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Question
Two dice are thrown. Find the probability of the event that the product of numbers on their upper faces is 12.
Solution
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n(S) = 36
Let A = event that the product of numbers on their upper faces is 12
A = {(2,6), (3, 4), (4,3), (6,2)}
n(A) = 4
`"Probability of the event A"="Number of favourable outcomes"/"Total number of outcomes"=4/36=1/9`
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