The probability of getting the sum as a prime number, when two dice are thrown together, is :

Correct — option (3), 5/12.

Step 1 — count all outcomes. Each die shows 1 to 6, so two dice give 6 × 6 = 36 equally likely ordered pairs.

Step 2 — list the prime sums. The sums run from 2 to 12, and the primes among them are 2, 3, 5, 7 and 11.

Step 3 — count the pairs for each prime sum.
Sum 2: (1, 1) → 1
Sum 3: (1, 2), (2, 1) → 2
Sum 5: (1, 4), (4, 1), (2, 3), (3, 2) → 4

Sum 7: (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3) → 6
Sum 11: (5, 6), (6, 5) → 2

Step 4 — add and divide.
Favourable pairs = 1 + 2 + 4 + 6 + 2 = 15
Probability = 15/36 = 5/12

The idea to remember: with two dice, count ordered pairs; (1, 2) and (2, 1) are two separate outcomes among the 36.

Why the others are wrong

• (1)
  1/2 — 1/2 = 18/36 is the probability of an odd sum: the sums 3, 5, 7, 9 and 11 come from 2 + 4 + 6 + 4 + 2 = 18 pairs.

  Odd is not the same as prime. The odd sum 9 (= 3 × 3) is not prime, and the even sum 2 is. Dropping 9's 4 pairs and adding 2's 1 pair gives 15 pairs, or 5/12.

• (2)
  1/3 — 1/3 = 12/36 is the probability that the sum is a multiple of 3: the sums 3, 6, 9 and 12 come from 2 + 5 + 4 + 1 = 12 pairs.

  Prime sums come from 15 pairs, so the probability asked for is 15/36 = 5/12.

• (4)
  7/12 — 7/12 = 21/36 is the probability that the sum is not prime, the complement of the event asked.

  The sums 4, 6, 8, 9, 10 and 12 come from 3 + 5 + 5 + 4 + 3 + 1 = 21 pairs, and 21 + 15 = 36. The probability of a prime sum is 1 − 7/12 = 5/12.

Classical probability: when all outcomes are equally likely, the probability of an event = favourable outcomes ÷ total outcomes.

For two dice the equally likely outcomes are the 36 ordered pairs, not the 11 possible sums. The number of pairs giving a sum s is 6 − |s − 7|, so sum 7 has 6 pairs while sums 2 and 12 have 1 each.

The complement rule, P(not A) = 1 − P(A), links an event to its opposite: a prime sum has probability 5/12, a non-prime sum 7/12.

A prime number has exactly two factors, 1 and itself, so 2 is prime and 1 is not.

Key facts

• Two dice have 6 × 6 = 36 equally likely ordered outcomes, with sums from 2 to 12.
• The number of ways to get sum s with two dice is 6 − |s − 7|; sum 7 has 6 ways, sums 2 and 12 have 1 each.
• The prime numbers up to 12 are 2, 3, 5, 7 and 11; 1 is not a prime number.
• Classical probability = favourable outcomes ÷ total outcomes, when every outcome is equally likely.
• P(not A) = 1 − P(A); with two dice, P(sum not prime) = 1 − 5/12 = 7/12.

Common traps

• Counting unordered pairs, so that (1, 2) and (2, 1) become one outcome. The total of 36 counts ordered pairs, so the favourable outcomes must be ordered too.
• Leaving out 2 because it is even, or taking in 9 because it is odd. 2 is prime; 9 = 3 × 3 is not.
• Treating the 11 possible sums as equally likely. Sum 7 comes from 6 pairs and sum 2 from 1, so a count of sums does not give the probability.

UnlockIAS will link similar questions from RAS Pre 2023 and 2021 here once those papers are published on this site.

• practice — not a real PYQ

  Two dice are thrown together. What is the probability that the sum of the numbers is a perfect square?

  • (a)7/36
  • (b)1/12
  • (c)1/9
  • (d)2/11
  Answer(1) — The square sums possible are 4 (3 pairs) and 9 (4 pairs), so 7/36. Option (2) counts sum 4 alone; option (3) counts sum 9 alone; option (4) treats the 11 possible sums as equally likely.
• practice — not a real PYQ

  Two dice are thrown together. What is the probability that the sum of the numbers is greater than 9?

  • (a)1/6
  • (b)1/12
  • (c)5/36
  • (d)3/11
  Answer(1) — Sums 10, 11 and 12 come from 3 + 2 + 1 = 6 pairs, so 6/36 = 1/6. Option (2) counts sum 10 alone; option (3) leaves out sum 12; option (4) treats the 11 possible sums as equally likely.