A random variable $X$ has the following. -probability distribution Then $\mathrm{P}(\mathrm{X}\gt2)$ is…

A random variable $X$ has the following. -probability distribution
Then $\mathrm{P}(\mathrm{X}\gt2)$ is equal to
  1. $\frac{7}{12}$
  2. $\frac{23}{36}$
  3. $\frac{1}{36}$
  4. $\frac{1}{6}$

Solution

$\begin{aligned} & \mathrm{k}^2+2 \mathrm{k}+\mathrm{k}+2 \mathrm{k}+5 \mathrm{k}^2=1 \\ & \therefore \quad 6 \mathrm{k}^2+5 \mathrm{k}=1 \\ & \therefore \quad 6 \mathrm{k}^2+5 \mathrm{k}-1=0 \\ & \therefore \quad 6 \mathrm{k}^2+6 \mathrm{k}-\mathrm{k}-1=0 \\ & \therefore \quad(6 k-1)(k+1)=0 \\ & \therefore \quad k=\frac{1}{6} \quad \ldots[\because k=-1 \text { is not possible }] \\ & \therefore \quad P(X\gt2)=1-P(X \leq 2) \\ & =1-\left(\mathrm{k}^2+2 \mathrm{k}\right) \\ & =1-\left(\frac{1}{36}+\frac{2}{6}\right) \\ & =1-\frac{13}{36} \\ & =\frac{23}{36} \end{aligned}$

Asked in: MHT CET 2024 (04 May Shift 1)

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