Question

Two symmetrical dice are thrown. The probability of throwing a doublet such that their sum is greater than $9$ is

• A. $\frac{1}{36}$
• B. $\frac{1}{18}$
• C. $\frac{1}{9}$
• D. $\frac{1}{72}$

Answer 1

Answer: (B)

$\frac{1}{18}$

Throwing a doublet means same number on both dice.
So, if the sum is a doublet and sum is greater than $9$, then the combinations possible are $(5,5),(6,6)$.
Hence, the required probability $=\frac{2}{36}=\frac{1}{18}$