Question

Two dice are thrown. What is the probability that the sum of the faces equals or exceeds $10?$

• A. $\frac{1}{12}$
• B. $\frac{1}{4}$
• C. $\frac{1}{3}$
• D. $\frac{1}{6}$

Answer 1

The correct option is D

$\frac{1}{6}$

Find the required probability:

Total number of events $=6^2=36$

Favourable events $=\left\{(6,4),(5,5),(4,6),(6,5),(5,6),(6,6)\right\}$

Number of favourable events $=6$

Required probability $=\frac{\mathrm{number}\mathrm{of}\mathrm{favorable}\mathrm{outcomes}}{\mathrm{total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{6}{36}$

$=\frac{1}{6}$

Therefore, the probability that the sum of the faces equals or exceeds $10$ is $\frac{1}{6}.$

Hence, option (D) is correct.