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A die is thrown two times and the sum of the scores appearing on the die is observed to be a multiple of 4. Then the conditional probability that the score 4 has appeared atleast once is :(1) \(\frac{1}{8}\)(2) \(\frac{1}{9}\)(3) \(\frac{1}{3}\)(4) \(\frac{1}{4}\) |
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Answer» (2) \(\frac{1}{9}\) A : Sum obtained is a multiple of 4. A = {(1, 3), (2, 2), (3, 1), (2, 6), (3, 5), (4, 4), (5, 3), (6, 2), (6, 6)} B : Score of 4 has appeared at least once. B = {(1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4), (4, 1), (4, 2), (4, 3), (4, 5), (4, 6)} Required probability = P\((\frac{B}{A})\) = \(\frac{P\,∩\,A}{P(A)}\) = (1/36)/(9/36) = 1/9 |
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