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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

#1
+2275
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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
+2275
+2

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