A die is thrown three times, if the first throw is a four, find the chance of getting `15` as the sum.

Explanation

If `1st` throw is `4` then the sum of numbers appearing on last `2nd` throws must be equal to `11`.

which means last `2` throws are `(6,5)` or `(5,6)`


There are `10` ways to get sum as `15` =`{(4,5,6),(4,6,5),...,(3,6,6)}`

Required probability `=frac{2}{10}=frac{1}{5}`