A random variable $x$ has the following probability distribution. Then value of $k$ is and $\mathrm{P}(3 \lt…

A random variable $x$ has the following probability distribution. Then value of $k$ is and $\mathrm{P}(3 \lt x \leq 6)$ has the value
  1. $\frac{1}{20}, \frac{3}{7}$
  2. $\frac{5}{21}, \frac{3}{7}$
  3. $\frac{1}{21}, \frac{3}{7}$
  4. $\frac{1}{20}, \frac{4}{7}$

Solution

$\begin{aligned} & \text {Since } \sum_{x=0}^8 \mathrm{P}(\mathrm{X}=x)=1 \\ & \therefore \quad \mathrm{k}+2 \mathrm{k}+3 \mathrm{k}+4 \mathrm{k}+4 \mathrm{k}+3 \mathrm{k}+2 \mathrm{k}+\mathrm{k}+\mathrm{k}=1 \\ & \Rightarrow \mathrm{k}=\frac{1}{21} \\ & \therefore \quad \mathrm{P}(3 \lt x \leq 6)=\mathrm{P}(\mathrm{X}=4)+\mathrm{P}(\mathrm{X}=5)+\mathrm{P}(\mathrm{X}=6) \\ &=\frac{4}{21}+\frac{3}{21}+\frac{2}{21} \\ &=\frac{9}{21}=\frac{3}{7}\end{aligned}$

Asked in: MHT CET 2024 (03 May Shift 1)

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