Let A be the event that the absolute differen | vedclass

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Question

$|x-y| < a \Rightarrow-a < x-y < a$

$\Rightarrow x-y < a 	ext { and } x-y > -a$

$P ( A )=\frac{\operatorname{ar}( OACDEG )}{( OB

Vedclass · Accepted Answer

$10$

Vedclass · Answer

$|x-y| < a \Rightarrow-a < x-y < a$

$\Rightarrow x-y < a 	ext { and } x-y > -a$

$P ( A )=\frac{\operatorname{ar}( OACDEG )}{( OBDF )}$

$=\frac{\operatorname{ar}( OBDF )-\operatorname{ar}( ABC )-\operatorname{ar}( EFG )}{\operatorname{ar}( OBDF )}$

$\Rightarrow \frac{11}{36}=\frac{(60)^2-\frac{1}{2}(60- a )^2-\frac{1}{2}(60- a )^2}{3600}$

$\Rightarrow \quad 1100=3600-(60- a )^2$

$\Rightarrow \quad(60- a )^2=2500 \Rightarrow 60- a =50$

$\Rightarrow \quad a =10$

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 $...............$.

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