Q. 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) $\frac{11}{36}$
Explanation
(a)
The prime numbers between 2 to12 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) $ie,\ 4$
Case IV When sum is 7, total cases are (1, 6), (2, 5), (3, 4), (6, 1), (5, 2), (4, 3) $ie,\ 6$
Case V When sum is 11, total cases are (5, 6), (6, 5) $ie,\ 2$
$\therefore\ Required\ probability = \frac{15}{36} = \frac{5}{12}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.