For a binomial distribution with mean 6 and variance 2, find the first two terms of the distribution.

Answer

To solve the problem, we need to find the first two terms of a binomial distribution given the mean and variance. Let's go through the solution step by step.

### Step 1: Understand the parameters of the binomial distribution
In a binomial distribution, the mean (μ) and variance (σ²) are given by:
- Mean: \( \mu = n \cdot p \)
- Variance: \( \sigma^2 = n \cdot p \cdot q \)

where:
- \( n \) = number of trials
- \( p \) = probability of success
- \( q \) = probability of failure (where \( q = 1 - p \))

Given:
- Mean \( \mu = 6 \)
- Variance \( \sigma^2 = 2 \)

### Step 2: Set up the equations
From the mean, we have:
1. \( n \cdot p = 6 \) (Equation 1)

From the variance, we have:
2. \( n \cdot p \cdot q = 2 \) (Equation 2)

### Step 3: Express \( q \) in terms of \( p \)
Since \( q = 1 - p \), we can substitute \( q \) in Equation 2:
\[ n \cdot p \cdot (1 - p) = 2 \]

### Step 4: Substitute \( n \) from Equation 1 into Equation 2
From Equation 1, we can express \( n \) as:
\[ n = \frac{6}{p} \]

Substituting \( n \) into Equation 2:
\[ \frac{6}{p} \cdot p \cdot (1 - p) = 2 \]
This simplifies to:
\[ 6(1 - p) = 2 \]

### Step 5: Solve for \( p \)
Expanding the equation:
\[ 6 - 6p = 2 \]
Rearranging gives:
\[ 6p = 4 \implies p = \frac{2}{3} \]

### Step 6: Find \( q \)
Now, we can find \( q \):
\[ q = 1 - p = 1 - \frac{2}{3} = \frac{1}{3} \]

### Step 7: Substitute \( p \) back to find \( n \)
Using \( p \) in Equation 1:
\[ n \cdot \frac{2}{3} = 6 \implies n = 6 \cdot \frac{3}{2} = 9 \]

### Step 8: Find the first two terms of the binomial distribution
The general formula for the \( (r+1)^{th} \) term in a binomial distribution is:
\[ T_{r+1} = \binom{n}{r} p^r q^{n-r} \]

**First Term (r = 0):**
\[ T_1 = \binom{9}{0} p^0 q^{9} = 1 \cdot \left(\frac{2}{3}\right)^0 \cdot \left(\frac{1}{3}\right)^9 = \left(\frac{1}{3}\right)^9 \]

**Second Term (r = 1):**
\[ T_2 = \binom{9}{1} p^1 q^{8} = 9 \cdot \left(\frac{2}{3}\right)^1 \cdot \left(\frac{1}{3}\right)^8 = 9 \cdot \frac{2}{3} \cdot \left(\frac{1}{3}\right)^8 = \frac{18}{3^9} \]

### Final Result
Thus, the first two terms of the binomial distribution are:
1. \( \left(\frac{1}{3}\right)^9 \)
2. \( \frac{18}{3^9} \)