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Question
The faces of a die bear numbers 0, 1, 2, 3, 4, 5. If the die is rolled twice, then find the probability that the product of digits on the upper face is zero.
Sum
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Solution
Sample space,
S = {(0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (0, 5),
(1, 0), (1, 1), (1, 2), (1, 3), (1, 4), (1, 5),
(2, 0), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5),
(3, 0), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5),
(4, 0), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5),
(5, 0), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5)}
∴ n(S) = 36
Let A be the event that the product of digits on the upper face is zero.
∴ A = {(0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (1, 0), (2, 0), (3, 0), (4, 0), (5, 0)}
∴ n(A) = 11
∴ P(A) = `("n"("A"))/("n"("S"))`
∴ P(A) = `11/36`
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