Question

An unbiased six-faced dice whose faces are marked with numbers 1, 2, 3, 4, 5, and 6 is rolled twice in succession and the number on the top face is recorded each time. The probability that the sum of the two recorded numbers is a prime number is ______

• A. \( \dfrac{3}{36} \)
• B. \( \dfrac{13}{36} \)
• C. \( \dfrac{15}{36} \)
• D. \( \dfrac{19}{36} \)

Hint:

List the sums from 2 to 12 that are prime, then count the dice pairs giving each of those sums.

Answer 1

The Correct Option is C

Rolling two dice in a row creates $6 \times 6 = 36$ equally likely pairs, and this problem asks what fraction of those pairs add up to a prime number.

1. List every sum from 2 to 12 and mark which ones are prime: sums 4, 6, 8, 9, 10, and 12 are not prime, while sums 2, 3, 5, 7, and 11 are prime.
2. Count how many pairs give each prime sum: sum 2 comes from 1 pair, sum 3 from 2 pairs, sum 5 from 4 pairs, sum 7 from 6 pairs, and sum 11 from 2 pairs. Adding these gives $1 + 2 + 4 + 6 + 2 = 15$ pairs out of 36 total.
3. Turn the count into a probability: $P = \dfrac{15}{36}$, which stays in this form since it matches one of the listed options directly.

The other three options, $\dfrac{3}{36}$, $\dfrac{13}{36}$, and $\dfrac{19}{36}$, all come from missing or double counting some of the five prime sums, so option C, $\dfrac{15}{36}$, is the correct probability.