Let X be a random variable such that the probability function of a distribution is given by P(X = 0) = 1/2,P(X = j) = 1/3ʲ(j = 1,2,3,...,∞ ). Then the mean of the distribution and P(X is positive and even) respectively are :

Solution

Maths sir · 4 steps

The mean is an arithmetico-geometric series. Shift it by one place, subtract, and it becomes a plain GP.

1. Check the distribution. The probabilities for j ≥ 1 add to (1/3)/(1 - 1/3) = 1/2. With P(X=0) = 1/2 the total is 1, so this is a valid distribution and the given values are used as they are.
2. Set up the mean. The X = 0 term adds nothing, so E(X) = S = 1/3 + 2/3² + 3/3³ + …. Write S/3 = 1/3² + 2/3³ + … and subtract it from S, lining up equal powers of 3.

   S - S/3 = 1/3 + 1/3² + 1/3³ + … = 1/2 ⇒ 2S/3 = 1/2 ⇒ S = 3/4

3. Positive and even. Take j = 2, 4, 6, …: P = 1/3² + 1/3⁴ + …. This GP has first term 1/9 and ratio 1/9, so P = (1/9)/(1 - 1/9) = 1/8.
4. So. The mean is 3/4 and P(X positive and even) = 1/8, which is option (b).

The idea to keep. For Σ j rʲ, multiply by r and subtract to get a GP, or use r/(1-r)² directly. For even terms only, square the ratio.

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