Probability of $x^2 \ge 4$ in uniform distribution I want to find the probability of $x^2+bx+1=0$ that has at least one real root. Also, $b$ is a uniform random variable on the interval $[-3,3]$.
I know the condition for this quadratic equation to have real roots is $b^2 \ge 4$.
The question is should I calculate the below integral to find the probability?
$$P(b^2 \ge 4)=\int_{-3}^{3}(b^2-4)db $$
 A: Actually, you need to restate $b^2\ge 4$ in terms of $b$. Then, you can use the integral with the density of $b$. 
$$P(b^2\ge 4)=P(|b|\ge 2)=P(b\le -2)+P(b\ge 2)=\int_{-3}^{-2}f_B(b)db+\int_{2}^3f_B(b)db$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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\begin{align}
\int_{-3}^{3}{1 \over 6}\bracks{b^{2} > 4}\dd b & =
{1 \over 3}\int_{0}^{3}\bracks{b^{2} > 4}\dd b =
{1 \over 3}\int_{0}^{3}\bracks{b > 2}\dd b
\\[5mm] & =
{1 \over 3}\int_{2}^{3}\dd b = \bbx{1 \over 3}
\end{align}
