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 binomial random variable \( X \) with a given mean and variance. Let's break it down step by step.

### Step 1: Identify the parameters of the binomial distribution
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 \( q = 1 - p \)

Given:
- Mean \( \mu = 6 \)
- Variance \( \sigma^2 = 3 \)

### Step 2: Set up the equations
From the mean:
\[ n \cdot p = 6 \quad (1) \]
From the variance:
\[ n \cdot p \cdot q = 3 \quad (2) \]

### Step 3: Substitute \( q \) in terms of \( p \)
Since \( q = 1 - p \), we can substitute this into equation (2):
\[ n \cdot p \cdot (1 - p) = 3 \quad (3) \]

### Step 4: Divide equation (1) by equation (3)
From equation (1):
\[ p = \frac{6}{n} \]
Substituting \( p \) into equation (3):
\[ n \cdot \frac{6}{n} \cdot \left(1 - \frac{6}{n}\right) = 3 \]
This simplifies to:
\[ 6 \left(1 - \frac{6}{n}\right) = 3 \]
\[ 6 - \frac{36}{n} = 3 \]
\[ \frac{36}{n} = 3 \]
\[ n = 12 \]

### Step 5: Find \( p \) and \( q \)
Now substituting \( n = 12 \) back into equation (1):
\[ 12 \cdot p = 6 \implies p = \frac{6}{12} = \frac{1}{2} \]
And since \( q = 1 - p \):
\[ q = 1 - \frac{1}{2} = \frac{1}{2} \]

### Step 6: Calculate \( P(X \geq 2) \)
To find \( P(X \geq 2) \):
\[ P(X \geq 2) = 1 - P(X = 0) - P(X = 1) \]

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

### Step 8: Combine the probabilities
Now, we can combine:
\[ P(X = 0) + P(X = 1) = \frac{1}{4096} + \frac{12}{4096} = \frac{13}{4096} \]
Thus,
\[ P(X \geq 2) = 1 - \left(\frac{13}{4096}\right) = \frac{4096 - 13}{4096} = \frac{4083}{4096} \]

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