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Let $X$ be a random variable with probability distribution function,
$P(X=x)=K\left(\frac{2}{5}\right)^x, x=1,2,3, \ldots \ldots$ Then, value of $K$ is
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$P(X=x)=K\left(\frac{2}{5}\right)^x, x=1,2,3, \ldots \ldots$ Then, value of $K$ is
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Verified Answer
The correct answer is:
$\frac{3}{2}$
Given, $P(X=x)=K(2 / 5)^x$
We know $\Sigma P_i=1$
$\therefore \quad K\left[\left(\frac{2}{5}\right)+\left(\frac{2}{5}\right)^2+\left(\frac{2}{5}\right)^3+\ldots\right]=1$
$\begin{aligned} & \Rightarrow \quad K\left(\frac{\frac{2}{5}}{1-2 / 5}\right)=1 \\ & \Rightarrow \quad K \times \frac{2}{3}=1 \Rightarrow K=3 / 2\end{aligned}$
We know $\Sigma P_i=1$
$\therefore \quad K\left[\left(\frac{2}{5}\right)+\left(\frac{2}{5}\right)^2+\left(\frac{2}{5}\right)^3+\ldots\right]=1$
$\begin{aligned} & \Rightarrow \quad K\left(\frac{\frac{2}{5}}{1-2 / 5}\right)=1 \\ & \Rightarrow \quad K \times \frac{2}{3}=1 \Rightarrow K=3 / 2\end{aligned}$
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