If X is a binomial variate with mean 10 and variance 5 then find `P(X le 1)`.

Answer

To solve the problem, we need to find \( P(X \leq 1) \) for a binomial random variable \( X \) with a mean of 10 and a variance of 5.

### Step-by-Step Solution:

1. **Identify the Mean and Variance Formulas**:
The mean \( \mu \) and variance \( \sigma^2 \) of a binomial distribution are given by:
\[ \mu = nP \quad \text{(1)} \]
\[ \sigma^2 = nPQ \quad \text{(2)} \]
where \( n \) is the number of trials, \( P \) is the probability of success, and \( Q = 1 - P \) is the probability of failure.

2. **Set Up the Equations**:
From the problem, we know:
\[ nP = 10 \quad \text{(from the mean)} \]
\[ nPQ = 5 \quad \text{(from the variance)} \]

3. **Express \( Q \) in Terms of \( P \)**:
Since \( Q = 1 - P \), we can substitute \( Q \) into equation (2):
\[ nP(1 - P) = 5 \quad \text{(3)} \]

4. **Substitute \( nP \) from Equation (1) into Equation (3)**:
From equation (1), we can express \( n \) as:
\[ n = \frac{10}{P} \]
Substitute this into equation (3):
\[ \frac{10}{P} P (1 - P) = 5 \]
Simplifying gives:
\[ 10(1 - P) = 5 \]
\[ 10 - 10P = 5 \]
\[ 10P = 5 \quad \Rightarrow \quad P = 0.5 \]

5. **Find \( Q \)**:
Now that we have \( P \):
\[ Q = 1 - P = 1 - 0.5 = 0.5 \]

6. **Find \( n \)**:
Substitute \( P \) back into equation (1) to find \( n \):
\[ nP = 10 \quad \Rightarrow \quad n(0.5) = 10 \quad \Rightarrow \quad n = 20 \]

7. **Calculate \( P(X \leq 1) \)**:
We need to find \( P(X \leq 1) = P(X = 0) + P(X = 1) \).

- **Calculate \( P(X = 0) \)**:
\[ P(X = 0) = \binom{20}{0} (0.5)^0 (0.5)^{20} = 1 \cdot 1 \cdot (0.5)^{20} = (0.5)^{20} \]

- **Calculate \( P(X = 1) \)**:
\[ P(X = 1) = \binom{20}{1} (0.5)^1 (0.5)^{19} = 20 \cdot (0.5) \cdot (0.5)^{19} = 20 \cdot (0.5)^{20} \]

8. **Combine the Probabilities**:
\[ P(X \leq 1) = P(X = 0) + P(X = 1) = (0.5)^{20} + 20 \cdot (0.5)^{20} = (0.5)^{20} (1 + 20) = 21 \cdot (0.5)^{20} \]

9. **Final Expression**:
\[ P(X \leq 1) = \frac{21}{2^{20}} \]

### Final Answer:
\[ P(X \leq 1) = \frac{21}{2^{20}} \]