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If $\mathrm{P}(\mathrm{X}=\mathrm{x})=\mathrm{k}\left(\frac{3}{8}\right)^{\mathrm{X}}, \mathrm{x}=1,2,3, \ldots$ is the probability distribution function of a discrete random váriable $\mathrm{X}$, then $\mathrm{k}=$
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Verified Answer
The correct answer is:
$\frac{5}{3}$
$\because \quad P(X=x)=k\left(\frac{3}{8}\right)^X ; X=1,2,3, \ldots$
Now, $P(X=1)+P(X=2)+P(X=3)+\ldots . .=1$
$\begin{aligned}
& \Rightarrow \quad k\left(\frac{3}{8}\right)^1+k\left(\frac{3}{8}\right)^2+k\left(\frac{3}{8}\right)^3+\ldots .=1 \\
& \Rightarrow \quad k\left[\left(\frac{3}{8}\right)^1+\left(\frac{3}{8}\right)^2+\left(\frac{3}{8}\right)^3+\ldots .\right]=1 \\
& \Rightarrow \quad k \cdot \frac{\frac{3}{8}}{1-\frac{3}{8}}=1 \Rightarrow k=\frac{5}{3}
\end{aligned}$
Now, $P(X=1)+P(X=2)+P(X=3)+\ldots . .=1$
$\begin{aligned}
& \Rightarrow \quad k\left(\frac{3}{8}\right)^1+k\left(\frac{3}{8}\right)^2+k\left(\frac{3}{8}\right)^3+\ldots .=1 \\
& \Rightarrow \quad k\left[\left(\frac{3}{8}\right)^1+\left(\frac{3}{8}\right)^2+\left(\frac{3}{8}\right)^3+\ldots .\right]=1 \\
& \Rightarrow \quad k \cdot \frac{\frac{3}{8}}{1-\frac{3}{8}}=1 \Rightarrow k=\frac{5}{3}
\end{aligned}$
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