Question

Let $A$ be the event that the absolute difference between two randomly choosen real numbers in the sample space $[0,60]$ is less than or equal to a If $P(A)=\frac{11}{36}$, then a is equal to _____

Answer 1

Correct Answer: 27

The correct answer is 50.
$|x-y|<a\Rightarrow -a<x-y<a$
$\Rightarrow x-y<a\text{and}x-y>-a$

$P(A)=\frac{\mathrm{ar}(OACDEG)}{(OBDF)}$
$=\frac{\mathrm{ar}(OBDF)-\mathrm{ar}(ABC)-\mathrm{ar}(EFG)}{\mathrm{ar}(OBDF)}$
$\Rightarrow \frac{11}{36}=\frac{(60{)}^2-\frac{1}{2}(60-a{)}^2-\frac{1}{2}(60-a{)}^2}{3600}$
$\Rightarrow 1100=3600-(60-a{)}^2$
$\Rightarrow (60-a{)}^2=2500\Rightarrow 60-a=50$
$\Rightarrow a=10$