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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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  • $\begingroup$ Not proposing to close, but just to link to the current choice of node for this same math question. In fact, this post here is good and came very close to being the node itself. $\endgroup$ – Lee David Chung Lin Feb 13 at 10:44
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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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  • $\begingroup$ Thank you for that, so for complex would the probability be 1 - 43/216 = 173/216 ? $\endgroup$ – Dexter Mar 6 '13 at 21:12
  • $\begingroup$ 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 $173/216$. $\endgroup$ – André Nicolas Mar 6 '13 at 21:30
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    $\begingroup$ Thank you for your help, I am able to finish this question now. $\endgroup$ – Dexter Mar 6 '13 at 21:32
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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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  • $\begingroup$ I am doing Introductory course in Probability & statistics, method you have suggested I have never heard of it. Sorry $\endgroup$ – Dexter Mar 6 '13 at 20:55
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The roots of the above quadratic equation will be real if the discriminant will be non-negative, $ i.e. $ if $$b^2-4ac\geq 0 \Rightarrow b^2\geq 4ac $$ It is clear that $ a,b,c\in \{1,2,3,4,5,6\} $ as they are determined by throwing the dice.
$ \therefore\ $ the total number of possible outcomes is $ 6\times 6 \times 6=216. $

Let us find the total no. of favorable cases for the above-required probability:

enter image description here

Since, $ b^2\geq4ac $ and since the maximum value of $ b^2 $ is 36, $ ac=10,11,12,\ldots,36 $ is not possible. So, the probability that the above equation has real root is $ \frac{43}{216} $ $ \therefore $ Required probability $= 1-\frac{43}{216} =\frac{173}{216}$

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