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  • The random variable X can take only the values 0, 1, 2. Given that P (X = 0) = P (X = 1) = p and that E(X2) = E[X], find the value of p.

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.

Answers (1)

Given-

X=0,1,2 and P(X) at X=0 and 1,
Let X=2, P(X) is x.

\\\Rightarrow p+p+x=1$ \\$\Rightarrow x=1-2 p$
The following distribution is obtained
\begin{array}{|l|l|l|l|} \hline \mathrm{X} & 0 & 1 & 2 \\ \hline \mathrm{P}(\mathrm{X}) & \mathrm{P} & \mathrm{P} & 1-2 \mathrm{p} \\ \hline \end{array}
\\\therefore E(X)=X P(X)$ \\$=0 \times \mathrm{P}+1 \times \mathrm{P}+2(1-2 \mathrm{P})$ \\$=P+2-4 P=2-3 P$
And E(X)^{2}=X^{2} P(X)$
\\=0 \times p+1 \times P+4 \times(1-2 P)$ \\$=\mathrm{P}+4-8 \mathrm{P}=4-7 \mathrm{P}$
And, as we know that E\left(X^{2}\right)=E(X)$
\\\Rightarrow 4-7 p=2-3 p$ \\$\Rightarrow 4 p=2$ \\$\Rightarrow p=\frac{1}{2}$

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