Answer - (C)
First throw is $4$. So, total no. of outcomes will be $6$ choices for second die and $6$ choices for the third die.
So, total no.of outcomes
$n(S) = 6 \cdot 6 = 36$
Now, to get $15$ as sum, given that first die has $4$ on its face, it means the sum of faces of dice in second and third throw should be $15-4=11$
There are exactly $2$ such outcomes.
$( 5,6)$ and $(6,5)$
Hence, no. of favourable outcomes $n(A) = 2$
Hence, required probability = $\frac{n(A)}{n(S)} = \frac{2}{36} = \frac{1}{18}$