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Two real numbers are chosen at random between 0 and 2. What is the probability that the sum of their squares is no more than 3? Express your answer as a common fraction in terms of pi.

 Jul 4, 2022

Best Answer 

 #1
avatar+2293 
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The area of the sucsesses is a quarter of a circle with a radius of \(\sqrt 3\), so the area is \(\pi \sqrt{3}^2 \times {1 \over 4} = {3 \over 4} \pi\)

 

The area of the total region is \(2 \times 2 = 4\)

 

So, the probability is \({{{3 \over 4} \pi }\over 4 } = \color{brown}\boxed{3 \pi \over 16}\)

 

Here is an image: 

 

 Jul 4, 2022
 #1
avatar+2293 
+1
Best Answer

The area of the sucsesses is a quarter of a circle with a radius of \(\sqrt 3\), so the area is \(\pi \sqrt{3}^2 \times {1 \over 4} = {3 \over 4} \pi\)

 

The area of the total region is \(2 \times 2 = 4\)

 

So, the probability is \({{{3 \over 4} \pi }\over 4 } = \color{brown}\boxed{3 \pi \over 16}\)

 

Here is an image: 

 

BuilderBoi Jul 4, 2022

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