Question

Two symmetrical dice are thrown. The probability that the product of the numbers on their upper most faces is $12$ is

• A. $\frac{1}{7}$
• B. $\frac{1}{8}$
• C. $\frac{1}{9}$
• D. $\frac{1}{6}$

Answer 1

Answer: (C)

$\frac{1}{9}$

The different cases in which the product of the numbers on their upper most faces is $12$ are $(3,4),(4,3),(2,6),(6,2)$
Number of cases $=4$
Total number of cases $=36$
The required probability $=\frac{4}{36}=\frac{1}{9}$