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I was doing some review on probability and came across the following exercise:

A quadratic equation $ax^2+bx+c=0$ is copied by a typist. However, the numbers standing for a, b and c are blurred and she can only see that they are integers of one digit. What is the probability that the equation she types has real roots?

Quadratic equations have real roots when the determinant is greater than or equal to $0$. Therefore $P(\text{real roots})=P(b^2-4ac\ge0)=P(b^2\ge4ac)$ My original thoughts were to find $1-P\left(\left(\frac{b}{2}\right)^2<4ac\right)$ where I would break it into cases where $b$ ranges from $0$ to $9$. I would then find out how many cases $4ac$ was larger than $b^2$ and divide it by $19^2*18$ (since $a\ne0$). When I found out it would be too tedious I thought of a possible geometric interpretation. Maybe the the probability could be expressed as ratios between volumes or something similar. When that seemed to over-complicate the problem I figured I was probably approaching it incorrectly.

Any hints or nudges in the general direction would be greatly appreciated.

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    P(b^2 \geq 4ac) = P(ac \leq 0) + P(b^2 \geq 4ac \, \vert ac >0) P(ac >0) might help you reduce the number of cases.2012-06-20

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In the case where $|4ac| > b^2$ the probability is 1/2 by changing the sign of $c$.

In the case where $|4ac| \leq b^2$ and $a \neq 0$, the probability is $1$.

In the case where $a=0$ one has to decide how to interpret the words "real roots". The simplest is to exclude equations of degree less than 2 (probability 1/19). If a single root of multiplicity one is allowed, or the equation $0=0$, then the answer will be different.

The main part of the problem is then to count the cases where $b^2 < 4|ac|$. If $|ac| > 20$ there is no constraint on $b$. There are a finite number of possibilities with $|ac| \leq 20$ and they can be accounted for by hand.

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    The denominator of 576 suggests that $b$ and $c$ may be required to be nonzero. For each value of $|ac| \leq 20$ the number of possible $b$ can be computed; add this up over the possible pairs $(a,c)$.2012-06-20