Let a random variable X take values 0, 1, 2, 3 with P(X=0)=P(X=1)=p, P(X=2)=P(X=3) and E(X²)=2E(X). Then the value of 8p−1 is :

Solution

Maths sir · 4 steps

Call the unknown probability q, then use two equations: the total is 1 and the given moment condition.

1. Name the unknowns. Let P(X=2) = P(X=3) = q. All probabilities add to 1, so 2p + 2q = 1.
2. Write both expectations. E(X) = 0 · p + 1 · p + 2q + 3q = p + 5q. Also E(X²) = 0 · p + 1 · p + 4q + 9q = p + 13q.

   p + 13q = 2(p + 5q) ⇒ 3q = p ⇒ q = p/3

3. Use the total. Put q = p/3 into 2p + 2q = 1. This gives 2p + 2p/3 = 8p/3 = 1, so p = 3/8.
4. So. 8p - 1 = 3 - 1 = 2, which is option (a).

The idea to keep. For a discrete distribution with unknown probabilities, you always get one equation free: the probabilities sum to 1. Each extra condition, such as one on E(X) or E(X²), gives you one more equation.

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