If $Q$~UNIF$(0,3)$, find the probability that the roots of the equation $g(t)=0$ are real, where $g(t)=4t^2+4Qt+Q+2$.
There was a similar question asked that I looked at, but I am still a little confused.
I know that real roots imply $b^2-4ac \ge 0$. So, we have $a=4, b=4Q,$ and $c=Q+2$.
This gave me:
$$16(Q-2)(Q+1)\ge 0$$
$$Q\ge 2, Q \le -1$$
From here, I'm not sure how to find the probability that the roots are real. $Q\le -1$ isn't defined on $(0,3)$, so we are just looking at $Q\ge 2$? Because it is continuous that means we don't have to distinguish between $\ge$ and $\gt$.
$$P(Q\gt 2)=1-(Q=2)$$
From the CDF of $X$~UNIF$(a,b)$:
$$F(x:a,b) =\begin{cases}0 & x\le a \\ \frac{x-a}{b-a} & {a\lt x \lt b} \\\ 1 & b\le x \\\end{cases}$$
Is it as simple as plugging in our $x,a,$ and $b$?
$$\frac{2-0}{3-0}=\frac23$$
$$P(Q\ge 2)=1-\frac23=\frac13?$$
I've never encountered finding a probability like this and wanted to be sure I was doing it right.
Best Answer
It seems just as simple as that. The event whose probability you are to find (the equation has real roots) is equivalent to $Q \le -1 \vee Q \ge 2$, and since $P(Q \le -1) = 0$ for the specified distribution, the answer is $P(Q \ge 2)$, which computes to $\frac{1}{3}$ using exactly the method you described.