A die is thrown three times. If the first throw is a five, the probability of getting 14 as the sum is

$\frac{1}{9}$

The correct answer is Option (1) → $\frac{1}{9}$

Given that the first throw is $5$.

To get total sum $14$, the sum of the second and third throws must be

$14-5=9$

Possible ordered pairs $(x,y)$ such that $x+y=9$ where $x,y\in\{1,2,3,4,5,6\}$ are

$(3,6),(4,5),(5,4),(6,3)$

Total possible outcomes for second and third throws $=6\times6=36$

Favourable outcomes $=4$

Probability $=\frac{4}{36}$

$=\frac{1}{9}$