Chapter 6 - Pg 290 Question#1 a P(1,2 = 1/36 P(2,3 = 1/18...

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Pg 290 Question #1 a) P(1,2) = 1/36 P(2,3) = 1/18, P(2,4) = 1/36, P(3,4) = 1/18, P(3.5) = 1/18 P(3,6) = 1/36, P(4,5) = 1/18, P(4,6) = 1/18, P(4,7) = 1/18, P(4,8) = 1/36 P(5,6) = 1/18, P(5,7) = 1/18, P(5,8) = 1/18, P(5,9) = 1/18, P(5,10) = 1/36, P(6,7) = 1/18, P(6,8) = 1/18, P(6,9) = 1/18, P(6,10) = 1/18, P(6,11) = 1/18, P(6,12) = 1/36 All other cases the probability is 0. b) P(i, i) = i/36 i=1, 2, 3, 4, 5, 6 P(i=1, j=2-6) = 1/36 P(i=2, j=3-6) = 1/36 P(i=3, j=4-6) = 1/36 P(i=4, j=5,6) = 1/36 P(5,6) = 1/36 Rest are all zeros. c) P(i, i) = 1/36 i = 1,2,3,4,5,6 P(i=1, j=2-6) = 1/18 P(i=2, j=3-6) = 1/18 P(i=3, j=4-6) = 1/18 P(i=4, j=5,6) = 1/18 P(5,6) = 1/18 1
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Pg 290 Question #2 a) Simply use the concepts of combinatorics for this and the third question: P(0, 0) = (8/13)(7/12) = 14/39 ≈ 0.359 P(1, 0) = P(0, 1) = (8/13)(5/12) = 10/39 ≈ 0.256 P(1, 1) = (5/13)(4/12) = 5/39 ≈ 0.128 b) P(0, 0, 0) = (8/13)(7/12)(6/11) = 28/143 ≈ 0.196 P(1, 0, 0) = P(0, 1, 0) = P(0, 0, 1) = (8/13)(7/12)(5/11) = 70/429 ≈ 0.163 P(1, 1, 0) = P(0, 1, 1) = P(1, 0, 1) = (8/13)(5/12)(4/11) = 40/429 ≈ 0.093 P(1, 1, 1) = (5/13)(4/12)(3/11) = 5/143 ≈ 0.035 2
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Pg 290 Question #3 In this problem we have 3 marked balls and 10 unmarked balls. a) P(0, 0) = P{1 st ball is unmarked}*P{2 nd ball unmarked | 1 st ball unmarked} = (10/13)(9/12) = 15/26 ≈ 0.577 P(1,0) = P(0, 1) = P{1 st ball is marked/un}*P{2 nd ball un/marked | 1 st ball marked/un} = (10/13)(3/12) = 5/26 ≈ 0.192 P(1, 1) = P{1 st ball is marked}*P{2 nd ball marked | 1 st ball marked} = (3/13)(2/12) = 1/26 ≈ 0.038 b) Using similar analysis as above we know have three cases: P(0, 0, 0) = (10/13)(9/12)(8/11) = 60/143 ≈ 0.42 P(1, 0, 0) = P(0, 1, 0) = P(0, 0, 1) = (10/13)(9/12)(3/11) = 45/286 ≈ 0.157
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