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(C) Given the probability distribution function $P(X=x)=K\left(\frac{2}{5}\right)^x$ for $x=1, 2, 3, \ldots$.We know that the sum of all probabilities

Vedclass · Accepted Answer

$\frac{3}{2}$

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(C) Given the probability distribution function $P(X=x)=K\left(\frac{2}{5}\right)^x$ for $x=1, 2, 3, \ldots$.We know that the sum of all probabilities in a distribution must be equal to $1$,i.e.,$\sum_{x=1}^{\infty} P(X=x) = 1$.Substituting the given function,we get $K \sum_{x=1}^{\infty} \left(\frac{2}{5}\right)^x = 1$.This is an infinite geometric series with the first term $a = \frac{2}{5}$ and common ratio $r = \frac{2}{5}$.The sum of an infinite geometric series is given by $S = \frac{a}{1-r}$.Therefore,$K \left( \frac{2/5}{1 - 2/5} \right) = 1$.$K \left( \frac{2/5}{3/5} \right) = 1$.$K \left( \frac{2}{3} \right) = 1$.Thus,$K = \frac{3}{2}$.

$X$	$\alpha$	$1$	$0$	$-3$
$P(X)$	$\frac{1}{3}$	$K$	$\frac{1}{6}$	$\frac{1}{4}$

Let $X$ be a random variable with probability distribution function,$P(X=x)=K\left(\frac{2}{5}\right)^x, x=1, 2, 3, \ldots$. Then,the value of $K$ is

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