A random variable $\mathrm{X}$ has following distribution $ \begin{array}{|l|l|l|l|l|l|l|} \hline…

A random variable $\mathrm{X}$ has following distribution $ \begin{array}{|l|l|l|l|l|l|l|} \hline \mathrm{X}=\mathrm{x} & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \mathrm{P}(\mathrm{X}=\mathrm{x}) & \mathrm{k} & 3 \mathrm{k} & 5 \mathrm{k} & 7 \mathrm{k} & 8 \mathrm{k} & \mathrm{K} \\ \hline \end{array} $ Then $\mathrm{P}(2 \leq \mathrm{x} < 5)=$
  1. $\frac{7}{25}$
  2. $\frac{3}{5}$
  3. $\frac{24}{25}$
  4. $\frac{23}{25}$

Solution

We have $\mathrm{k}+3 \mathrm{k}+5 \mathrm{k}+7 \mathrm{x}+8 \mathrm{k}+\mathrm{k}=1 \Rightarrow \mathrm{k}=\frac{1}{25}$ $\therefore \mathrm{P}(2 \leq \mathrm{x} \leq 5)=\frac{1}{25}(3+5+7)=\frac{15}{25}=\frac{3}{5}$

Asked in: MHT CET 2021 (22 Sep Shift 1)

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