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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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1 Answer 1

up vote 2 down vote accepted

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 $\frac{1}{4\pi}$.

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