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Question
Two unbiased dice are thrown. Find the probability that neither a doublet nor a total of 8 will appear
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Solution
We know that in a single throw of two dices, the total number of possible outcomes is (6 × 6) = 36.
Let S be the sample space.
Then n(S) = 36
Let E1 = event of getting neither a doublet nor a total of 8
Then E1' = event of getting either a doublet or a total of 8
∴ E1' = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (2, 6), (3, 5), (5, 3), (6, 2)}
i.e. n(E1') = 10
Thus, P(E1') = \[\frac{10}{36}\]
Hence, required probability P(E1) = 1- P(E1')
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