Let S = { w₁,w₂,......} be the sample space associated to a random experiment. Let P(wₙ) = P(wₙ₋₁)/2,n ≥ 2. Let A = { 2k + 3l:k,l ∈ N} and B = { wₙ:n ∈ A}. Then P(B) is equal to :

Solution

Maths sir · 3 steps

First fix P(w₁) from total probability 1, then list which n can be written as 2k+3l.

1. Find each probability. Each probability is half the one before, so P(wₙ) = P(w₁)/2ⁿ⁻¹. All probabilities must add to 1, which gives P(w₁)(1 + 1/2 + 1/4 + …) = 2P(w₁) = 1. So P(w₁) = 1/2 and P(wₙ) = 1/2ⁿ.
2. Build the set A. With k, l ≥ 1, the smallest value is 2 + 3 = 5. The number 6 is impossible, because 2k + 3l = 6 needs k = 0 or l = 0. Every odd n ≥ 5 is 2k + 3, and every even n ≥ 8 is 2k + 6. So A = {5, 7, 8, 9, 10, …}.

   P(B) = 1/2⁵ + (1/2⁷ + 1/2⁸ + …) = 1/32 + (1/2⁷)/(1 - 1/2) = 1/32 + 1/64

3. So. P(B) = 2/64 + 1/64 = 3/64, which is option (d).

The idea to keep. In an infinite sample space, the condition that all probabilities add to 1 fixes the first term. With sets like {2k+3l}, always check the small numbers one by one for gaps.

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