In a binomial distribution, the mean is `(2)/(3)` and the variance is `(5)/(9)`. What is the probability that X = 2?

Options

A. `(5)/(36)`
B. `(25)/(36)`
C. `(25)/(216)`
D. `(25)/(54)`

Answer

To solve the problem, we need to find the probability that \( X = 2 \) in a binomial distribution where the mean \( \mu \) is \( \frac{2}{3} \) and the variance \( \sigma^2 \) is \( \frac{5}{9} \).

### Step 1: Understand the Mean and Variance of a Binomial Distribution
In a binomial distribution, the mean \( \mu \) is given by:
\[ \mu = n \cdot p \]
and the variance \( \sigma^2 \) is given by:
\[ \sigma^2 = n \cdot p \cdot q \]
where \( q = 1 - p \).

### Step 2: Set Up the Equations
From the problem statement, we have:
1. \( n \cdot p = \frac{2}{3} \) (Equation 1)
2. \( n \cdot p \cdot q = \frac{5}{9} \) (Equation 2)

### Step 3: Substitute \( q \) in Terms of \( p \)
Since \( q = 1 - p \), we can rewrite Equation 2 as:
\[ n \cdot p \cdot (1 - p) = \frac{5}{9} \]

### Step 4: Divide Equation 1 by Equation 2
To eliminate \( n \), we divide Equation 1 by Equation 2:
\[ \frac{n \cdot p}{n \cdot p \cdot (1 - p)} = \frac{\frac{2}{3}}{\frac{5}{9}} \]
This simplifies to:
\[ \frac{1}{1 - p} = \frac{2}{3} \cdot \frac{9}{5} = \frac{6}{5} \]

### Step 5: Solve for \( p \)
Cross-multiplying gives:
\[ 5 = 6(1 - p) \]
Expanding this gives:
\[ 5 = 6 - 6p \implies 6p = 6 - 5 \implies 6p = 1 \implies p = \frac{1}{6} \]

### Step 6: Substitute \( p \) Back to Find \( n \)
Now substitute \( p \) back into Equation 1:
\[ n \cdot \frac{1}{6} = \frac{2}{3} \implies n = \frac{2}{3} \cdot 6 = 4 \]

### Step 7: Calculate \( q \)
Now we can find \( q \):
\[ q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6} \]

### Step 8: Find the Probability \( P(X = 2) \)
Using the binomial probability formula:
\[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]
For \( k = 2 \):
\[ P(X = 2) = \binom{4}{2} \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^{4-2} \]
Calculating \( \binom{4}{2} = 6 \):
\[ P(X = 2) = 6 \cdot \left(\frac{1}{6}\right)^2 \cdot \left(\frac{5}{6}\right)^2 \]
\[ = 6 \cdot \frac{1}{36} \cdot \frac{25}{36} = 6 \cdot \frac{25}{1296} = \frac{150}{1296} = \frac{25}{216} \]

### Final Answer
Thus, the probability that \( X = 2 \) is:
\[ \boxed{\frac{25}{216}} \]