The mean and variance of a random variable X having a binomial probability distribution are 6 and 3 respectively, then the probabiltiy `P(X ge 2)` is

Options

A. `(13)/(4096)`
B. `(4083)/(4096)`
C. `(3)/(1024)`
D. `(13)/(2048)`

Answer

To solve the problem, we need to find the probability \( P(X \geq 2) \) for a random variable \( X \) that follows a binomial distribution, given that the mean is 6 and the variance is 3.

### Step-by-Step Solution:

1. **Understand the Binomial Distribution Parameters**:
The mean \( \mu \) and variance \( \sigma^2 \) of a binomial distribution are given by:
\[ \mu = n \cdot p \]
\[ \sigma^2 = n \cdot p \cdot q \]
where \( q = 1 - p \).

2. **Set Up the Equations**:
From the problem, we have:
\[ n \cdot p = 6 \quad \text{(1)} \]
\[ n \cdot p \cdot q = 3 \quad \text{(2)} \]

3. **Substituting \( q \)**:
Since \( q = 1 - p \), we can substitute \( q \) in equation (2):
\[ n \cdot p \cdot (1 - p) = 3 \]

4. **Express \( n \) in terms of \( p \)**:
From equation (1), we can express \( n \) as:
\[ n = \frac{6}{p} \]
Substitute this into equation (2):
\[ \frac{6}{p} \cdot p \cdot (1 - p) = 3 \]
Simplifying gives:
\[ 6(1 - p) = 3 \]
\[ 6 - 6p = 3 \]
\[ 6p = 3 \implies p = \frac{1}{2} \]

5. **Find \( n \)**:
Substitute \( p \) back into equation (1):
\[ n \cdot \frac{1}{2} = 6 \implies n = 12 \]

6. **Calculate \( P(X \geq 2) \)**:
To find \( P(X \geq 2) \), we can use the complement:
\[ P(X \geq 2) = 1 - P(X < 2) = 1 - (P(X = 0) + P(X = 1)) \]

7. **Calculate \( P(X = 0) \) and \( P(X = 1) \)**:
Using the binomial probability formula:
\[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]
- For \( P(X = 0) \):
\[ P(X = 0) = \binom{12}{0} \left(\frac{1}{2}\right)^0 \left(\frac{1}{2}\right)^{12} = 1 \cdot 1 \cdot \frac{1}{4096} = \frac{1}{4096} \]
- For \( P(X = 1) \):
\[ P(X = 1) = \binom{12}{1} \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^{11} = 12 \cdot \frac{1}{2} \cdot \frac{1}{2048} = \frac{12}{4096} \]

8. **Combine the Probabilities**:
\[ P(X < 2) = P(X = 0) + P(X = 1) = \frac{1}{4096} + \frac{12}{4096} = \frac{13}{4096} \]
Therefore,
\[ P(X \geq 2) = 1 - P(X < 2) = 1 - \frac{13}{4096} = \frac{4096 - 13}{4096} = \frac{4083}{4096} \]

### Final Answer:
\[ P(X \geq 2) = \frac{4083}{4096} \]