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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 6? Express your answer as a common fraction in terms of pi.

 Aug 5, 2022

Best Answer 

 #1
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The success area is 1/4th of a circle with a radius of \(\sqrt 6\), so the area of this is \((\sqrt{6})^2 \pi \div 4 = {3 \pi \over 2}\)

 

The total region is \(2 \times 2 = 4\)

 

So, the probability is \({{3 \pi \over 2} \over 4} = \color{brown}\boxed{3 \pi \over 8}\)

 Aug 6, 2022
 #1
avatar+2448 
0
Best Answer

The success area is 1/4th of a circle with a radius of \(\sqrt 6\), so the area of this is \((\sqrt{6})^2 \pi \div 4 = {3 \pi \over 2}\)

 

The total region is \(2 \times 2 = 4\)

 

So, the probability is \({{3 \pi \over 2} \over 4} = \color{brown}\boxed{3 \pi \over 8}\)

BuilderBoi Aug 6, 2022

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