Q47 / 160 · TS EAMCET 2015 · Mathematics

Let $X$ be binomial variate with parameters $n=6$ and $p$. If $4 P(X=4)=P(X=2)$, then $p$ equals

  • A
    $1 / 2$
  • Correct answer
    B
    $1 / 3$
  • C
    $1 / 4$
  • D
    $1 / 6$

Solution

We have, $n=6$. It is given that, $$ \begin{aligned} & 4 P(X=4)=P(X=2) \\ & \Rightarrow \quad 4 \cdot{ }^6 C_4 p^4 q^2={ }^6 C_2 p^2 q^4 \\ & {\left[\because P(X=r)={ }^n C_r p^r q^{n-r}\right]} \\ & \Rightarrow \quad 4 \times \frac{6 \times 5}{2 \times 1} p^2=\frac{6 \times 5}{2 \times 1} q^2 \\ & \Rightarrow \quad 4 p^2=q^2 \\ & \Rightarrow \quad 2 p=q \\ & \Rightarrow \quad 2 p=1-p \quad[\because q=1-p] \\ & \Rightarrow \quad p=\frac{1}{3} \\ & \end{aligned} $$

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

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