A : If the difference between the mean and variance of a binomial distribution for 5 trials is `5//9` then the distribution o is `(2//3+1//3)^(5)`
R : The binomial distribution with parameters n, p is `(q+p)^(n)`

Options

A. `((2)/(5)+(3)/(5))^(5)`
B. `((2)/(3)+(1)/(3))^(5)`
C. `((1)/(3)+(2)/(3))^(5)`
D. `((3)/(4)+(1)/(4))^(5)`

Answer

To solve the problem step by step, we need to find the parameters of a binomial distribution given that the difference between the mean and variance is \( \frac{5}{9} \) for 5 trials.

### Step 1: Identify the parameters of the binomial distribution
The binomial distribution has two parameters:
- \( n \): number of trials
- \( p \): probability of success in each trial

From the problem, we know that:
- \( n = 5 \)

### Step 2: Write the formulas for mean and variance
The mean \( \mu \) and variance \( \sigma^2 \) of a binomial distribution are given by:
- Mean: \( \mu = n \cdot p \)
- Variance: \( \sigma^2 = n \cdot p \cdot q \) where \( q = 1 - p \)

### Step 3: Set up the equation based on the given information
We are given that the difference between the mean and variance is \( \frac{5}{9} \):
\[ \mu - \sigma^2 = \frac{5}{9} \]
Substituting the formulas for mean and variance:
\[ n \cdot p - n \cdot p \cdot q = \frac{5}{9} \]

### Step 4: Substitute \( q \) in terms of \( p \)
Since \( q = 1 - p \), we can rewrite the equation:
\[ n \cdot p - n \cdot p \cdot (1 - p) = \frac{5}{9} \]
This simplifies to:
\[ n \cdot p - n \cdot p + n \cdot p^2 = \frac{5}{9} \]
Thus, we have:
\[ n \cdot p^2 = \frac{5}{9} \]

### Step 5: Substitute \( n = 5 \) into the equation
Now substituting \( n = 5 \):
\[ 5 \cdot p^2 = \frac{5}{9} \]
Dividing both sides by 5:
\[ p^2 = \frac{1}{9} \]

### Step 6: Solve for \( p \)
Taking the square root of both sides:
\[ p = \frac{1}{3} \]

### Step 7: Find \( q \)
Using the relation \( q = 1 - p \):
\[ q = 1 - \frac{1}{3} = \frac{2}{3} \]

### Step 8: Write the binomial distribution
The binomial distribution can be expressed as:
\[ (q + p)^n = \left(\frac{2}{3} + \frac{1}{3}\right)^5 = 1^5 = 1 \]

### Final Answer
Thus, the binomial distribution is:
\[ \left(\frac{2}{3} + \frac{1}{3}\right)^5 \]