A biased die is marked with numbers 2, 4, 8, 16, 32, 32 on its faces and the probability of getting a face with mark n is 1/n. If the die is thrown thrice, then the probability, that the sum of the numbers obtained is 48, is :

Solution

Maths sir · 3 steps

Fix the probability of getting a 32 first, then list every way three throws can add to 48.

1. Probability of each number. P(2) = 1/2, P(4) = 1/4, P(8) = 1/8, P(16) = 1/16. The mark 32 is on two faces, each with probability 1/32, so P(32) = 2/32 = 1/16. These add up to 1, as they must.
2. Ways to make 48. The numbers available are 2, 4, 8, 16 and 32. Every triple adding to 48 is a rearrangement of 32 + 8 + 8 or of 16 + 16 + 16. No triple using a 2 or a 4 can reach 48.

   Outcome	Orders	Probability of one order	Total
   32, 8, 8	3	1/16 × 1/8 × 1/8 = 1/2¹⁰	3/2¹⁰ = 12/2¹²
   16, 16, 16	1	(1/16)³ = 1/2¹²	1/2¹²

   P(sum = 48) = 12/2¹² + 1/2¹² = 13/2¹²

3. So. The required probability is 13/2¹², which is option (d).

The idea to keep. When a value appears on more than one face, add the face probabilities first. Then list every combination that gives the target sum and count its orders.

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