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14.Probability
hard
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 $...............$.
A
$100$
B
$0.1$
C
$15$
D
$10$
(JEE MAIN-2023)
Solution
$|x-y| < a \Rightarrow-a < x-y < a$
$\Rightarrow x-y < a \text { and } x-y > -a$
$P ( A )=\frac{\operatorname{ar}( OACDEG )}{( OBDF )}$
$=\frac{\operatorname{ar}( OBDF )-\operatorname{ar}( ABC )-\operatorname{ar}( EFG )}{\operatorname{ar}( OBDF )}$
$\Rightarrow \frac{11}{36}=\frac{(60)^2-\frac{1}{2}(60- a )^2-\frac{1}{2}(60- a )^2}{3600}$
$\Rightarrow \quad 1100=3600-(60- a )^2$
$\Rightarrow \quad(60- a )^2=2500 \Rightarrow 60- a =50$
$\Rightarrow \quad a =10$
Standard 11
Mathematics
