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

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