If the difference between the mean and variance of a Binomial variate is 5/9 then find the probability for the event of 2 success when the experiment is conducted 5 times.

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Understand the given information
We are given that the difference between the mean and variance of a Binomial variate is \( \frac{5}{9} \). We need to find the probability of getting exactly 2 successes when the experiment is conducted 5 times.

### Step 2: Define the parameters of the Binomial distribution
For a Binomial distribution, the mean \( \mu \) and variance \( \sigma^2 \) are defined as:
- Mean: \( \mu = n \cdot p \)
- Variance: \( \sigma^2 = n \cdot p \cdot q \) where \( q = 1 - p \)

Here, \( n \) is the number of trials, and \( p \) is the probability of success.

### Step 3: Set up the equation based on the given difference
From the problem, we have:
\[ \mu - \sigma^2 = \frac{5}{9} \]
Substituting the expressions for mean and variance:
\[ n \cdot p - n \cdot p \cdot (1 - p) = \frac{5}{9} \]

### Step 4: Simplify the equation
Factoring out \( n \cdot p \):
\[ n \cdot p \cdot (1 - (1 - p)) = \frac{5}{9} \]
This simplifies to:
\[ n \cdot p^2 = \frac{5}{9} \]

### Step 5: Substitute \( n = 5 \)
Since the experiment is conducted 5 times, we set \( n = 5 \):
\[ 5 \cdot p^2 = \frac{5}{9} \]
Dividing both sides by 5:
\[ p^2 = \frac{1}{9} \]

### Step 6: Solve for \( p \)
Taking the square root of both sides:
\[ p = \frac{1}{3} \]
Then, calculate \( q \):
\[ q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3} \]

### Step 7: Calculate the probability of getting exactly 2 successes
Using the Binomial probability formula:
\[ P(X = x) = \binom{n}{x} p^x q^{n-x} \]
For \( n = 5 \) and \( x = 2 \):
\[ P(X = 2) = \binom{5}{2} \left(\frac{1}{3}\right)^2 \left(\frac{2}{3}\right)^{5-2} \]

### Step 8: Calculate \( \binom{5}{2} \)
\[ \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \]

### Step 9: Substitute values into the probability formula
\[ P(X = 2) = 10 \cdot \left(\frac{1}{3}\right)^2 \cdot \left(\frac{2}{3}\right)^3 \]
Calculating further:
\[ = 10 \cdot \frac{1}{9} \cdot \frac{8}{27} \]
\[ = 10 \cdot \frac{8}{243} = \frac{80}{243} \]

### Final Answer
Thus, the probability of getting exactly 2 successes when the experiment is conducted 5 times is:
\[ \boxed{\frac{80}{243}} \]