Answer:
[tex]31.8\%[/tex]
Step-by-step explanation:
The area of the circle is [tex]A=\pi r^2=\pi(4)^2=16\pi[/tex]
The area of the triangle is [tex]A=\frac{bh}{2}=\frac{8*4}{2}=\frac{32}{2}=16[/tex]
Hence, the probability of a randomly selected point within the circle falls in the red shaded area is [tex]\frac{16}{16\pi}=\frac{1}{\pi}\approx0.318\approx31.8\%[/tex]
