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

Step 1: Set up the sample space.
Each roll of the dice gives one of 6 outcomes, so two rolls in succession give \( 6 \times 6 = 36 \) equally likely ordered outcomes. The sum of the two recorded numbers can be any value from 2 to 12.

Step 2: Pick out the prime sums.
A prime number greater than 1 has no divisors other than 1 and itself. Among the possible sums 2 through 12, the prime values are 2, 3, 5, 7, and 11.

Step 3: Count the outcomes for each prime sum.
Sum 2: only (1,1), so 1 way. Sum 3: (1,2), (2,1), so 2 ways. Sum 5: (1,4), (2,3), (3,2), (4,1), so 4 ways. Sum 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1), so 6 ways. Sum 11: (5,6), (6,5), so 2 ways.

Step 4: Add up the favourable outcomes.
Total favourable outcomes \( = 1 + 2 + 4 + 6 + 2 = 15 \).

Final Answer:
The probability is the favourable count over the total count.
\[ \boxed{P(\text{prime sum}) = \dfrac{15}{36}} \]