A dies is thrown three times and the sum of three numbers obtained is $15$. The probability of first throw being $3$ is:

This problem requires calculating the conditional probability of the first throw being 3, given that the sum of all three throws is 15. When the first die is fixed at 3, the remaining two dice must sum to 12, which can only be achieved by the outcome (6, 6); thus, there is exactly 1 favourable outcome. The total number of outcomes where three dice sum to 15 is 10, consisting of the permutations of (6, 6, 3), (6, 5, 4), and (5, 5, 5). The probability is therefore 1 divided by 10, yielding 1/10.

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