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(C) For a probability distribution,the sum of all probabilities must be equal to $1$. Given $P(X=k) = \frac{3^{ck}}{k!}$ for $k = 1, 2, 3, \ldots$,we

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$\log_3(\log_e 2)$

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(C) For a probability distribution,the sum of all probabilities must be equal to $1$. Given $P(X=k) = \frac{3^{ck}}{k!}$ for $k = 1, 2, 3, \ldots$,we have: $\sum_{k=1}^{\infty} P(X=k) = \sum_{k=1}^{\infty} \frac{(3^c)^k}{k!} = 1$. Recall the Taylor series expansion for the exponential function: $e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!} = 1 + \sum_{k=1}^{\infty} \frac{x^k}{k!}$. Therefore,$\sum_{k=1}^{\infty} \frac{x^k}{k!} = e^x - 1$. Substituting $x = 3^c$,we get: $e^{3^c} - 1 = 1$,which implies $e^{3^c} = 2$. Taking the natural logarithm on both sides: $3^c = \log_e 2$. Taking the logarithm base $3$ on both sides: $c = \log_3(\log_e 2)$. Thus,the correct option is $C$.

$X$	$1$	$2$	$3$	$4$	$5$
$P(X=x)$	$K$	$2K$	$3K$	$2K$	$K$

$X$	$0$	$2$	$4$	$6$	$8$
$P(X)$	$a$	$2a$	$a+b$	$2b$	$3b$

$X$	$-3$	$-1$	$0$	$1$	$3$	$5$	$7$	$9$
$F(X=x)$	$0.1$	$0.3$	$0.5$	$0.65$	$0.75$	$0.85$	$0.90$	$1$

If the probability function of a random variable $X$ is given by $P(X=k) = \frac{3^{ck}}{k!}$ for $k = 1, 2, 3, \ldots$ (where $c$ is a constant),then $c =$

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