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 $p$.
Answer & explanation
Correct answer: option 1
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
|
$X$ |
0 |
1 |
2 |
3 |
|
$P(X)$ |
$p$ |
$p$ |
$k$ |
$k$ |
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:
|
$x_i$ |
$p_i$ |
$p_ix_i$ |
$p_i{x_i}^2$ |
|
0 |
$p$ |
0 |
0 |
|
1 |
$p$ |
$p$ |
$p$ |
|
2 |
$\frac{1}{2}- p$ |
$1- 2p$ |
$2-4p$ |
|
3 |
$\frac{1}{2}- p$ |
$\frac{3}{2}-3p$ |
$\frac{9}{2}- 9p$ |
|
Total |
|
$\frac{5}{2}-4p$ |
$\frac{13}{2}-12p$ |
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 $\frac{3}{8}$