A die is thrown three times and the sum of three numbers obtained is 15. The probability of first throw being 4 is

1/18 1/5 4/5 17/18

1/5

If first throw is four, then sum of numbers appearing on last two throws must be equal to eleven. That means last two throws are (6, 5) or (5, 6) Now there are 10 ways to get the sum as 15. [(5, 5, 5) (4, 5, 6) (3,6,6)] ⇒ Required probability = $\frac{2}{10}=\frac{1}{5}$