A random variable $X$ has its range $\{-1,0,1\}$. If its mean is 0.2 and $P(X=0)=0.2$, then $P(X=1)=$

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Aicharya · Accepted Answer

According to given information, $\begin{array}{ccc} \hline X=-1 & X=0 & X=1 \ \hline a & b & c \ \hline \end{array}$ Also, given mean $=0.2$ $\begin{array}{ll}\Rightarrow & \frac{-a+c}{a+b+c}=\frac{2}{10} \quad \ldots (i) \ \text {and } & \frac{b}{a+b+c}=\frac{2}{10} \quad \ldots (ii) \end{array}$ Now, $\quad 1+\frac{c-a}{a+b+c}=\frac{2}{10}+1$ $\Rightarrow \quad \frac{a+b+c+c-a}{a+b+c}=\frac{12}{10}$ $\Rightarrow \quad \frac{b+2 c}{a+b+c}=\frac{12}{10}$ $\Rightarrow \quad 2\left(\frac{c}{a+b+c}\right)+\frac{2}{10}=\frac{12}{10} \quad$ [by Eq. (ii)] $\Rightarrow \quad 2 P(X=1)=\frac{12}{10}-\frac{2}{10}=1$ $\Rightarrow \quad 2 P(X=1)=1$ $\Rightarrow \quad P(X=1)=\frac{1}{2}=0.5$

A random variable \(X\) has its range \(\{-1,0,1\}\). If its mean is 0.2 and \(P(X=0)=0.2\), then \(P(X=1)=\)

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