Q84 / 160 · TS EAMCET 2005 · Mathematics

For a binomial variate $X$ with $n=6$, if $P(X=2)=9 P(X=4)$, then its variance is

  • A
    $\frac{8}{9}$
  • B
    $\frac{1}{4}$
  • Correct answer
    C
    $\frac{9}{8}$
  • D
    4

Solution

Given that $n=6$ and $\begin{aligned} & P(X=2)=9 P(X=4) \\ \Rightarrow \quad{ }^6 C_2 p^2 q^4 & =9 \cdot{ }^6 C_4 p^4 q^2 \end{aligned}$ $\begin{aligned} \Rightarrow \quad 9 p^2 & =q^2 \\ p & =\frac{1}{3} q \end{aligned}$ $\because$ We know that $p+q=1$ $\begin{aligned} & \Rightarrow \quad \frac{q}{3}+q=1 \\ & \Rightarrow \quad q=\frac{3}{4} \text { and } p=\frac{1}{4} \\ & \therefore \text { variance }=n p q=6 \cdot \frac{1}{4} \cdot \frac{3}{4}=\frac{9}{8} \end{aligned}$

Hence, the correct answer is option: 'C'.

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