Question

Two uniform dice marked $1$ to $6$ are thrown together. The probability that the score on the two dice is at the most $7$ is

• A. $\frac{5}{12}$
• B. $\frac{7}{12}$
• C. $\frac{3}{4}$
• D. $\frac{1}{2}$

Answer 1

The correct option is B $\frac{7}{12}$

The score on the two dice is at the most $7$ is $=1-$ probability of score on the two dice being greater than $7$
The different cases in which the score on the two dice is greater than 7 $=(2,6),(6,2),(3,5),(5,3),(4,4),(3,6),(6,3),(4,5),(5,4),(4,6),(6,4),(5,5),(5,6),(6,5),(6,6)$
Number of Cases $=15$
Total number of cases $=36$
The probability that the score on the two dice is at the most 7 is $=1-\frac{15}{36}=\frac{21}{36}=\frac{7}{12}$