Let A be the event that the absolute differen | vedclass

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(D) Let the two numbers be $x$ and $y$ where $0 \le x, y \le 60$. The total area of the sample space is $60 	imes 60 = 3600$.The condition is $|x - y

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(D) Let the two numbers be $x$ and $y$ where $0 \le x, y \le 60$. The total area of the sample space is $60 	imes 60 = 3600$.The condition is $|x - y| \le a$,which implies $-a \le x - y \le a$.The area of the region where $|x - y| > a$ is the sum of the areas of two right-angled triangles with legs $(60 - a)$.Area of region $|x - y| > a = \frac{1}{2}(60 - a)^2 + \frac{1}{2}(60 - a)^2 = (60 - a)^2$.Thus,the area of the region where $|x - y| \le a$ is $3600 - (60 - a)^2$.Given $P(A) = \frac{3600 - (60 - a)^2}{3600} = \frac{11}{36}$.Multiplying by $3600$,we get $3600 - (60 - a)^2 = 1100$.$(60 - a)^2 = 3600 - 1100 = 2500$.$60 - a = 50 \Rightarrow a = 10$.

Let $A$ be the event that the absolute difference between two randomly chosen 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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