Q270 / 270 · NDA 2016 (Phase 2) · Mathematics

Let the random variable $X$ follow $B(6, p)$. If $16 P(X=4)=$ $P(X=2)$, then what is the value of $p$ ?

  • A
    $\frac{1}{3}$
  • B
    $\frac{1}{4}$
  • Correct answer
    C
    $\frac{1}{5}$
  • D
    $\frac{1}{6}$

Solution

$\begin{aligned} & \text { X follows } B(6, p)=16 P(x=4)=P(x=2) \\ & \Rightarrow 16{ }^6 C_4 \cdot p^4(1-p)^{6-4}={ }^6 C_2 \cdot p^2(1-p)^{6-2} \\ & \Rightarrow 16 \times \frac{6!}{4!2!} p^4(1-p)^2=\frac{6!}{2!4!} p^2(1-p)^4 \\ & \Rightarrow 16 p^2=(1-p)^2 \Rightarrow 16 p^2=1+p^2-2 p \\ & \Rightarrow 15 p^2+2 p-1=0 \Rightarrow 15 p^2+5 p-3 p-1=0 \\ & \Rightarrow p=\frac{1}{5},-\frac{1}{3} \end{aligned}$ As $\mathrm{P} \gt 0 \Rightarrow \mathrm{P}=\frac{1}{5}$

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

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