Probability for all pairs $(x,y)$ with $x,y\in{[0,1]}$ that $\theta= \cos^{-1}\frac{-y^2+y-x^2}{\sqrt{x^2+(y-1)^2}\sqrt{x^2+y^2}} < 90^\circ$

Question

As the title says, the probability for all pairs $(x,y)$ with $x,y\in[0,1]$ that $$\theta= \cos^{-1}\left(\frac{-y^2+y-x^2}{\sqrt{x^2+(y-1)^2}\sqrt{x^2+y^2}}\right)<90^\circ$$. Or, phrased differently, the conditions on $x$ and $y$ such that $\theta<90^\circ$. By nature of the inverse cosine function I know that both the numerator and denumerator need to be greater than zero and I know the probability will involve $\pi$ somehow... Note that the distribution of $x$ and $y$ is random and occurs with equal chance. Thank you for your contributions.

Answer 1

The numerator $-y^2+y−x^2$ is positive inside a circle centered at $(0,1/2)$, with radius $1/2$. Hence your probability is half the area of that circle: $\pi/8$.

Answer 2

Note that the denominator is always positive, so we simply need to find when the numerator will be positive. Thus, we let $-y^2+y-x^2 > 0$, and since $x$ is non-negative and $\max(\sqrt{y-y^2})=1/2<1$, we can just solve for it to get $0 \leq x < \sqrt{y-y^2}\leq 1$. Thus we have that the probability is $\frac{\mu_{success}}{\mu_{total}} = \frac{\mu(\{(x,y)\in\mathbb{R}^2|0\leq x < \sqrt{y-y^2}\})}{\mu(\{(x,y)\in\mathbb{R}^2|0\leq x,y \leq 1\})} = \frac{\int_0^1\int_0^{\sqrt{y-y^2}}dxdy}{\int_0^1\int_0^1dxdy} = \pi/8.$

See Aretino's answer for the geometric view, noting that $y-y^2 = (\frac{1}{2})^2-(\frac{1}{2}-y)^2$.