To find \(P(|X - 4| \le 2)\) for a binomial distribution \(X \sim B(n=8, p=\frac{1}{2})\), we need to calculate the probability that \(X\) takes values within 2 of the mean, specifically \(P(2 \le X \le 6)\).
The probability mass function (PMF) for a binomial distribution is given by:
\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]
Here, \(n = 8\) and \(p = \frac{1}{2}\), so we can calculate:
- \[
P(X = 2) = \binom{8}{2} \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^{6} = \frac{28}{256} = \frac{7}{64}
\]
- \[
P(X = 3) = \binom{8}{3} \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^{5} = \frac{56}{256} = \frac{7}{32}
\]
- \[
P(X = 4) = \binom{8}{4} \left(\frac{1}{2}\right)^4 \left(\frac{1}{2}\right)^{4} = \frac{70}{256} = \frac{35}{128}
\]
- \[
P(X = 5) = \binom{8}{5} \left(\frac{1}{2}\right)^5 \left(\frac{1}{2}\right)^{3} = \frac{56}{256} = \frac{7}{32}
\]
- \[
P(X = 6) = \binom{8}{6} \left(\frac{1}{2}\right)^6 \left(\frac{1}{2}\right)^{2} = \frac{28}{256} = \frac{7}{64}
\]
Thus,
\[
P(2 \le X \le 6) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)
\]
Calculating this gives:
\[
P(2 \le X \le 6) = \frac{7}{64} + \frac{7}{32} + \frac{35}{128} + \frac{7}{32} + \frac{7}{64}
\]
Simplifying the terms by having a common denominator, we have:
\[
P(2 \le X \le 6) = \frac{14}{128} + \frac{28}{128} + \frac{35}{128} + \frac{28}{128} + \frac{14}{128} = \frac{119}{128}
\]
Therefore, the probability is \(\frac{119}{128}\), which corresponds to the correct option.