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 .If then is equal to ______ . [2023]

Question

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 ______ . [2023]

Smart Achievers · Accepted Answer

(10)

$| x - y | \leq a$

$\Rightarrow - a \leq x - y \leq a \Rightarrow x - y \leq a and x - y \geq - a$

$P (A) = \frac{ar (O A C D E G)}{ar (O B D F)} = \frac{ar (O B D F) - ar (∆ A B C) - ar (∆ E F G)}{ar (O B D F)}$

$\Rightarrow \frac{11}{36} = \frac{60^{2} - \frac{1}{2} {(60 - a)}^{2} - \frac{1}{2} {(60 - a)}^{2}}{3600}$

$\Rightarrow 1100 = 3600 - {(60 - a)}^{2}$

$\Rightarrow {(60 - a)}^{2} = 2500 \Rightarrow 60 - a = 50 \Rightarrow a = 10$

Solution

Q.
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 ______ . [2023]

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Solution

Q. 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)=1136, then a is equal to ______ . [2023]

Ans. (10) |x-y|≤a ⇒ -a≤x-y≤a ⇒ x-y≤a and x-y≥-a P(A)=ar(OACDEG)ar(OBDF)=ar(OBDF)-ar(∆ABC)-ar(∆EFG)ar(OBDF) ⇒ 1136=602-12(60-a)2-12(60-a)23600 ⇒ 1100=3600-(60-a)2 ⇒ (60-a)2=2500 ⇒ 60-a=50 ⇒ a=10

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Q.
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 ______ . [2023]