A random variable $\mathrm{X}$ has the following probability distribution Then $F(4)=$

A random variable $\mathrm{X}$ has the following probability distribution Then $F(4)=$
  1. $\frac{3}{10}$
  2. $\frac{1}{10}$
  3. $\frac{7}{10}$
  4. $\frac{4}{5}$

Solution

$\begin{aligned} & \text { Here } \mathrm{k}+2 \mathrm{k}+2 \mathrm{k}+3 \mathrm{k}+\mathrm{k}^2+2 \mathrm{k}^2+7 \mathrm{k}^2+\mathrm{k}=1 \\ & \therefore 10 \mathrm{k}^2+9 \mathrm{k}-1=0 \Rightarrow(10 \mathrm{k}-1)(\mathrm{k}+1)=0 \\ & \therefore \mathrm{k}=\frac{1}{10} \quad \ldots(\mathrm{k} \geq 0) \\ & \therefore \mathrm{F}(4)=\mathrm{k}+2 \mathrm{k}+2 \mathrm{k}+3 \mathrm{k}=8 \mathrm{k}=\frac{8}{10}=\frac{4}{5}\end{aligned}$

Asked in: MHT CET 2021 (20 Sep Shift 1)

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