If X follows a binomial distribution with parameters `n=8` and `p=1//2`, then `p(|X-4|le2)` equals

Options

A. `121//128`
B. `119//128`
C. `117//128`
D. `115//128`

Answer

To solve the problem, we need to find \( P(|X-4| \leq 2) \) for a binomial random variable \( X \) with parameters \( n = 8 \) and \( p = \frac{1}{2} \).

### Step-by-Step Solution:

1. **Understanding the Absolute Value Condition**:
The expression \( |X - 4| \leq 2 \) can be rewritten as:
\[ -2 \leq X - 4 \leq 2 \]
This implies:
\[ 2 \leq X \leq 6 \]
Therefore, we need to find:
\[ P(2 \leq X \leq 6) \]

2. **Using the Binomial Probability Formula**:
The probability mass function for a binomial distribution is given by:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
For our case, \( n = 8 \) and \( p = \frac{1}{2} \). Thus, the formula simplifies to:
\[ P(X = k) = \binom{8}{k} \left(\frac{1}{2}\right)^8 \]

3. **Calculating the Required Probabilities**:
We need to calculate \( P(X = 2) \), \( P(X = 3) \), \( P(X = 4) \), \( P(X = 5) \), and \( P(X = 6) \):
\[ P(X = 2) = \binom{8}{2} \left(\frac{1}{2}\right)^8 \]
\[ P(X = 3) = \binom{8}{3} \left(\frac{1}{2}\right)^8 \]
\[ P(X = 4) = \binom{8}{4} \left(\frac{1}{2}\right)^8 \]
\[ P(X = 5) = \binom{8}{5} \left(\frac{1}{2}\right)^8 \]
\[ P(X = 6) = \binom{8}{6} \left(\frac{1}{2}\right)^8 \]

4. **Calculating Each Probability**:
Now we calculate each of these:
- \( P(X = 2) = \binom{8}{2} \left(\frac{1}{2}\right)^8 = 28 \cdot \frac{1}{256} = \frac{28}{256} \)
- \( P(X = 3) = \binom{8}{3} \left(\frac{1}{2}\right)^8 = 56 \cdot \frac{1}{256} = \frac{56}{256} \)
- \( P(X = 4) = \binom{8}{4} \left(\frac{1}{2}\right)^8 = 70 \cdot \frac{1}{256} = \frac{70}{256} \)
- \( P(X = 5) = \binom{8}{5} \left(\frac{1}{2}\right)^8 = 56 \cdot \frac{1}{256} = \frac{56}{256} \)
- \( P(X = 6) = \binom{8}{6} \left(\frac{1}{2}\right)^8 = 28 \cdot \frac{1}{256} = \frac{28}{256} \)

5. **Summing the Probabilities**:
Now we sum these probabilities:
\[ P(2 \leq X \leq 6) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) \]
\[ = \frac{28 + 56 + 70 + 56 + 28}{256} = \frac{238}{256} \]

6. **Simplifying the Result**:
To simplify \( \frac{238}{256} \):
\[ \frac{238}{256} = \frac{119}{128} \]

### Final Answer:
Thus, the final probability is:
\[ P(|X - 4| \leq 2) = \frac{119}{128} \]