Find the constant C, so that
`F(x)=C((2)/(3))^(x),x=1,2,3………………` is the p.d.f of a discrete random variable X.

Answer

To find the constant \( C \) such that

\[ F(x) = C \left( \frac{2}{3} \right)^x, \quad x = 1, 2, 3, \ldots \]

is the probability density function (p.d.f.) of a discrete random variable \( X \), we need to ensure that the sum of the probabilities equals 1. This is a fundamental property of probability distributions.

### Step-by-Step Solution:

1. **Set up the equation for the p.d.f.**:
We know that the sum of the probabilities must equal 1:
\[ \sum_{x=1}^{\infty} F(x) = 1 \]
Substituting \( F(x) \):
\[ \sum_{x=1}^{\infty} C \left( \frac{2}{3} \right)^x = 1 \]

2. **Factor out the constant \( C \)**:
\[ C \sum_{x=1}^{\infty} \left( \frac{2}{3} \right)^x = 1 \]

3. **Recognize the series as a geometric series**:
The series \( \sum_{x=1}^{\infty} \left( \frac{2}{3} \right)^x \) is a geometric series where the first term \( a = \frac{2}{3} \) and the common ratio \( r = \frac{2}{3} \).

4. **Use the formula for the sum of an infinite geometric series**:
The sum \( S \) of an infinite geometric series can be calculated using the formula:
\[ S = \frac{a}{1 - r} \]
For our series:
\[ S = \frac{\frac{2}{3}}{1 - \frac{2}{3}} = \frac{\frac{2}{3}}{\frac{1}{3}} = 2 \]

5. **Substitute the sum back into the equation**:
Now substituting back into our equation:
\[ C \cdot 2 = 1 \]

6. **Solve for \( C \)**:
\[ C = \frac{1}{2} \]

### Final Answer:
Thus, the constant \( C \) is

\[ \boxed{\frac{1}{2}} \]