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.
+2666 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:
+2666 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:
