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If the range of a random variable $X$ is $\{0,1,2,3,4, \ldots \ldots\}$ with $P(X=k)=\frac{(k+1) a}{3^k}$ for $k \geq 0$, then $a$ is equal to
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Verified Answer
The correct answer is:
$\frac{4}{9}$
Given that
$P(X=k)=\frac{(k+1) a}{3^k} \text { for } x \in(0,1,2, \ldots \infty)$
As we know that
$P(0)+P(1)+P(2)+\ldots \infty=1$
$\begin{aligned}
S & =a\left(1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+\ldots \infty\right) \\
\frac{1}{3} S & =a\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\ldots \infty\right) \\
\hline S-\frac{1}{3} S & =a\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\ldots \infty\right) \\
\Rightarrow \quad \frac{2}{3} S & =a\left(\frac{1}{1-\frac{1}{3}}\right) \\
\Rightarrow \quad \frac{2}{3} S & =\frac{3 a}{2} \\
\Rightarrow \quad S & =\frac{9 a}{4}
\end{aligned}$
From equation (i)
$\begin{array}{llrl}
\Rightarrow & & \frac{9 a}{4} & =1 \\
\Rightarrow & a & =\frac{4}{9}
\end{array}$
$P(X=k)=\frac{(k+1) a}{3^k} \text { for } x \in(0,1,2, \ldots \infty)$
As we know that
$P(0)+P(1)+P(2)+\ldots \infty=1$
$\begin{aligned}
S & =a\left(1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+\ldots \infty\right) \\
\frac{1}{3} S & =a\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\ldots \infty\right) \\
\hline S-\frac{1}{3} S & =a\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\ldots \infty\right) \\
\Rightarrow \quad \frac{2}{3} S & =a\left(\frac{1}{1-\frac{1}{3}}\right) \\
\Rightarrow \quad \frac{2}{3} S & =\frac{3 a}{2} \\
\Rightarrow \quad S & =\frac{9 a}{4}
\end{aligned}$
From equation (i)
$\begin{array}{llrl}
\Rightarrow & & \frac{9 a}{4} & =1 \\
\Rightarrow & a & =\frac{4}{9}
\end{array}$
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