If for a binomial distribution the mean is 6 and the standard deviation is `sqrt2`, then P(X=r)=

Options

A. `""^(9) C_(r) ((2)/(3))^(r) ((1)/(3))^(9-r)`
B. `""^(9) C_(r) ((1)/(3))^(r) ((2)/(3))^(9-r)`
C. `""^(12) C_(r) ((2)/(3))^(r) ((1)/(3))^(12-r)`
D. `""^(12) C_(r) ((1)/(3))^(r) ((2)/(3))^(9-r)`

Answer

To solve the problem, we need to find \( P(X = r) \) for a binomial distribution given that the mean is 6 and the standard deviation is \( \sqrt{2} \).

### Step-by-Step Solution:

1. **Identify the Mean and Standard Deviation**:
- The mean \( \mu \) of a binomial distribution is given by:
\[ \mu = n \cdot p \]
where \( n \) is the number of trials and \( p \) is the probability of success.
- We are given that \( \mu = 6 \).

2. **Identify the Variance**:
- The standard deviation \( \sigma \) is given by:
\[ \sigma = \sqrt{n \cdot p \cdot q} \]
where \( q = 1 - p \).
- We are given that \( \sigma = \sqrt{2} \).
- Therefore, the variance \( \sigma^2 \) is:
\[ \sigma^2 = n \cdot p \cdot q = 2 \]

3. **Set Up the Equations**:
- From the mean, we have:
\[ n \cdot p = 6 \quad \text{(1)} \]
- From the variance, we have:
\[ n \cdot p \cdot q = 2 \quad \text{(2)} \]

4. **Express \( q \) in Terms of \( p \)**:
- Since \( q = 1 - p \), we can substitute \( q \) in equation (2):
\[ n \cdot p \cdot (1 - p) = 2 \]
- From equation (1), we can express \( n \) as:
\[ n = \frac{6}{p} \]
- Substitute \( n \) into the modified equation (2):
\[ \frac{6}{p} \cdot p \cdot (1 - p) = 2 \]
- Simplifying gives:
\[ 6(1 - p) = 2 \]
- Thus:
\[ 6 - 6p = 2 \quad \Rightarrow \quad 6p = 4 \quad \Rightarrow \quad p = \frac{2}{3} \]

5. **Find \( q \)**:
- Now, using \( p \):
\[ q = 1 - p = 1 - \frac{2}{3} = \frac{1}{3} \]

6. **Find \( n \)**:
- Substitute \( p \) back into equation (1):
\[ n \cdot \frac{2}{3} = 6 \quad \Rightarrow \quad n = 6 \cdot \frac{3}{2} = 9 \]

7. **Calculate \( P(X = r) \)**:
- The probability \( P(X = r) \) for a binomial distribution is given by:
\[ P(X = r) = \binom{n}{r} p^r q^{n-r} \]
- Substitute \( n = 9 \), \( p = \frac{2}{3} \), and \( q = \frac{1}{3} \):
\[ P(X = r) = \binom{9}{r} \left(\frac{2}{3}\right)^r \left(\frac{1}{3}\right)^{9 - r} \]

### Final Answer:
\[ P(X = r) = \binom{9}{r} \left(\frac{2}{3}\right)^r \left(\frac{1}{3}\right)^{9 - r} \]