The mean and variance of a random variable X having binomial distribution are 4 and 2 respectively. Then `P(x gt 6)=`

Options

A. `(1)/(32)`
B. `(37)/(256)`
C. `(9)/(256)`
D. `(1)/(256)`

Answer

To solve the problem, we need to find \( P(X > 6) \) for a binomial distribution with given mean and variance.

### Step 1: Understand the Mean and Variance of a Binomial Distribution
For a binomial distribution with parameters \( n \) (number of trials) and \( p \) (probability of success), the mean \( \mu \) and variance \( \sigma^2 \) are given by:
- Mean: \( \mu = n \cdot p \)
- Variance: \( \sigma^2 = n \cdot p \cdot q \)
where \( q = 1 - p \) (probability of failure).

### Step 2: Set Up the Equations
From the problem, we know:
- Mean \( \mu = 4 \)
- Variance \( \sigma^2 = 2 \)

This gives us the equations:
1. \( n \cdot p = 4 \) (1)
2. \( n \cdot p \cdot q = 2 \) (2)

### Step 3: Express \( q \) in Terms of \( p \)
From equation (1), we can express \( q \):
\[ q = 1 - p \]

### Step 4: Substitute \( q \) into the Variance Equation
Substituting \( q \) into equation (2):
\[ n \cdot p \cdot (1 - p) = 2 \]
Now substituting \( n \cdot p = 4 \) from equation (1):
\[ 4 \cdot (1 - p) = 2 \]

### Step 5: Solve for \( p \)
Solving the equation:
\[ 4 - 4p = 2 \]
\[ 4p = 2 \]
\[ p = \frac{1}{2} \]

### Step 6: Find \( q \) and \( n \)
Now, substituting \( p \) back to find \( q \):
\[ q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2} \]

Now substitute \( p \) back into equation (1) to find \( n \):
\[ n \cdot \frac{1}{2} = 4 \implies n = 4 \cdot 2 = 8 \]

### Step 7: Calculate \( P(X > 6) \)
We need to find \( P(X > 6) \):
\[ P(X > 6) = P(X = 7) + P(X = 8) \]

Using the binomial probability formula:
\[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]

Calculating \( P(X = 7) \):
\[ P(X = 7) = \binom{8}{7} \left(\frac{1}{2}\right)^7 \left(\frac{1}{2}\right)^{1} = \binom{8}{7} \left(\frac{1}{2}\right)^8 = 8 \cdot \frac{1}{256} = \frac{8}{256} \]

Calculating \( P(X = 8) \):
\[ P(X = 8) = \binom{8}{8} \left(\frac{1}{2}\right)^8 \left(\frac{1}{2}\right)^{0} = 1 \cdot \frac{1}{256} = \frac{1}{256} \]

### Step 8: Combine the Probabilities
Now, combine the probabilities:
\[ P(X > 6) = P(X = 7) + P(X = 8) = \frac{8}{256} + \frac{1}{256} = \frac{9}{256} \]

### Final Answer
Thus, the probability that \( X > 6 \) is:
\[ \boxed{\frac{9}{256}} \]