Two dies are rolled. If both dices have six faces numbered 1, 2, 3, 5, 7 and 11, then the probability that the sum of the numbers on the top faces is less than or equal to 8 is :

Solution

Maths sir · 3 steps

Fix the first die one face at a time and count how many faces on the second keep the sum at 8 or less.

1. Total outcomes. Each die has 6 faces, so there are 6 × 6 = 36 ordered outcomes, all equally likely. Order matters: (1, 7) and (7, 1) are different outcomes.
2. Count favourable. For each value a on the first die, the second die must show a face b with b ≤ 8 - a, chosen from 1, 2, 3, 5, 7, 11.

   First die a	Need b ≤	Allowed b	Count
   1	7	1, 2, 3, 5, 7	5
   2	6	1, 2, 3, 5	4
   3	5	1, 2, 3, 5	4
   5	3	1, 2, 3	3
   7	1	1	1
   11	-3	none	0

3. So. Favourable outcomes = 5 + 4 + 4 + 3 + 1 + 0 = 17, so the probability is 17/36, option (d).

The idea to keep. With unusual dice faces, do not guess. Fix one die, list the allowed faces of the other, and count ordered pairs out of 6 × 6 = 36.

Answer from the NTA final answer key. Solution by PrepWiser.