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Question
In a single throw of three dice, find the probability of getting a total of 17 or 18.
Solution
If three dices are thrown simultaneously, then all the possible outcomes = 63 = 216
∴ Total number of possible outcome, n(S) = 216
Let A = event of getting a sum of numbers on three dice as 17 or 18
Then the favourable outcomes are given as
A = {(6, 6, 5), (6, 5, 6), (5, 6, 6), (6, 6, 6)}
Number of favourable outcomes, n(A) = 4
Hence, required probability, P(A) = P (sum of the numbers on three dices as 17 or 18) = \[\frac{4}{216} = \frac{1}{54}\]
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