We know that probability of having $ \sin (a x)>0$ for a random $x$ is $\frac{1}{2}$.
Can we say something about the probability of the following condition? $$-\frac{1}{4}\leq \sin (a x)\leq \frac{1}{2}$$
Here, $a$ and $x$ are continuous and $x>0$ and $a>0$.
The function $$x\mapsto\sin(a x)\qquad(-\infty<x<\infty)$$ is periodic with period $T:={2\pi\over a}$. When we assume that $x$ is uniformly distributed modulo $T$ then we may assume as well that $a=1$, and that $x$ is uniformly distributed modulo $2\pi$.
Draw the sin curve in the $x$-interval $[0,2\pi]$ representing a full period, and measure the total length $L$ of the subintervals where $-{1\over4}<\sin x<{1\over2}$. The probability $p$ you are after then is ${L\over2\pi}$. In this way you obtain $$p={L\over2\pi}={2\arcsin{1\over2}+2\arcsin{1\over4}\over2\pi}=0.247097\ .$$