The probability distribution of a discrete random variable $\mathrm{X}$ is Find the value of $\mathrm{P}(2 <…

The probability distribution of a discrete random variable $\mathrm{X}$ is Find the value of $\mathrm{P}(2 < \mathrm{X} < 6)$
  1. $\frac{4}{21}$
  2. $\frac{1}{21}$
  3. $\frac{10}{21}$
  4. $\frac{4}{7}$

Solution

We have $\mathrm{K}+2 \mathrm{~K}+3 \mathrm{~K}+4 \mathrm{~K}+5 \mathrm{~K}+6 \mathrm{~K}=1 \Rightarrow \mathrm{K}=\frac{1}{21}$ $\begin{aligned} & \therefore \mathrm{P}(2 < \mathrm{x} < 6)=\mathrm{P}(\mathrm{x}=3)+\mathrm{P}(\mathrm{x}=4)+\mathrm{P}(\mathrm{x}=5) \\ & =\frac{3}{21}+\frac{4}{21}+\frac{5}{21}=\frac{12}{21}=\frac{4}{7} \end{aligned}$

Asked in: MHT CET 2021 (23 Sep Shift 1)

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