**Question: **Two dices are rolled. If both dices have six faces numbered $1,2,3,5,7$ and 11 , then the probability that the sum of the numbers on the top faces is less than or equal to 8 is :

$\frac{4}{9}$

$\frac{17}{36}$

$\frac{5}{12}$

$\frac{1}{2}$

Correct Option: , 2

**Solution:**
$n(E)=5+4+4+3+1=17$

So, $P(E)=\frac{17}{36}$