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Probability — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSProbability5 Marks
Question
The random variable $X$ can take only the values $0, 1, 2.$ Given that $P(X = 0) = P (X = 1) = p$ and that $E(X^2) = E[X], $ find the value of $p.$
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Answer

Since, $X = 0, 1, 2$ and $P (X)$ at $X = 0$ and $1$ is $p,$ let at $X = 2, P (X)$ is $x.$
$\Rightarrow p + p + x = 1$
$\Rightarrow x = 1 – 2p$
We get, the following distribution
$X$ $0$ $1$ $2$
$P(X)$ $p$ $q$ $1 - 2p$
$\therefore\text{E}[\text{X}]=\sum\text{XP}(\text{X})$
$=0\cdot\text{p}+1\cdot\text{p}+2(1-2\text{p})$
$=\text{p}+2-4\text{p}$
$=2-3\text{p}$
And $\text{E}(\text{X}^2)=\sum\text{X}^2\text{P}(\text{X})$
$=0\cdot\text{p}+1\cdot\text{p}+4\cdot(1-2\text{p})$
$=\text{p}+4-8\text{p}$
$=2-7\text{p}$
Also, given that $\text{E}(\text{X}^2)=\text{E}[\text{X}]$
$\Rightarrow4-7\text{p}=2-3\text{p}$
$\Rightarrow4\text{p}=2$
$\Rightarrow\text{p}=\frac{1}{2}$

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