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

$ |x-y|< a \Rightarrow-a< x-y< a $
$ \Rightarrow x-y< a \text { and } x-y >-a$

$ P ( A )=\frac{\operatorname{ar}( OACDEG )}{( OBDF )} $
$ =\frac{\operatorname{ar}( OBDF )-\operatorname{ar}( ABC )-\operatorname{ar}( EFG )}{\operatorname{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$