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:

• A. $\frac{17}{36}$
• B. $\frac{4}{9}$
• C. $\frac{5}{12}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $\frac{17}{36}$

$n\left(S\right)=36$
Possible ordered pair :
$(1,1),(1,2),(1,3),(1,5),(1,7)(2,1),(2,2),(2,3),(2,5),(3,1),(3,2),(3,3),(3,5),(5,1),(5,2),(5,3),(7,1)$
Required number of ordered pair $=17$
Hence, the required probability
$=\frac{17}{36}$