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Question
Two unbiased dice are thrown. Find the probability that the sum of the numbers obtained on the two dice is neither a multiple of 2 nor a multiple of 3
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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 E2 be the event of getting the sum of the numbers obtained on the two dice is neither a multiple of 2 nor a multiple of 3. Then,
E2' = event of getting the sum of the numbers obtained on the two dice is either a multiple of 2 or a multiple of 3.
∴ E2' = {(1,1), (1,2), (2, 1), (1, 3), (2, 2), (3, 1), (1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (2, 6), (3, 5), (4, 4), (5, 3), (6, 2), (3, 6),
(4, 5), (5, 4), (6, 3), (4, 6), (5, 5), (6, 4), (6, 6)}
i.e. n(E2') = 24
Thus, P(E2') = \[\frac{24}{36} = \frac{2}{3}\]
Hence, required probability P(E2) = 1 - P(E2')
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