Sum
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.
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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`
Concept: Basic Ideas of Probability
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