#### Question

The random variable X can take only the values 0, 1, 2, 3. Give that P(X = 0) = P(X = 1) = p and P(X = 2) = P(X = 3) such that `Sigmap_i x_i^2 = 2Sigmap_ix_i`. Find the value of p

#### Solution

It is given that the random variable X can take only the values 0, 1, 2, 3.

Given:

P(X = 0) = P(X = 1) = p

P(X = 2) = P(X = 3)

Let P(X = 2) = P(X = 3) = q

Now,

P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 1

`=> p + p + (q + q) = 1`

`=> q = (1- 2p)/2`

Since, `Sigmap_ix_i^2 = 2Sigmap_ix_i`

`=> 0 + p(1)^2 + ((1-2p)/2)(2^2 + 3^2) = 2[0 + p + ((1-2p)/2) (2 + 3)]`

`=> p + 13/2 (1- 2p) = 2[p + 5/2(1-2p)]`

`=> p + 13/2 - 13p = 2p + 5 - 10p`

`=> 13/2 - 12p = -8p + 5`

`=> 4p = 3/2`

`=> p = 3/8`

Is there an error in this question or solution?

Solution The Random Variable X Can Take Only the Values 0, 1, 2, 3. Give that P(X = 0) = P(X = 1) = P and P(X = 2) = P(X = 3) Such that `Sigmap_I X_I^2 = 2sigmap_Ix_I`. Find the Value of P Concept: Variance of a Random Variable.