Determine a binomial distribution whose mean is 9 and variance is 6.

Answer

To determine a binomial distribution with a mean of 9 and a variance of 6, we can follow these steps:

### Step 1: Define the Binomial Distribution Parameters
Let \( X \) be a binomial random variable with parameters \( n \) (number of trials) and \( p \) (probability of success). The mean \( \mu \) and variance \( \sigma^2 \) of a binomial distribution are given by:
- Mean: \( \mu = np \)
- Variance: \( \sigma^2 = npq \) where \( q = 1 - p \)

### Step 2: Set Up the Equations
From the problem, we know:
1. \( np = 9 \) (Equation 1)
2. \( npq = 6 \) (Equation 2)

### Step 3: Substitute \( q \) in Terms of \( p \)
Since \( q = 1 - p \), we can substitute \( q \) in Equation 2:
\[ np(1 - p) = 6 \]
Substituting \( np = 9 \) from Equation 1 into this gives:
\[ 9(1 - p) = 6 \]

### Step 4: Solve for \( p \)
Now, we can solve for \( p \):
\[ 9 - 9p = 6 \]
\[ 9p = 3 \]
\[ p = \frac{1}{3} \]

### Step 5: Find \( q \)
Now that we have \( p \), we can find \( q \):
\[ q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3} \]

### Step 6: Substitute \( p \) Back to Find \( n \)
Using Equation 1 (\( np = 9 \)):
\[ n \cdot \frac{1}{3} = 9 \]
Multiplying both sides by 3:
\[ n = 27 \]

### Step 7: Write the Binomial Distribution
Now that we have \( n \), \( p \), and \( q \), we can write the binomial distribution:
\[ P(X = r) = \binom{n}{r} p^r q^{n-r} \]
Substituting the values:
\[ P(X = r) = \binom{27}{r} \left(\frac{1}{3}\right)^r \left(\frac{2}{3}\right)^{27 - r} \]

### Final Answer
The required binomial distribution is:
\[ P(X = r) = \binom{27}{r} \left(\frac{1}{3}\right)^r \left(\frac{2}{3}\right)^{27 - r} \]

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