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Clearly, n(S) = 6 × 6 = 36 Let E be the event that the sum of the numbers on the two faces is divisible by 4 or 6. Then, E = {(1, 3), (1, 5), (2, 2), (2, 4), (2,6)(3,1)(3, 3), (3, 5), (4, 2), (4, 4), (5,1), (5, 3), (6, 2), (6,6)} ∴ n(E) = 14 Hence, P(E) = n(E)/n(S) = 14/36 = 7/18
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