For a binomial variate $X$ with $n=6$ if $P(X=4)=\frac{135}{2^{12}}$, then its variance is

For a binomial variate $X$ with $n=6$ if $P(X=4)=\frac{135}{2^{12}}$, then its variance is
  1. $\frac{8}{9}$
  2. $\frac{1}{4}$
  3. 4
  4. $\frac{9}{8}$

Solution

$\begin{aligned} & \text { Given, } P(X=4)=\frac{135}{2^{12}} \\ & \Rightarrow{ }^6 C_4 p^4 q^2=\frac{135}{2^{12}} \\ & \Rightarrow 15 p^4 q^2=\frac{135}{2^{12}} \\ & \Rightarrow p^4 q^2=\frac{3^2}{2^{12}} \\ & \Rightarrow p^2 q=\frac{3}{2^6} \\ & \Rightarrow p^2(1-p)=\frac{3}{64} \\ & \Rightarrow p^2(1-p)=\left(\frac{1}{4}\right)^2 \cdot\left(1-\frac{1}{4}\right) \\ & \Rightarrow p=\frac{1}{4} \\ & \text { and } q=1-\frac{1}{4}=\frac{3}{4} \\ & \text { Variance }=n p q \\ & \quad=6 \times \frac{1}{4} \times \frac{3}{4} \\ & =\frac{9}{8}\end{aligned}$

Asked in: MHT CET 2023 (10 May Shift 1)

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