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