A random variable $X$ has the following probability distribution Then $\mathrm{p}(x \geq 2)$ is equal to

A random variable $X$ has the following probability distribution
Then $\mathrm{p}(x \geq 2)$ is equal to
  1. $\frac{35}{36}$
  2. $\frac{34}{36}$
  3. $\frac{33}{36}$
  4. $\frac{31}{36}$

Solution

$\begin{aligned} & \text { Since } \sum_{x=1}^5 \mathrm{P}(\mathrm{X}=x)=1 \\ & \mathrm{k}^2+2 \mathrm{k}+\mathrm{k}+2 \mathrm{k}+5 \mathrm{k}^2=1 \\ & \Rightarrow 6 \mathrm{k}^2+5 \mathrm{k}-1=0 \\ & \Rightarrow(\mathrm{k}+1)(6 \mathrm{k}-1)=0 \\ & \Rightarrow \mathrm{k}=\frac{1}{6} \quad \ldots[\because \mathrm{k} \geq 0]\end{aligned}$ $\begin{aligned} & \mathrm{P}(\mathrm{X} \geq 2)=\mathrm{P}(\mathrm{X}=2)+\mathrm{P}(\mathrm{X}=3)+\mathrm{P}(\mathrm{X}=4) \\ &+\mathrm{P}(\mathrm{X}=5)\end{aligned}$ $\begin{aligned} & =\frac{2}{6}+\frac{1}{6}+\frac{2}{6}+\frac{5}{36} \\ & =\frac{12+6+12+5}{36} \\ & =\frac{35}{36}\end{aligned}$

Asked in: MHT CET 2024 (15 May Shift 1)

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