A random variable $\mathrm{X}$ has the following probability distribution…

A random variable $\mathrm{X}$ has the following probability distribution \begin{array}{|r|l|l|l|l|l|l|l|l|l|} \hline \mathrm{x} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline \mathrm{P}(\mathrm{X}=\mathrm{x}) & \mathrm{K} & 2 \mathrm{~K} & 3 \mathrm{~K} & 4 \mathrm{~K} & 4 \mathrm{~K} & 3 \mathrm{~K} & 2 \mathrm{~K} & \mathrm{~K} & \mathrm{~K} \\ \hline \end{array} Then $\mathrm{P}(3 < \mathrm{x} \leq 6)=$
  1. $\frac{3}{7}$
  2. $\frac{4}{7}$
  3. $\frac{13}{21}$
  4. $\frac{8}{21}$

Solution

We know that $\begin{aligned} & \mathrm{k}+2 \mathrm{k}+3 \mathrm{k}+4 \mathrm{k}+4 \mathrm{k}+3 \mathrm{k}+2 \mathrm{k}+\mathrm{k}+\mathrm{k}=1 \Rightarrow 21 \mathrm{k}=1 \\ & \Rightarrow \mathrm{k}=\frac{1}{21} \end{aligned}$ When $x=4, P=4 k=\frac{4}{21}$, When $x=5, P=3 k=\frac{3}{21}$, When $\mathrm{x}=6, \mathrm{P}=2 \mathrm{k}=\frac{2}{21}$ $\therefore \mathrm{P}(3 < \mathrm{x} \leq 6)==\frac{4+3+2}{21}=\frac{9}{21}=\frac{3}{7}$

Asked in: MHT CET 2021 (23 Sep Shift 2)

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