<s>[INST] Probability, Uniform Distribution. Suppose $A$ is a random number chosen uniformly from the interval $(-5,6)$,what is the probability that the quadratic equation $x^2 + Ax + 1 = 0$, has no real root?
This is my approach, 
Using the quadratic equation we know,
$b^2 - 4ac < 0$ (has no real roots)
Therefore solving we get $A^2 - 4 < 0$,
hence $-2 < A < 2$ 
Now we know the $p.d.f$ of a uniform distribution is as follows. 
Therefore finally we get, 
$P(-2<x<2)= 1/(6+5)$ or $1/11$
Now knowing this is a uniform distribution we can convert the above into a $c.d.f$ 
using integration therefore arriving at the formula on Wikipedia. However since it's uniform distribution why can't we just multiply $1/11$ by $5$ ? Since $[-2,2]$ has $5$ distinct values. Using the formula for $c.d.f$ on Wikipedia I get $4/11$.
 [/INST] The first part looks fine - $x^2 + Ax + 1 = 0$ having no real solution means $(x + A/2)^2 = A^2/4 - 1$ < 0, which implies $|A| < 2$. 
But then you're done! Since $A$ is uniformy distributed on $(-5,6)$, and since $(-2,2)$ lies fully within that interval, the probabily is simply the ratio of the interval lengths, i.e. $$
  P(A \in [-2,2]) = \frac{2 - (-2)}{6 - (-5)} = \frac{4}{11} \text{.}
$$
You can also derive that using the CDF, of course. The CDF of the uniform distribution on $(-5,6)$ is $$
  F(x) = \begin{cases}
    0 &\text{if $x < -5$} \\
    \frac{x+5}{11} &\text{if $x \in [-5,6]$} \\
    1 &\text{if $x \geq 6$.} \\
  \end{cases}
$$
Therefore, $$\begin{eqnarray}
  P(A \in (-2,2)) &=&  P((X \leq 2) \setminus (X < -2)) = P(X \leq 2) - P(X \leq -2) \\
  &=& \frac{2+5}{11} - \frac{-2+5}{11} = \frac{4}{11} \text{.}
\end{eqnarray}$$
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