Probability that a quadratic equation with random coefficients has two real solutions? In the following second order equation $ax^2+2bx+1.5=0$ where $a$ and $b$ are given by random points $(a,b)$ in the $[0,2]\times[0,1]$ rectangle, what is the probability of having two real solutions?
I'm a little lost here. I tried integrating $4b^2-6a$ with $a=0\to 2$ and $b=0\to 1$ as limits but the integral comes up negative.
I created a simulation of the problem using matlab and the probability is 0.11 but I want to find a way to solve it on paper and not with using matlab.
Any thoughts?
 A: The quadratic formula tells us this equation has two real solutions exactly when the discriminant $4b^2 - 6a$ is positive.  This describes a set of points $R$ in the $(a,b)$ plane, shown in blue here:

When the joint density of $(a,b)$ is $f$, then (by definition of pdf) the probability of any event (like $R$) is given by its integral $\int_R f(a,b)da\ db$.  Because the distribution is uniform, this is the same as finding the area of the shaded region as a proportion of the total area (equal to $2$), often written as a double integral like
$$\frac{1}{2}\int_0^1 \int_{a\lt 4b^2/6,\ 0\le a\le 2} da\ db.$$
However, we can reduce this to a single integral: the shaded region is bounded by the parabola $a = 2 b^2/3$ on the right, $a=0$ on the left, and extends from $b=0$ to $b=1$.  Its area therefore is $\int_0^1 2b^2/3\ db = 2/9$.  That amounts to $1/9 = 0.1111\ldots$ of the total area.

Edit (see comments).  In case this is unclear, we can proceed more formally.  The uniform distribution function $f$ is obtained by knowing (a) it is constant on the rectangle $[0,2]\times[0,1]$ and (b) is zero outside this set.  From (a) and the fact that any PDF must integrate to unity forces $f(a,b)=1/2$ inside the rectangle, whence
$$\eqalign{
f(a,b) = &1/2, &0 \le a \le 2, 0 \le b \le 1\\
         &0 &\text{otherwise}.
}$$
Integration is defined in terms of characteristic functions: the integral over an event $E$ with respect to a measure $d\mu$, written, $\int \cdots \int_E d\mu$, equals $\int \cdots \int I_E(x) d\mu(x)$ where the multiple integral is taken over all possible values of $x$ and $I_E(x) = 1$ when $x\in E$ and $I_E(x)=0$ otherwise.  The figure immediately shows that the solution is
$${\Pr}_f[(a,b)\in R] = \int \int_R f(a,b)\ da \ db$$
and the original double integral expression follows immediately from this expression by the definition of integration.  For more about this, consult any textbook on measure theory and integration or--for a less formal approach--consult any advanced calculus text that covers multiple integration.  End of edit.

This is essentially problem #50 from Fred Mosteller's Fifty Challenging Problems in Probability:

What is the probability that the quadratic equation $x^2 + 2bx + c = 0$ has real roots?

Solving it requires proposing some "reasonable" probability distribution for $(b,c)$.  Mosteller chooses a set of uniform distributions over a sequence of rectangles that grows without bound and takes the limit.
