I've ran across this problem in my textbook, really don't know how to solve it.
Answer: .3973
What I've tried:
Since the rabbit's location is uniformly distributed, and depends on both an x and y coordinate I get f(x,y) = $\frac{1}{40*80}=\frac{1}{3200}$
The area of ellipse is $\pi(20)(30)$ = 1885
the area of the triangle is (.5)(40)(40) = 800
meaning the total prohibited area is 1885+800 = 2685.
This means that there's a $\frac{2685}{3200}$ probability of Jack the rabbit getting eating or hit with a sling shot. Or $\frac{3200-2685}{3200}$ = $\frac{515}{3200}$ = .161
I know this is clearly wrong, and the technique isn't the best. I'd assume the proper/best technique invokes double integrals. However, when I consider this I'm unsure how to set the limits of integration, without summing multiple double integrals.
$\int_0^{30}$ $\int_{0}^{20}$ $\frac{1}{3200}$ dydx to represent the area that's safe below the ellipses.
$\int_{30}^{40}$ $\int_{0}^{40}$ $\frac{1}{3200}$ dydx to represents the area that's safe between the ellipses and the triangle.
$\int_{40}^{80}$ $\int_{0}^{40}$ $\frac{1}{3200}$ dydx to represent the safe area above the triangle.
Computing these and summing these results in;
$\int_0^{30}$ $\int_{0}^{20}$ $\frac{1}{3200}$ dydx + $\int_{40}^{80}$ $\int_{0}^{40}$ $\frac{1}{3200}$ dydx + $\int_{40}^{80}$ $\int_{0}^{40}$ $\frac{1}{3200}$ dydx = $\frac{3}{16}$ + $\frac{1}{8}$ + $\frac{1}{2}$ = $\frac{13}{16}$ = .8125
I really have no idea how else to do this, but I'm assuming my limits of integration are way off.
The area of the entire ellipse is about $1885$, but only a quarter of the ellipse matters. $$\frac{\frac{1}{4} \pi \cdot 20 \cdot 30 + 800}{3200} \approx 0.3973. $$ (I think there's a mistake, though; 39% is the probability that the rabbit is not safe, not the probability that it is safe.)