Question

When 2 dice are thrown, it is observed that the sum of the numbers appeared on the top faces of both the dice is a prime number. Then the probability of having a multiple of 3 among the pair of numbers thus obtained is:

• A. \(\frac{8}{15}\)
• B. \(\frac{11}{36}\)
• C. \(\frac{5}{9}\)
• D. \(\frac{5}{12}\)

Hint:

Calculate the total prime outcomes first and then determine how many of these include a multiple of 3.

Answer 1

The Correct Option is A

Prime sums possible from two dice are 2, 3, 5, 7, and 11. We identify the outcomes where the sum is prime and calculate the scenarios where at least one number is a multiple of 3: \[ \text{Prime sums pairs:} (2,3,5,7,11) \] \[ \text{Total prime outcomes:} 15 \] \[ \text{Multiple of 3 in outcomes:} (1,2), (2,1), (3,2), (2,3), (4,3), (3,4) \] \[ \text{Favorable outcomes:} 8 \] \[ P(\text{Multiple of 3} \mid \text{Prime Sum}) = \frac{8}{15} \]