Q80 / 160 · TS EAMCET 2024 (10 May Shift 2) · Mathematics

If $\mathrm{X} \sim B(6, p)$ is a binomial variate and $\frac{\mathrm{P}(\mathrm{X}=4)}{\mathrm{P}(\mathrm{X}=2)}=\frac{1}{9}$, then $p=$

  • A
    $\frac{1}{2}$
  • B
    $\frac{1}{9}$
  • C
    $\frac{1}{3}$
  • Correct answer
    D
    $\frac{1}{4}$

Solution

$\mathrm{X} \sim \mathcal{B}(\sigma, p)$ is binomial variate $\Rightarrow n=6 ; \mathbb{P}(\mathbb{X}=x)={ }^6 \mathrm{C}_x p^x q^{6-x}$ $\begin{aligned} & \frac{\mathrm{P}(\mathrm{X}=4)}{\mathrm{P}(\mathrm{X}=2)}=\frac{1}{9} \Rightarrow \frac{{ }^6 \mathrm{C}_4 p^4 q^2}{{ }^6 \mathrm{C}_2 p^2 q^4}=\frac{1}{9} \\ & \Rightarrow 9 p^2=q^2 \Rightarrow 9 p^2=(\mathbb{1}-p)^2 \Rightarrow 8 p^2+2 p-1=0 \\ & \Rightarrow p=\frac{-2 \pm \sqrt{4+32}}{16}\end{aligned}$ $p=-\frac{1}{2}, \frac{1}{4} \Rightarrow p=\frac{1}{4} \quad[\because p$ cannot be negative $]$

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

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