Question

Two dice are thrown simultaneously. What is the probability that the sum of the numbers appearing on the dice is a prime number?

• A. \(\frac{5}{12}\)
• B. \(\frac{7}{12}\)
• C. \(\frac{1}{2}\)
• D. \(\frac{13}{36}\)
• E.

Hint:

To quickly find the number of ways to get a sum \(S\) on two dice:
• If \(S \le 7\), ways \(= S - 1\).
• If \(S > 7\), ways \(= 13 - S\). Using this: \(1(2) + 2(3) + 4(5) + 6(7) + 2(11) = 15\).

Answer 1

The Correct Option is A

Concept: When two dice are rolled, the total number of possible outcomes is: \[ 6 \times 6 = 36 \] We need to find the probability that the sum of the numbers is a prime number. The possible sums for two dice range from \(2\) to \(12\). The prime numbers in this range are: \[ 2, 3, 5, 7, 11 \]

Step 1: Find the number of outcomes for each prime sum.
• Sum \(=2\): \((1,1)\) \(\rightarrow\) 1 way
• Sum \(=3\): \((1,2), (2,1)\) \(\rightarrow\) 2 ways
• Sum \(=5\): \((1,4), (4,1), (2,3), (3,2)\) \(\rightarrow\) 4 ways
• Sum \(=7\): \((1,6), (6,1), (2,5), (5,2), (3,4), (4,3)\) \(\rightarrow\) 6 ways
• Sum \(=11\): \((5,6), (6,5)\) \(\rightarrow\) 2 ways

Step 2: Calculate total favourable outcomes. \[ \text{Total favourable} = 1 + 2 + 4 + 6 + 2 = 15 \]

Step 3: Find the probability. \[ P = \frac{\text{Favourable outcomes}}{\text{Total outcomes}} = \frac{15}{36} \] Simplifying by dividing both numerator and denominator by 3: \[ P = \frac{5}{12} \]