Let a random variable \(\mathrm{X}\) have a Binomial distribution with mean 8 and variance 4 . If \(P(X \leq 2)=\frac{K}{2^{16}}\), then \(K\) is

Let \(\mathrm{X} \sim \mathrm{B}(\mathrm{n}, \mathrm{p})\)
According to the given conditions, Mean \(=n p=8\) and variance \(=n p q=4\) \(\Rightarrow \mathrm{p}=\mathrm{q}=\frac{1}{2}\) and \(\mathrm{n}=16\)
\(P(X \leq 2)=\frac{K}{2^{16}}\)
\(\Rightarrow \mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2)=\frac{\mathrm{K}}{2^{16}}\)
\(\therefore { }^{16} \mathrm{C}_0\left(\frac{1}{2}\right)^0\left(\frac{1}{2}\right)^{16}+{ }^{16} \mathrm{C}_1\left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^{15} \) \( +{ }^{16} \mathrm{C}_2\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^{14}=\frac{\mathrm{K}}{2^{16}}\)
\(\therefore \frac{1+16+120}{2^{16}}=\frac{K}{2^{16}} \)
\( \therefore K=137\)