Question

Two dice are tossed. The probability that the total score is a prime number is:

• A. $\frac{1}{12}$
• B. $\frac{5}{12}$
• C. $\frac{7}{12}$
• D. $\frac{11}{12}$

Answer 1

Answer: (B)

$\frac{5}{12}$

Clearly, $n\left(S\right)=(6\times 6)=36.$

Let E = Event that the sum is a prime number.

Then E= { (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4,3),(5, 2), (5, 6), (6, 1), (6, 5) }

n(E) = 15.

Hence, the probability that the total score is a prime number is

$P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{15}{36}=\frac{5}{12}$