<s>[INST] What is the probability that the quadratic equation $ax^2+x+1=0$ has two real roots? A number $a$ is chosen at random within the interval $(-1, 1)$. What is the probability that the quadratic equation $ax^2+x+1=0$ has two real roots?
For it to have its real roots, we must guarantee that $1-4a \geq 0$, or $a\leq \frac{1}{4}$.
It is no longer clear to me what I have to do.
 [/INST] $$a \in(-1,1) \land a \leq \frac{1}{4} \iff a \in(-1,\frac{1}{4}]$$
because $(-1,1) \cup (-\infty,\frac{1}{4}] = (-1,\frac{1}{4}]$
We can therefore, divide the length of both intervals to get the probability:
$$P(a \in(-1,\frac{1}{4}]) = \frac{|\frac{1}{4} - \left(-1\right)|}{|1 - \left(-1\right)|} = \frac{\frac{5}{4}}{2} = \frac{5}{8}$$
The fact that $(-1,\frac{1}{4}]$ doesn't include $-1$ and $\left(-1,1\right)$ doesn't include either of its endpoints doesn't make a difference because they are single points that can be considered insignificant compared to the infinite number of reals in the intervals.
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