A random variable X takes the values `0,1,2,3,...,` with prbability `PX(=x)=k(x+1)((1)/(5))^x`, where k is a constant, then `P(X=0)` is.

Options

A. `(2)/(25)`
B. `(4)/(25)`
C. `(9)/(25)`
D. `(16)/(25)`

Answer

To solve the problem, we need to find the probability \( P(X=0) \) where the probability mass function is given by:

\[ P(X=x) = k(x+1)\left(\frac{1}{5}\right)^x \]

### Step 1: Find the value of \( k \)

To find \( k \), we need to ensure that the total probability sums to 1:

\[ \sum_{x=0}^{\infty} P(X=x) = 1 \]

Substituting the expression for \( P(X=x) \):

\[ \sum_{x=0}^{\infty} k(x+1)\left(\frac{1}{5}\right)^x = 1 \]

### Step 2: Simplify the series

We can factor out \( k \):

\[ k \sum_{x=0}^{\infty} (x+1)\left(\frac{1}{5}\right)^x = 1 \]

The series \( \sum_{x=0}^{\infty} (x+1) r^x \) can be evaluated using the formula:

\[ \sum_{x=0}^{\infty} (x+1) r^x = \frac{1}{(1-r)^2} \]

For our case, \( r = \frac{1}{5} \):

\[ \sum_{x=0}^{\infty} (x+1) \left(\frac{1}{5}\right)^x = \frac{1}{\left(1 - \frac{1}{5}\right)^2} = \frac{1}{\left(\frac{4}{5}\right)^2} = \frac{25}{16} \]

### Step 3: Substitute back to find \( k \)

Now substituting this back into our equation:

\[ k \cdot \frac{25}{16} = 1 \]

Solving for \( k \):

\[ k = \frac{16}{25} \]

### Step 4: Calculate \( P(X=0) \)

Now we can find \( P(X=0) \):

\[ P(X=0) = k(0+1)\left(\frac{1}{5}\right)^0 = k \cdot 1 \cdot 1 = k \]

Substituting the value of \( k \):

\[ P(X=0) = \frac{16}{25} \]

### Final Answer

Thus, the probability \( P(X=0) \) is:

\[ \boxed{\frac{16}{25}} \]