For a binimial variate X with n=6, if `P(X=4)=(135)/(2^(12))` then its variance is

Options

A. `8//9`
B. `1//4`
C. `9//8`
D. `4`

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript.

### Step 1: Understand the Problem
We are given a binomial random variable \( X \) with \( n = 6 \) and the probability \( P(X = 4) = \frac{135}{2^{12}} \). We need to find the variance of this binomial distribution.

### Step 2: Use the Binomial Probability Formula
The probability of getting exactly \( r \) successes in \( n \) trials in a binomial distribution is given by:
\[ P(X = r) = \binom{n}{r} p^r q^{n-r} \]
where \( q = 1 - p \).

In our case, we have:
- \( n = 6 \)
- \( r = 4 \)
- \( P(X = 4) = \frac{135}{2^{12}} \)

### Step 3: Substitute Values into the Formula
Substituting the known values into the formula:
\[ P(X = 4) = \binom{6}{4} p^4 q^{2} \]
Calculating \( \binom{6}{4} \):
\[ \binom{6}{4} = \binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15 \]
Thus, we have:
\[ \frac{135}{2^{12}} = 15 p^4 q^2 \]

### Step 4: Simplify the Equation
Dividing both sides by 15:
\[ p^4 q^2 = \frac{135}{15 \times 2^{12}} = \frac{9}{2^{12}} \]

### Step 5: Express \( q \) in Terms of \( p \)
Since \( q = 1 - p \), we can substitute \( q \) into the equation:
\[ p^4 (1 - p)^2 = \frac{9}{2^{12}} \]

### Step 6: Take the Square Root
Taking the square root of both sides gives:
\[ p^2 (1 - p) = \frac{3}{2^6} = \frac{3}{64} \]

### Step 7: Rearrange the Equation
Rearranging gives:
\[ p^2 (1 - p) = \frac{3}{64} \]

### Step 8: Solve for \( p \)
We can express \( 1 - p \) as \( q \) and rewrite:
\[ p^2 q = \frac{3}{64} \]
We can try \( p = \frac{1}{4} \):
\[ q = 1 - p = 1 - \frac{1}{4} = \frac{3}{4} \]
Substituting \( p \) and \( q \):
\[ \left(\frac{1}{4}\right)^2 \left(\frac{3}{4}\right) = \frac{1}{16} \cdot \frac{3}{4} = \frac{3}{64} \]
This satisfies the equation.

### Step 9: Calculate the Variance
Now that we have \( p \) and \( q \):
- \( p = \frac{1}{4} \)
- \( q = \frac{3}{4} \)
- \( n = 6 \)

The variance \( \sigma^2 \) of a binomial distribution is given by:
\[ \sigma^2 = n p q \]
Substituting the values:
\[ \sigma^2 = 6 \cdot \frac{1}{4} \cdot \frac{3}{4} = 6 \cdot \frac{3}{16} = \frac{18}{16} = \frac{9}{8} \]

### Final Answer
The variance of the binomial distribution is \( \frac{9}{8} \).

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