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Question
Three fair dice are thrown simultaneously. What is the probability that the sum of numbers on their tops, is at least 6?
Options
`5/108`
`1/24`
`103/108`
`17/18`
Solution
`103/108`
Explanation:
We get sum of numbers on top of three dice less than 6 in following ways.
(1, 1, 1), (1, 1, 2), (1, 1, 3), (1, 2, 1), (1, 3, 1), (2, 1, 1), (3, 1, 1), (2, 2, 1), (2, 1, 2), (1, 2, 2).
∴ n (E) = 10
n(S) = 6 × 6 × 6 = 216
Now, probability of getting sum less than 6
`= 10/216 = 5/108`
∴ Probability of getting sum at least 6
`= 1 - 5/108 = 103/108`
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