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InterviewSolution
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Two unbiased dice were thrown. Find the probability that neither a doublet nor a total of 10 will appear. |
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Answer» Let S be the sample space. Then, n(S) = 36. Let `E_(1)` = event that a doublet appears, and `E_(2) =` event of getting a total of 10. Then, `E_(1) = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)},` and `E_(2) = {(4, 6), (5, 5,), (6, 4)}.` `therefore (E_(1) nn E_(2)) = {(5, 5)}.` Thus, `n(E_(1)) = 6, m(E_(2)) = 3 and n(E_(1) nn E_(2)) = 1.` `therefore P(E_(1)) = (n(E_(1)))/(n(S)) = 6/36 = 1/6, P(E_(2)) = (n(E_(2)))/(n(S)) = 3/36 = 1/12` and `P(E_(1) nn E_(2)) = (n(E_(1) nn E_(2)))/(n(S)) = 1/36.` `therefore` P(getting a doublet or a total of 10) `= P(E_(1) or E_(2)) = P(E_(1) uu E_(2))` `= P(E_(1)) + P(E_(2)) - P(E_(1) nn E_(2))` `= (1/6 + 1/ 12 - 1/36) = 8/36 = 2/9.` `therefore` P(getting neither a doublet nor a total of 10) `= P(bar(E_(1)) and bar(E_(2))) = P(bar(E_(1)) nn bar(E_(2)))` `= P(bar(E_(1) uu E_(2))) = 1 - P(E_(1) uu E_(2)) = (1 - 2/9) = 7/9.` Hence, the required probability is `7/9`. |
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