Probability of $x^2 \ge 4$ in uniform distribution

Question

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?

$$P(b^2 \ge 4)=\int_{-3}^{3}(b^2-4)db $$

Answer 1

Actually, you need to restate $b^2\ge 4$ in terms of $b$. Then, you can use the integral with the density of $b$. $$P(b^2\ge 4)=P(|b|\ge 2)=P(b\le -2)+P(b\ge 2)=\int_{-3}^{-2}f_B(b)db+\int_{2}^3f_B(b)db$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}