Let $X,Y$ be random variables with joint density given by: $f(x,y) = c\mathbb{I}\{x^2+y^2\leq 1\}$ meaning uniformly distributed over the unit circle. First calculate $c$, then what is $\mathbb{P}(X^2 + Y^2 \leq \frac{1}{4})$.
Since the area of the unit circle is $\pi$ and all values are equally likely within, $c = \frac{1}{\pi}$.
$\mathbb{P}(X^2+Y^2 \leq \frac{1}{4})$ has an area of $\pi(\frac{1}{4})^2 = \frac{\pi}{16}$ and thus the probability of the area is $\frac{1}{16}$
No, not quite. The inequality $$X^2 + Y^2 \le 1/4$$ corresponds to a disk of radius $1/2$, because the equation for a circle is $$x^2 + y^2 = r^2$$ where $r$ is the radius. To see this another way, note that $X^2 + Y^2 = 1/4$ for $(X,Y) = (1/2, 0)$.
Therefore, the area of such a disk is $\pi(1/2)^2 = \pi/4$, not $\pi/16$.