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Question :

The coefficients a,b,c of the quadratic equation $ax^2+bx+c=0$ are determined by throwing a die three times and reading off the value shown on the uppermost face of each die. i.e, if you throw a 1, 5 and 3 respectively, the equation is $1x^2+5x+3=0$.

Find the probabilities that the quadratic equation you obtain :

  1. has real roots;
  2. has complex roots;
  3. has equal roots.

Thank you for your attention.

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2 Answers 2

up vote 2 down vote accepted

We show the path to the answers using somewhat tedious listing. The symmetry between $a$ and $c$ could be used to cut down on the work. We use the fact that the quadratic has real solutions if and only if the discriminant $b^2-4ac$ is $\ge 0$, and equal solutions if and only if $b^2-4ac=0$.

Equality is easiest. This happens if and only if $b^2=4ac$. That forces $b=2$, $4$, or $6$. If $b=2$, we need $a=c=1$, so the only configuration is $(1,2,1)$. If $b=4$, we want $ac=4$, which can happen in $3$ ways, $(1,4,4)$, $(4,4,1)$, and $(2,4,2)$. Finally, if $b=6$, we want $ac=9$, which only happens with the configuration $(3,6,3)$. Each configuration has probability $\frac{1}{6^3}$, so the required probability is $\frac{5}{216}$.

For real solutions , we want $b^2\ge 4ac$. That cannot happen if $b=1$. If $b=2$, it can only happen if $ac=1$, giving a contribution of $\frac{1}{216}$. If $b=3$, we want $ac\le 2$, which can happen in $3$ ways, for a contribution of $\frac{3}{216}$.

We leave the cases $b=4$ and $b=5$ to you. For $b=6$, we want $ac\le 9$. Let us list the ways. With $a=1$, $b$ can have $6$ values. With $a=2$ there are $4$. With $a=3$ there are $3$. With $a=4$ there are $2$. And there are $1$ each for $a=7$ and $a=6$. That gives a contribution of $\frac{17}{216}$.

For complex, one could say that the probability is $1$, since every real number is in particular a complex number. But what is probably intended is complex and non-real. Then the required probability is $1$ minus the probability the root(s) are real.

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Thank you for that, so for complex would the probability be 1 - 43/216 = 173/216 ? –  Dexter Mar 6 '13 at 21:12
I also got $43/216$ for real. Slippage in this sort of thing is possible, one has to write down possibilities neatly and systematically. For a larger problem, I would look for shortcuts, but here it is not really worthwhile. Yes, for complex non-real, probability is $173/216$. –  André Nicolas Mar 6 '13 at 21:30
Thank you for your help, I am able to finish this question now. –  Dexter Mar 6 '13 at 21:32

Hint: Given the discrete random coefficients, $a, b, c \sim U(1, 6)$, the discriminant of the quadratic equation, $\displaystyle \Delta = b^2 -4ac$, determines whether or not you get real or complex values and whether or not they are distinct. If $\Delta > 0$ then x is a distinct real with two values., $\Delta = 0$, then x is real with one distinct value, otherwise x is complex with two distinct values.

Thus you are looking for $\mathbb{P}(\Delta < 0)$, $\mathbb{P}(\Delta = 0)$ and $\mathbb{P}(\Delta > 0)$. Which can be determined by the method of moment generating functions.

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I am doing Introductory course in Probability & statistics, method you have suggested I have never heard of it. Sorry –  Dexter Mar 6 '13 at 20:55

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