A die is thrown three times and the sum of the three numbers is found to be 15. The probability that the first throw was a four is

A die is thrown three times and the sum of the three numbers is found to be 15. This is possible only when the combinations are - (3,6,6), (4,5,6), and (5,5,5)

When the combination is (3,6,6), there are 3 cases possible
When the combination is (4,5,6), there are 6 cases possible
When the combination is (5,5,5), there is only 1 possible case.

Thus, there are a total of 10 cases when the sum of the numbers is 15.

We need to find the probability that the first throw was a four. If the first throw was a four, then the second throw must be either 5 or 6, and the third throw must be either 6 or 5, respectively. Thus, there are 2 cases possible.

Probability that the sum is 15, and the first throw is 4 = $$\dfrac{2}{10}=\dfrac{1}{5}$$