Suppose that I roll $3$ dice and write down the outcome as the coefficients $a,b,c$ in the polynomial $ax^2+bx+c$ respectively. What is the probability that this polynomial has no real roots?
So, I have to count the number of triples $(a,b,c)$ such that $b^2 < 4ac$, where $a,b,c \in \{1,2,3,4,5,6\}$. I'm not sure how I can do that. Please give me a hint first. This problem is from a high school probability course, so I think it must have a very basic solution.
On
Following up on Berci's answer:
You can count it out simply and arrive at the answer: $\frac{89}{108}$.
Added later: I had made an error in my counting! The answer is $\frac{43}{6^3}$
Here are the 43 possibilities: [[1,2,1,0],[1,3,1,5],[1,4,1,12],[1,5,1,21],[1,6,1,32],[ 1,3,2,1],[1,4,2,8],[1,5,2,17],[1,6,2,28],[1,4,3,4],[1,5,3,13], [1,6,3,24],[1,4,4,0],[1,5,4,9],[1,6,4,20],[1,5,5,5],[1,6,5,16] ,[1,5,6,1],[1,6,6,12],[2,3,1,1],[2,4,1,8],[2,5,1,17],[2,6,1,28 ],[2,4,2,0],[2,5,2,9],[2,6,2,20],[2,5,3,1],[2,6,3,12],[2,6,4,4 ],[3,4,1,4],[3,5,1,13],[3,6,1,24],[3,5,2,1],[3,6,2,12],[3,6,3, 0],[4,4,1,0],[4,5,1,9],[4,6,1,20],[4,6,2,4],[5,5,1,5],[5,6,1, 16],[6,5,1,1],[6,6,1,12]]
Added even more later [I need sleep!] Those 43 values give real roots. So the probability of no real roots is $1-43/216 = 173/216$.
On
Also following up on @Berci's answer, here is a variation that you can do on some spreadsheet. First you set up a 6x6 table with $4ac$ in each cell. For each cell you want the number of b's that are integer, $\leq6$ and such that $b^2<4ac$. As you are working with integers you can replace that with $b^2\leq 4ac-1$, i.e. $b\leq\sqrt{4ac-1}$. That number (of b's) turns out to be the minimum of 6 and $\lfloor \sqrt{4ac-1}\rfloor$ (largest integer $\leq \sqrt{4ac-1}$. Computing that for each cell yields:
\begin{array}{cccccc} 1 & 2 & 3 & 3 & 4 & 4 \\ 2 & 3 & 4 & 5 & 6 & 6 \\ 3 & 4 & 5 & 6 & 6 & 6 \\ 3 & 5 & 6 & 6 & 6 & 6 \\ 4 & 6 & 6 & 6 & 6 & 6 \\ 4 & 6 & 6 & 6 & 6 & 6 \\ \end{array}
which then sums up to 173, thus the probability of 173/216.
To follow up on @nayrb's question, this generalizes to $n$-face dice in a straightforward way :
\begin{align*} p(n)=&\ \sum_{a=1}^n\sum_{b=1}^n \min(n,\lfloor \sqrt{4ac-1}\rfloor) \end{align*}
The limit for $n\rightarrow\infty$ is the continuous case :
\begin{align*} p(\infty)=&\ \int_{a=0}^1\int_{b=0}^1 \min(1,\sqrt{4ac})= \frac{31-6\log 2}{36}\approx0.745587 \end{align*}
On
A three-dimensional lattice would be preferable for investigating this, but we can get by with a two-dimensional grid. Since we wish to count the ordered triples for which $ \ b^2 \ < \ 4ac \ $ , we can set up a lattice for $ \ a \ $ and $ \ c \ $ running from 1 to 6 (contained in the light green square), and plot level curves for $ \ \frac{1}{4}b^2 \ $ with $ \ b \ $ also running from 1 to 6 . At each integer value of $ \ b \ $ , we can now easily count the number of lattice points "below" or on each level curve to find the number of combinations which do produce a discriminant value $ \ b^2 - 4ac \ \ge \ 0 \ $ . (The red diagonal line is added to indicate the symmetry in the choices for for $ \ a \ $ and $ \ c \ $ , which can simplify the counting.) These are then to be discarded from the count of desired outcomes.
We find that the number of combinations for each value of $ \ b \ $ is
b = 1 : 36
b = 2 : 36 - 1 = 35
b = 3 : 36 - 3 = 33
b = 4 : 36 - 8 = 28
b = 5 : 36 - 14 = 22
b = 6 : 36 - 17 = 19
As found by the other solvers here, the total number of combinations leading to complex-valued zeroes is 173 , making the probability $ \ \frac{173}{216} \ . $
If you don't have any further idea, probably the easiest is to draw a $6\times 6$ table of values $4ac$ for all possible $a,c$ pairs. (It will be symmetric.) Then you can conclude the number of $b$'s for a given pair $(a,c)$, and sum up all these.