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In a Binomial distribution, if $p=q$ and $n \geq 4$, then $2^n P(X=5)=$
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Verified Answer
The correct answer is:
${ }^n C_5$
Given, $p=q$
$\begin{aligned} & \because \quad p+q=1 \\ & \Rightarrow \quad p=q=\frac{1}{2}\end{aligned}$
Now, $2^n P(X=5)$
$\begin{aligned} & =2^n \cdot{ }^n C_5(P)^{n-5}(q)^5=2^n \cdot{ }^n C_5\left(\frac{1}{2}\right)^{n-5} \cdot\left(\frac{1}{2}\right)^5 \\ & =2^n \cdot{ }^n C_5\left(\frac{1}{2}\right)^n={ }^n C_5\end{aligned}$
$\begin{aligned} & \because \quad p+q=1 \\ & \Rightarrow \quad p=q=\frac{1}{2}\end{aligned}$
Now, $2^n P(X=5)$
$\begin{aligned} & =2^n \cdot{ }^n C_5(P)^{n-5}(q)^5=2^n \cdot{ }^n C_5\left(\frac{1}{2}\right)^{n-5} \cdot\left(\frac{1}{2}\right)^5 \\ & =2^n \cdot{ }^n C_5\left(\frac{1}{2}\right)^n={ }^n C_5\end{aligned}$
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