Q79 / 160 · AP EAMCET 2024 (20 May Shift 2) · Mathematics

For a binomial variate $X \sim B(n, p)$ the difference between the mean and variance is 1 and the difference between their squares is 11 . If the probability of $P(x=2)=m\left(\frac{5}{6}\right)^n$ and $n=36$ then $m: n=$

  • A
    $6: 5$
  • Correct answer
    B
    $7: 10$
  • C
    $36: 1$
  • D
    $42: 25$

Solution

Given that difference of mean variance is 1 $\therefore n p-n p q=1 \Rightarrow n p(1-q)=1 \Rightarrow n p^2=1$ $\qquad ....\mathrm{(i)}$ Also given $(n p)^2-(n p q)^2=11$ $\qquad ....\mathrm{(ii)}$ $n^2 p^2-n^2 p^2 q^2=11 \Rightarrow n^2 p^2\left(1-q^2\right)=11$ Divide (ii) by (i), we get $\begin{aligned} & \frac{n^2 p^2\left(1-q^2\right)}{n p^2}=11 \\ & \Rightarrow n\left(1-q^2\right)=11 \Rightarrow 1-q^2=\frac{11}{36} \qquad (\because n=36) \\ & \Rightarrow q=\frac{5}{6} \Rightarrow p=\frac{1}{6} \end{aligned}$
Now, $p(x=2)=m\left(\frac{5}{6}\right)^n$ $\begin{aligned} & { }^{36} C_2\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^{34}=m\left(\frac{5}{6}\right)^{36} \\ & \Rightarrow \frac{36 \times 35}{2} \times\left(\frac{1}{6}\right)^2 \times\left(\frac{5}{6}\right)^{34}=m\left(\frac{5}{6}\right)^{36} \\ & \Rightarrow \frac{36 \times 7}{10} \times \frac{5}{6} \times \frac{5}{6} \times\left(\frac{5}{6}\right)^{34}=m\left(\frac{5}{6}\right)^{36} \\ & =\frac{36 \times 7}{10} \times\left(\frac{5}{6}\right)^{36}=m\left(\frac{5}{6}\right)^{36} \therefore m=\frac{36 \times 7}{10} \end{aligned}$
So, $m: n=\frac{36 \times 7}{10}: 36=7: 10$

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

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