If \(X \sim B\left(8, \frac{1}{2}\right)\), then \(P(|x-4| \leq 2)=\)

We have \(\mathrm{n}=8, \mathrm{p}=\frac{1}{2} \quad \Rightarrow \mathrm{q}=\frac{1}{2}\)
To find \(\mathrm{P}(|\mathrm{X}-4| \leq 2) \quad \Rightarrow \mathrm{X}=2,3,4,5,6\)
Hence required probability
\(={ }^{8} C_{2}\left(\frac{1}{2}\right)^{6}\left(\frac{1}{2}\right)^{2}+{ }^{8} C_{3}\left(\frac{1}{2}\right)^{5}\left(\frac{1}{2}\right)^{3}+{ }^{8} C_{4}\left(\frac{1}{2}\right)^{4}\) \(\left(\frac{1}{2}\right)^{4}+{}^{8} C_{5}\left(\frac{1}{2}\right)^{3}\left(\frac{1}{2}\right)^{5}+{ }^{8} C_{6}\left(\frac{1}{2}\right)^{2}\left(\frac{1}{2}\right)^{6} \)
\( =\left(\frac{1}{2}\right)^{8}\left[{ }^{8} C_{2}+{ }^{8} C_{3}+{ }^{8} C_{4}+{ }^{8} C_{5}+{ }^{8} C_{6}\right] \)
\( =\frac{1}{256}[28+56+70+56+28]=\frac{238}{256}=\frac{119}{128}\)