Q80 / 160 · TS EAMCET 2024 (10 May Shift 1) · Mathematics

The mean of a binomial variate $X \sim B(n, p)$ is 1 . If $n\gt2$ and $P(X=2)=\frac{27}{128}$, then the variance of the distribution is

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

Solution

Mean $=n p=1$ (given), $p+q=1$, Variance $=n p q$ $n\gt2$ Check for $n=3, \quad p=\frac{1}{3}, q=\frac{2}{3}$ $P(X=2)=3 C_2(p)^2 \times(q)^1=3 \times \frac{1}{9} \times \frac{2}{3}=\frac{2}{9} \neq \frac{27}{128}$ $n=3$ (Not possible) Check for $n=4 \Rightarrow p=\frac{1}{4}, q=\frac{3}{4}$ $P(X=2)=4 C_2 \times(p)^2 \times(q)^2=6 \times \frac{1}{16} \times \frac{9}{16}=\frac{27}{128}$ $\therefore$ So for $n=4$, variance $=n p q=4 \times \frac{1}{4} \times \frac{3}{4}=\frac{3}{4}$.

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

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