I want to check that there are not any glaring issues with my solution to the following question:
Let $Y$ be a random variable uniformly distributed over $(0, 3)$. Consider the following polynomial equation in $x$ with random coefficients: $x^{2} + 2Yx + 4Y − 3$.
(a) What is the probability that the product of the roots of the equation is positive?
(b) What is the probability that the equation has real roots?
(a) I plugged the values of the coefficients into the quadratic formula to get the roots of the equation in terms of Y. Multiplying them together I get:
\begin{align*}
\frac{-2Y+\sqrt{(2Y)^{2}-4(4Y-3)}}{2}\cdot\frac{-2Y-\sqrt{(2Y)^{2}-4(4Y-3)}}{2} = 4Y-3
\end{align*}
And so $4Y-3 > 0$ when $Y > 3/4$. So the probability the product of our polynomial's roots is positive is the following: $\int_{\frac{3}{4}}^{3} \frac{1}{3} dy=(1-(1/4))=3/4$.
(b) The equation has real roots when $4Y^{2}-4(4Y-3) \geq 0$. In other words, when $Y^{2}-4Y+3 \geq 0$. The roots of this quadratic polynomial are 1 and 3, and the polynomial is greater than or equal to zero on the intervals $(-\infty,1]$ and $[3, \infty)$. So the probability that our original polynomial has real roots is:
$\int_{0}^{1} \frac{1}{3} dy+\int_{3}^{3} \frac{1}{3} dy=1/3$