A random variable has the following probability distribution The then value of $p$ is

A random variable has the following probability distribution
The then value of $p$ is
  1. $\frac{1}{10}$
  2. $\frac{1}{30}$
  3. $\frac{1}{100}$
  4. $\frac{3}{20}$

Solution

$\begin{array}{ll} & \text { Here, } \\ & 0+2 p+2 p+3 p+p^2+2 p^2+7 p^2+2 p=1 \\ \therefore \quad & 9 p+10 p^2=1 \\ \therefore \quad & 10 p^2+9 p-1=0 \\ \therefore \quad & 10 p^2+10 p-p-1=0 \\ \therefore \quad & 10 p(p+1)-1(p+1)=0 \\ \therefore \quad & p=\frac{1}{10} \text { or } p=-1 \\ \quad & \text { But } 0 \leq p \leq 1 \\ \therefore \quad & p=\frac{1}{10}\end{array}$

Asked in: MHT CET 2024 (04 May Shift 1)

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