Q88 / 150 · MHT CET 2024 (09 May Shift 1) · Mathematics

If the mean and the variance of Binomial variate $X$ are 2 and 1 respectively, then the probability that X takes a value greater than or equal to one is

  • A
    $\frac{1}{16}$
  • B
    $\frac{9}{16}$
  • C
    $\frac{3}{4}$
  • Correct answer
    D
    $\frac{15}{16}$

Solution

$\begin{aligned} & \text { Mean }=\mathrm{np}=2 \text { and variance }=\mathrm{npq}=1 \\ & \therefore \quad \mathrm{q}=\frac{1}{2} \end{aligned}$ $\begin{aligned} & \text { Also, } p=1-q=1-\frac{1}{2}=\frac{1}{2} \\ & \mathrm{n}=4 \end{aligned}$ $\begin{aligned} \therefore \quad P(X \geq 1) & =1-P(X=0) \\ & =1-{ }^4 C_0 p^0 q^4 \\ & =1-(1)\left(\frac{1}{2}\right)^0\left(\frac{1}{2}\right)^4 \\ & =1-\frac{1}{16} \\ & =\frac{15}{16} \end{aligned}$

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

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