JEE Main 2023 — Probability Question with Solution

Given,

$A$ be the event that the absolute difference between two randomly chosen real numbers in the sample space $\left[0,60\right]$ is less than or equal to $a$,

So, $\left|x-y\right|<a\Rightarrow -a<x-y<a$

$\Rightarrow x-y<a$ and $x-y>-a$

Now on plotting the diagram we get,

Now from diagram probability is given by,

$\mathrm{P}\left(\mathrm{A}\right)=\frac{\text{area}\left(\mathrm{OACDEG}\right)}{\text{area}\left(\mathrm{OBDF}\right)}$

$\Rightarrow \mathrm{P}\left(\mathrm{A}\right)=\frac{\text{area}\left(\mathrm{OBDF}\right)-\text{area}\left(\mathrm{ABC}\right)-\text{area}\left(\mathrm{EFG}\right)}{\text{area}\left(\mathrm{OBDF}\right)}$

$\Rightarrow \frac{11}{36}=\frac{{\left(60\right)}^2-\frac{1}{2}{\left(60-a\right)}^2-\frac{1}{2}{\left(60-a\right)}^2}{3600}$

$\Rightarrow 1100=3600-{\left(60-a\right)}^2$

$\Rightarrow {\left(60-a\right)}^2=2500\Rightarrow 60-a=50$

$\Rightarrow a=10$