Let a random variable X have a binomial distribution with mean 8 and variance r. If `P(X le 2 ) = (k)/(2^(16))`, then k is equal to

Options

A. 17
B. 121
C. 1
D. 137

Answer

To solve the problem step by step, we will use the properties of the binomial distribution and the given information about the mean and variance.

### Step 1: Understand the Binomial Distribution
A random variable \( X \) that follows a binomial distribution can be represented as \( X \sim B(n, p) \), where:
- \( n \) is the number of trials,
- \( p \) is the probability of success on each trial.

### Step 2: Use the Mean and Variance Formulas
For a binomial distribution:
- The mean \( \mu \) is given by \( \mu = np \).
- The variance \( \sigma^2 \) is given by \( \sigma^2 = np(1-p) \).

From the problem, we know:
- Mean \( \mu = 8 \)
- Variance \( \sigma^2 = r \)

### Step 3: Set Up the Equations
From the mean:
\[ np = 8 \tag{1} \]
From the variance:
\[ np(1-p) = r \tag{2} \]

### Step 4: Express \( p \) in Terms of \( n \)
From equation (1), we can express \( p \) as:
\[ p = \frac{8}{n} \tag{3} \]

### Step 5: Substitute \( p \) into the Variance Equation
Substituting equation (3) into equation (2):
\[ n \left(\frac{8}{n}\right) \left(1 - \frac{8}{n}\right) = r \]
This simplifies to:
\[ 8 \left(1 - \frac{8}{n}\right) = r \]
\[ 8 - \frac{64}{n} = r \tag{4} \]

### Step 6: Find \( n \) and \( p \)
We need to find \( n \) such that the variance \( r = 4 \) (as stated in the video transcript):
\[ 8 - \frac{64}{n} = 4 \]
Solving for \( n \):
\[ 8 - 4 = \frac{64}{n} \]
\[ 4 = \frac{64}{n} \]
\[ n = \frac{64}{4} = 16 \]

Now substitute \( n = 16 \) back into equation (3) to find \( p \):
\[ p = \frac{8}{16} = \frac{1}{2} \]

### Step 7: Calculate \( P(X \leq 2) \)
We need to find \( P(X \leq 2) \):
\[ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \]
Using the binomial probability formula:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
Calculating each term:
- For \( k = 0 \):
\[ P(X = 0) = \binom{16}{0} \left(\frac{1}{2}\right)^0 \left(\frac{1}{2}\right)^{16} = 1 \cdot 1 \cdot \frac{1}{65536} = \frac{1}{65536} \]
- For \( k = 1 \):
\[ P(X = 1) = \binom{16}{1} \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^{15} = 16 \cdot \frac{1}{2} \cdot \frac{1}{32768} = \frac{16}{65536} \]
- For \( k = 2 \):
\[ P(X = 2) = \binom{16}{2} \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^{14} = 120 \cdot \frac{1}{4} \cdot \frac{1}{16384} = \frac{120}{65536} \]

### Step 8: Combine the Probabilities
Now, add these probabilities:
\[ P(X \leq 2) = \frac{1}{65536} + \frac{16}{65536} + \frac{120}{65536} = \frac{137}{65536} \]

### Step 9: Express in the Required Form
The problem states that \( P(X \leq 2) = \frac{k}{2^{16}} \). Since \( 65536 = 2^{16} \):
\[ P(X \leq 2) = \frac{137}{2^{16}} \]
Thus, \( k = 137 \).

### Final Answer
The value of \( k \) is:
\[ \boxed{137} \]