# Let $x,y,z$ be independent uniform distribution on $(0,\pi)$. What's the probability that $z\leq \cos^2(x)\sin^2(y)$?

Let $x,y,z$ be independent uniformly distributed on $(0,\pi)$. What's the probability that $z\leq \cos^2(x)\sin^2(y)$?

I think applying trigonometry formula maybe useful. However I don't know how to deal with the problem.

Can anyone give a quick solution?

Thanks!

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What you need to do here is integrate the joint probability density function (which is simply the product of marginal densities here due to independence) over the event $\{z\leq\cos^2{(x)}\sin^2{(y)}\}$ which gives you: $$\int_0^{\pi}\int_0^{\pi}\int_{0}^{\cos^2{(x)}\sin^2{(y)}}\frac{1}{\pi^3}dzdydx=\int_0^{\pi}\int_0^{\pi}\frac{\cos^2{(x)}\sin^2{(y)}}{\pi^3}dydx$$ $$=\frac{1}{\pi^3}\int_0^{\pi}\cos^2{(x)}dx\int_0^{\pi}\sin^2{(y)}dy$$
Now we note that those integrals have the same value (draw a picture) and so $$\int_0^{\pi}\cos^2{(x)}dx=\frac{1}{2}(\int_0^{\pi}\cos^2{(x)}dx+\int_0^{\pi}\sin^2{(x)}dx)$$ $$=\frac{1}{2}\int_0^{\pi}(\cos^2{(x)}+\sin^2{(x)})dx=\frac{\pi}{2}$$ to give a final answer of $\frac{1}{4\pi}$.