = 504 512 . 2) (a) There are 365 × 365 × 365 × 365 = 365 4 possible combinations of 4 birthdays. There are 365 ways that all four friends can have the same birthday (i.e. they all have a birthday on Jan. 1, they all have a birthday on Jan. 2, etc., all the way through Dec. 31). Thus, P(all the same birthday) = 365 365 4 = 1 365 3 = 1 48 , 627 , 125 . (b) There are 365 × 364 × 363 × 362 combinations of 4 birthdays where each day is different. Thus, P(no one shares a birthday) = 365 × 364 × 363 × 362 365 4 0 . 9836. (c) The probability that none of the friends share a birthday was found in part (b). The comple- ment of no one sharing a birthday is that at least 2 of them have a common birthday. Thus, P(at least 2 share a birthday) = 1 - P(no one shares a birthday) = 1 - 365 × 364 × 363 × 362 365 4 0 . 0164. 3) Let A = pick white pair, B = pick blue pair, C = pick black pair, and D = pick yellow pair, and “1” and “2” denote the first and second pick respectively. (a) P(B2 | A1) = 2 7 (b) P(A2 | D1) = 3 7 (c) P(B2 | B1) = 1 7 (d) P(D2 B1) = P(D2 | B1) × P(B1) = P(B1) × P(D2 | B1) = ( 2 8 )( 1 7 ) = 2 56 = 1 28 . (e) P(B2 B1) = P(B2 | B1) × P(B1) = P(B1) × P(B2 | B1) = ( 2 8 )( 1 7 ) = 2 56 = 1 28 . (f) P(A2 A1) = P(A2 | A1) × P(A1) = P(A1) × P(A2 | A1) = ( 3 7 )( 2 6 ) = 6 42 = 1 7 . 4) For all four parts, let A = animal is a cat, B = animal is black. (a) P(A) = P(A B) + P(A B 0 ) = 5% + 35% = 40%. (b) P(A) = P(A B) + P(A B 0 ), so P(A B 0 ) = P(A) - P(A B) = 50% - 2% = 48%. (c) In this case, we already know that the animal is black, so event B has already occurred. So we want to know what the probability of A is given that B occurred. P(A | B) = P ( A B ) P ( B ) = 0 . 03 0 . 30 = 1 10 , or 10%. (d) In this case, we want to know the probability that an animal is black given that it is a cat (i.e. A has already occurred, and we want probability of B given that A occurred). P(A 0 ) = 70%, or 0.70. So P(A) = 1 - P(A 0 ) = 1 - 0.70 = 0.30. P(B | A) = P ( B A ) P ( A ) = 0 . 10 0 . 30 = 1 3 . 1
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2 MATH 121 DISCUSSION WEEK 10 SOLUTIONS 5) Given: P(A B) = 0.24, P(A B 0 ) = 0.26, P(A 0 B) = 0.16. (a) See separate attachment for Venn diagram. (b) P(A) = P(A B) + P(A B 0 ) = 0.24 + 0.26 = 0.50. (c) P(B) = P(A B) + P(A 0 B) = 0.24 + 0.16 = 0.40. (d) First, recall that P(A B) = P(A) + P(B) - P(A B) = 0.50 + 0.40 - 0.24 = 0.66. By DeMorgan, P(A 0 B 0 ) = P((A B) 0 ) = 1 - P(A B) = 1.00 - 0.66 = 0.34.
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