In a Binomial distribution,if p=q and n geq 4 | vedclass

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(D) Given,$p=q$. Since $p+q=1$,we have $p=q=\frac{1}{2}$. Now,the probability mass function for a Binomial distribution is $P(X=k) = { }^n C_k p^k q^{

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${ }^n C_5$

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(D) Given,$p=q$. Since $p+q=1$,we have $p=q=\frac{1}{2}$. Now,the probability mass function for a Binomial distribution is $P(X=k) = { }^n C_k p^k q^{n-k}$. Substituting $k=5$ and $p=q=\frac{1}{2}$: $P(X=5) = { }^n C_5 \left(\frac{1}{2}\right)^5 \left(\frac{1}{2}\right)^{n-5} = { }^n C_5 \left(\frac{1}{2}\right)^n$. Therefore,$2^n P(X=5) = 2^n \cdot { }^n C_5 \left(\frac{1}{2}\right)^n = { }^n C_5$.

In a Binomial distribution,if $p=q$ and $n \geq 4$,then $2^n P(X=5)=$

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