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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
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Verified Answer
The correct answer is:
$137$
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 $\mathrm{n}=16$
$P(X \leq 2)=\frac{K}{2^{16}}$
$\Rightarrow \mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2)=\frac{\mathrm{K}}{2^{16}}$
$\begin{aligned}
\therefore \quad{ }^{16} \mathrm{C}_0\left(\frac{1}{2}\right)^0\left(\frac{1}{2}\right)^{16}+{ }^{16} \mathrm{C}_1\left(\frac{1}{2}\right)^1 & \left(\frac{1}{2}\right)^{15} \\
& +{ }^{16} \mathrm{C}_2\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^{14}=\frac{\mathrm{K}}{2^{16}}
\end{aligned}$
$\begin{array}{ll}
\therefore & \frac{1+16+120}{2^{16}}=\frac{K}{2^{16}} \\
\therefore & K=137
\end{array}$
According to the given conditions, Mean $=n p=8$ and variance $=n p q=4$ $\Rightarrow \mathrm{p}=\mathrm{q}=\frac{1}{2}$ and $\mathrm{n}=16$
$P(X \leq 2)=\frac{K}{2^{16}}$
$\Rightarrow \mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2)=\frac{\mathrm{K}}{2^{16}}$
$\begin{aligned}
\therefore \quad{ }^{16} \mathrm{C}_0\left(\frac{1}{2}\right)^0\left(\frac{1}{2}\right)^{16}+{ }^{16} \mathrm{C}_1\left(\frac{1}{2}\right)^1 & \left(\frac{1}{2}\right)^{15} \\
& +{ }^{16} \mathrm{C}_2\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^{14}=\frac{\mathrm{K}}{2^{16}}
\end{aligned}$
$\begin{array}{ll}
\therefore & \frac{1+16+120}{2^{16}}=\frac{K}{2^{16}} \\
\therefore & K=137
\end{array}$
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