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(C) For a probability mass function (p.m.f.),the sum of all probabilities must be equal to $1$,i.e.,$\sum P(X) = 1$.Given $P(X) = k \binom{4}{x}$ for

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$\frac{1}{16}$

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(C) For a probability mass function (p.m.f.),the sum of all probabilities must be equal to $1$,i.e.,$\sum P(X) = 1$.Given $P(X) = k \binom{4}{x}$ for $x = 0, 1, 2, 3, 4$.Calculating each probability:$P(X=0) = k \binom{4}{0} = k 	imes 1 = k$$P(X=1) = k \binom{4}{1} = k 	imes 4 = 4k$$P(X=2) = k \binom{4}{2} = k 	imes 6 = 6k$$P(X=3) = k \binom{4}{3} = k 	imes 4 = 4k$$P(X=4) = k \binom{4}{4} = k 	imes 1 = k$Summing these probabilities:$k + 4k + 6k + 4k + k = 1$$16k = 1$$k = \frac{1}{16}$

$X=x$	$0$	$1$	$2$	$3$
$P(X=x)$	$\frac{1}{3}$	$\frac{1}{2}$	$0$	$\frac{1}{6}$

If the probability mass function (p.m.f.) is given by $P(X) = k \binom{4}{x}$ for $x = 0, 1, 2, 3, 4$ and $k > 0$,and $P(X) = 0$ otherwise,then the value of $k$ is:

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