The range of random variable `x={1, 2, 3….}` and the probabilities are given by `P(x=k)=(3^(CK))/(K!)` (k = 1, 2, 3, ……) and C is a constant. Then C =

Options

A. `log_(3)(log2)`
B. `(1)/(2)log(log2)`
C. `(log_(e)(log2))/("log"_(3)^(e ))`
D. `log_(2)(log3)`

Answer

To solve the problem, we need to find the value of the constant \( C \) such that the probabilities \( P(x=k) = \frac{3^{Ck}}{k!} \) form a valid probability distribution. This means that the sum of all probabilities must equal 1.

### Step-by-step Solution:

1. **Write the Probability Function:**
The probability function is given as:
\[ P(x=k) = \frac{3^{Ck}}{k!} \]
for \( k = 1, 2, 3, \ldots \)

2. **Sum the Probabilities:**
We need to find the sum of all probabilities:
\[ \sum_{k=1}^{\infty} P(x=k) = \sum_{k=1}^{\infty} \frac{3^{Ck}}{k!} \]

3. **Recognize the Series:**
The series \( \sum_{k=0}^{\infty} \frac{x^k}{k!} \) is the Taylor series expansion for \( e^x \). Therefore, we can rewrite our sum:
\[ \sum_{k=1}^{\infty} \frac{3^{Ck}}{k!} = e^{3^C} - 1 \]
(Note: We start from \( k=1 \) so we subtract the \( k=0 \) term which is 1.)

4. **Set the Sum Equal to 1:**
For the probabilities to sum to 1, we set:
\[ e^{3^C} - 1 = 1 \]
This simplifies to:
\[ e^{3^C} = 2 \]

5. **Solve for \( C \):**
Taking the natural logarithm of both sides:
\[ 3^C = \ln(2) \]
Now, we can solve for \( C \):
\[ C = \frac{\ln(2)}{3} \]

### Final Answer:
Thus, the value of the constant \( C \) is:
\[ C = \frac{\ln(2)}{3} \]