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If $\mathrm{P}(\mathrm{X}=\mathrm{x})=c\left(\frac{2}{3}\right)^x ; \mathrm{x}=1,2,3,4, \ldots$ is a probability distribution function of a random variable $\mathrm{X}$, then the value of $c$ is
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The correct answer is:
$\frac {1}{2}$
$P(X=x)=c\left(\frac{2}{3}\right)^x$
$\begin{aligned} & \Sigma P=1 \\ & \therefore \quad c\left(\frac{2}{3}\right)+c\left(\frac{2}{3}\right)^2+c\left(\frac{2}{3}\right)^3+\ldots . .=1 \\ & \Rightarrow c\left[\frac{2}{3}+\left(\frac{2}{3}\right)^2+\left(\frac{2}{3}\right)^3+\ldots \ldots\right]=1 \\ & \Rightarrow c\left[\frac{\frac{2}{3}}{1-\frac{2}{3}}\right]=1 \Rightarrow 2 c=1 \\ & \therefore \quad c=\frac{1}{2} .\end{aligned}$
$\begin{aligned} & \Sigma P=1 \\ & \therefore \quad c\left(\frac{2}{3}\right)+c\left(\frac{2}{3}\right)^2+c\left(\frac{2}{3}\right)^3+\ldots . .=1 \\ & \Rightarrow c\left[\frac{2}{3}+\left(\frac{2}{3}\right)^2+\left(\frac{2}{3}\right)^3+\ldots \ldots\right]=1 \\ & \Rightarrow c\left[\frac{\frac{2}{3}}{1-\frac{2}{3}}\right]=1 \Rightarrow 2 c=1 \\ & \therefore \quad c=\frac{1}{2} .\end{aligned}$
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