The random variable X can take only the values 0, 1, 2, 3. Given that $P(X = 0) = P(X = 1) = p$ and $P(X = 2) = P(X = 3)$ such that $∑p_i{x_i}^2=2∑p_ix_i$, find the value of $p$.

$\frac{3}{8}$

The correct answer is Option (1) → $\frac{3}{8}$

Given $P(X = 0) = P(X = 1) = p$ and $P(X = 2) = P(X = 3) = k$ (say)

The probability distribution of the random variable X is

We know that $Σp_i = 1$

$⇒ p+p+k+k=1⇒2p+ 2k=$

$⇒p+k=\frac{1}{2}⇒k=\frac{1}{2}-p$.

We construct the following table:

Given $∑p_i{x_i}^2=2∑p_ix_i$

$⇒\frac{13}{2}-12p=2(\frac{5}{2}-4p)⇒\frac{13}{2}-12p=5-8p⇒-4p=5-\frac{13}{2}$

$⇒-4p=-\frac{3}{2}⇒p=\frac{3}{8}$

Hence, the value of $p$ is $\frac{3}{8}$