In a Binomial distribution, if ' $n$ ' is the number of trials and the mean and variance are 4 and 3…
In a Binomial distribution, if ' $n$ ' is the number of trials and the mean and variance are 4 and 3 respectively, then $2^{32} p\left(X=\frac{n}{2}\right)=$
${ }^{16} C_8\left(3^8\right)$
${ }^{12} C_6\left(2^6\right)$
${ }^{32} C_{16}\left(3^{16}\right)$
${ }^{16} C_7\left(3^9\right)$
Solution
Let $X$ be the binomial variate for which mean $=4$ and variance $=3$, then $n p=4$ and $n p q=3$
$\Rightarrow q=\frac{3}{4}$
$\therefore \quad P=(1-q)=\left(1-\frac{3}{4}\right)=\frac{1}{4}$ and $n p=4$
$\Rightarrow n=\frac{4}{1} \times 4=16$
Thus, $n=16, p=\frac{1}{4}$ and $q=\frac{3}{4}$
Hence, the binomial distribution
$2^{32} P\left(X=\frac{n}{2}\right)=2^{32} \cdot{ }^{16} C_{\frac{16}{2}} \cdot\left(\frac{1}{4}\right)^{\frac{16}{2}}\left(\frac{3}{4}\right)^{16-\frac{16}{2}}$
$=2^{32} \cdot{ }^{16} C_8\left(\frac{1}{4}\right)^8 \times\left(\frac{3}{4}\right)^8=2^{32} \cdot{ }^{16} C_8 \frac{(3)^8}{(4)^{16}}$
$={ }^{16} C_8(3)^8 \quad\left[\because(4)^{16}=\left(2)^{32}\right]\right.$