If in a distribution probability of a random variable X taking the particular value x is `""^(n)C_(x)p^(x)q^(n-x)` where `p+q=1`, and `x=1,2,3, . . . .,n,` find the its mean.

Answer

To find the mean of the given distribution, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Probability Distribution**:
The probability of a random variable \( X \) taking the value \( x \) is given by:
\[ P(X = x) = \binom{n}{x} p^x q^{n-x} \]
where \( p + q = 1 \) and \( x = 1, 2, \ldots, n \).

2. **Mean Formula**:
The mean (or expected value) of a discrete random variable is calculated using the formula:
\[ E(X) = \sum_{x=1}^{n} x P(X = x) \]
Substituting the expression for \( P(X = x) \):
\[ E(X) = \sum_{x=1}^{n} x \cdot \binom{n}{x} p^x q^{n-x} \]

3. **Simplify the Expression**:
We can use the identity for combinations:
\[ x \cdot \binom{n}{x} = n \cdot \binom{n-1}{x-1} \]
Thus, we can rewrite our expression as:
\[ E(X) = \sum_{x=1}^{n} n \cdot \binom{n-1}{x-1} p^x q^{n-x} \]

4. **Factor Out Constants**:
Since \( n \) is a constant, we can factor it out of the summation:
\[ E(X) = n \sum_{x=1}^{n} \binom{n-1}{x-1} p^x q^{n-x} \]

5. **Change the Index of Summation**:
Let \( y = x - 1 \). Then when \( x = 1 \), \( y = 0 \) and when \( x = n \), \( y = n - 1 \). The summation becomes:
\[ E(X) = n \sum_{y=0}^{n-1} \binom{n-1}{y} p^{y+1} q^{(n-1)-y} \]
This can be rewritten as:
\[ E(X) = n p \sum_{y=0}^{n-1} \binom{n-1}{y} p^y q^{(n-1)-y} \]

6. **Apply the Binomial Theorem**:
The summation \( \sum_{y=0}^{n-1} \binom{n-1}{y} p^y q^{(n-1)-y} \) is equal to \( (p + q)^{n-1} \) by the Binomial Theorem. Since \( p + q = 1 \):
\[ \sum_{y=0}^{n-1} \binom{n-1}{y} p^y q^{(n-1)-y} = 1^{n-1} = 1 \]

7. **Final Calculation**:
Therefore, we have:
\[ E(X) = n p \cdot 1 = n p \]

### Conclusion:
The mean of the distribution is:
\[ \text{Mean} = n p \]