Question

Two dice are thrown together. Then the probability that the sum of numbers appearing on them is a prime number, is

• A. $\frac{5}{12}$
• B. $\frac{7}{18}$
• C. $\frac{13}{36}$
• D. $\frac{11}{36}$

Answer 1

Answer: (A)

The prime numbers between $2$ to $12$ are $2,3,5,7,11$
Case I When sum is $2$, total cases are $(1,1)$ ie, $1$
Case II When sum is $3$, total cases are $(1,2)$, $(2,1)$ ie, $2$
Case III When sum is $5$, total cases are $(1,4),(2,3),(4,1),(3,2)$ i e, $4$
Case IV When sum is $7$, total cases are $(1,6),(2,5),(3,4),(6,1),(5,2),(4,3)$ i e, $6$
Case V When sum is $11$, total cases are $(5,6)$, $(6,5)$ i e, $2$
$\therefore$ Required probability $=\frac{15}{36}=\frac{5}{12}$