If the probability that the random variable X takes values x is given by P(X=x)=k(x+1) 3⁻ˣ, x=0,1,2,3, …, where k is a constant, then P(X ≥ 2) is equal to :

Solution

Maths sir · 4 steps

First fix k so the probabilities add to 1, then take the first two values away from 1.

1. Total must be 1. Write r = 1/3. Adding all probabilities gives k(1 + 2r + 3r² + 4r³ + …) = 1. This bracket is an arithmetico-geometric series.
2. Sum the series. Call the bracket S. Then S - rS = 1 + r + r² + … = 1/(1-r), so S = 1/(1-r)².

   S = 1/(1-1/3)² = 9/4 ⇒ k = 4/9

3. Use the complement. P(X=0) = k = 4/9 = 12/27 and P(X=1) = k × 2 × 1/3 = 8/27. Together they give P(X ≤ 1) = 20/27.
4. So. P(X ≥ 2) = 1 - 20/27 = 7/27, which is option (c).

The idea to keep. For an infinite distribution, find the constant from total probability = 1, and get 'at least' answers from the complement. Remember Σ (n+1) rⁿ = 1/(1-r)².

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