If P(X=x)=cleft(frac{2}{3}right)^x for x=1, 2 | vedclass

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(C) For a probability distribution, the sum of all probabilities must be equal to $1$, i.e.,$\sum_{x=1}^{\infty} P(X=x) = 1$.Given $P(X=x) = c\left(\f

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(C) For a probability distribution, the sum of all probabilities must be equal to $1$, i.e.,$\sum_{x=1}^{\infty} P(X=x) = 1$.Given $P(X=x) = c\left(\frac{2}{3}\right)^x$.Substituting the values of $x$: $c\left(\frac{2}{3}\right) + c\left(\frac{2}{3}\right)^2 + c\left(\frac{2}{3}\right)^3 + \ldots = 1$.Taking $c$ as a common factor: $c \left[ \frac{2}{3} + \left(\frac{2}{3}\right)^2 + \left(\frac{2}{3}\right)^3 + \ldots \right] = 1$.The expression inside the bracket is an infinite geometric series with first term $a = \frac{2}{3}$ and common ratio $r = \frac{2}{3}$.The sum of an infinite geometric series is given by $S = \frac{a}{1-r}$.So, $c \left[ \frac{2/3}{1 - 2/3} \right] = 1$.$c \left[ \frac{2/3}{1/3} \right] = 1$.$c(2) = 1$.Therefore, $c = \frac{1}{2}$.

$X = x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{10}$	$k$	$\frac{1}{5}$	$2k$	$\frac{3}{10}$	$k$

$X = x$	$0$	$2$	$4$	$6$	$8$	$10$
$P(X = x)$	$0$	$k$	$2k$	$5k^2$	$2k^2$	$3k$

If $P(X=x)=c\left(\frac{2}{3}\right)^x$ for $x=1, 2, 3, 4, \ldots$ is a probability distribution function of a random variable $X$, then the value of $c$ is

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