If the probability that the random variable X takes the value 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 ≥ 3) is equal to

Solution

Maths sir · 4 steps

First fix k by making all probabilities add to 1, then use the complement: subtract the first three.

1. Total must be 1. The probabilities over x = 0, 1, 2, … must add up to 1. The sum Σ (x+1)rˣ is a standard series with r = 1/3.

   Σₓ₌₀^∞(x+1)(1/3)ˣ = 1/(1-1/3)² = 9/4

2. Find k. So k × 9/4 = 1, which gives k = 4/9.
3. First three values. P(X=0) = 4/9 = 12/27, P(X=1) = 4/9 × 2/3 = 8/27 and P(X=2) = 4/9 × 3/9 = 4/27. Together they add to 24/27 = 8/9.
4. So. P(X ≥ 3) = 1 - 8/9 = 1/9, which is option (a).

The idea to keep. When the tail of a distribution runs to infinity, do not add the tail. Find the constant from the total of 1, then subtract the few finite terms from 1.

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