Q83 / 160 · TS EAMCET 2007 · Mathematics

The mean and standard deviation of a binomial variate $X$ are 4 and $\sqrt{3}$ respectively. Then $P(X \geq 1)$ is equal to

  • A
    $1-\left(\frac{1}{4}\right)^{16}$
  • Correct answer
    B
    $1-\left(\frac{3}{4}\right)^{16}$
  • C
    $1-\left(\frac{2}{3}\right)^{16}$
  • D
    $1-\left(\frac{1}{3}\right)^{16}$

Solution

Mean $=n p=4$, variance $=n q p=3$ On solving, we get $q=\frac{3}{4}, n=16, p=\frac{1}{4}$ Now, $\begin{aligned} p(X \geq 1) & =1-p(X=0)=1-{ }^n C_0 p^0 q^{n-0} \\ & =1-\left(\frac{3}{4}\right)^{16}\end{aligned}$

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

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