Question

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

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

Answer 1

The correct option is C $\frac{1}{9}$

When both the dice display the same number on the top face, we call it a doublet.
Cases in which a doublet having sum less than $9$ are$(1,1),(2,2),(3,3),(4,4)$
Total Cases $=36$
The required probability $=\frac{4}{36}=\frac{1}{9}$