Calculating the probability of a function being real when it has random variables

Question

I'm trying to calculate the probability of

$ax^2+bx+c$ root being a real number when the variables $a,$ $b,$ and $c$ values are all randomized by throwing a standard die.

I got to the point where I can get the probability by calculating the chance of $b^2-4ac>0$, but I'm not sure how I can conveniently carry on from here and my attempts at doing it by hand (finding every possible real occurrence and calculating them by the total outcomes) have failed me.

In other words, the values of $a$, $b$ and $c$ are within $\{1,2,3,4,5,6\}$ and each of the three variables is randomly picked from that list with no special weighting (so $\frac 16$ chance to get any of the $6$ values).

The computation depends on what you mean by 'randomized by throwing a standard dice'. — copper.hat, Jan 14 '18 at 19:23
When you edit, please note that "dice" is plural...the singular is "die". If you meant a pair of dice (or more than $2$) you should say that. — lulu, Jan 14 '18 at 19:24

score 1 · Accepted Answer · answered Jan 14 '18 at 19:32

1

There are $43$ triples $(a,b,c)$ with $b^2 - 4 a c \ge 0$, so the probability is $43/6^3$. Note that you do want $\ge$, not just $>$, unless you just want two distinct roots. The easiest way to compute these by hand, I think, is to consider the possible values of $ac$.

answered Jan 14 '18 at 19:32

Robert Israel

448,999

Is the only legitimate way to really do it by hand? – Grak Jan 14 '18 at 19:43
@Grak : Probably there are some slick ways to do it, but maybe some of them won't be easy to find. – Michael Hardy Jan 14 '18 at 19:48
Fair enough, thanks. – Grak Jan 14 '18 at 19:56
Well, the actual formula is $\sum_{b=1}^6 \sum_{a=1}^{\min(6,[\frac {b^2}{4}])} (c=[\frac {b^2}{4a}])$ but... well, the only good way to do that for me was by hand. – fleablood Jan 14 '18 at 20:21
Are you asking whether it's legitimate to do it by computer? As far as I'm concerned, yes it is. – Robert Israel Jan 14 '18 at 22:29

score 1 · Answer 2 · answered Jan 14 '18 at 20:17

We need $b^2 - 4ac \ge 0$.

If $b= 6$ then $ac \le 9$. If $a =1$ then $c = 1...6$. If $a=2$ then $c=1...4$. If $c=3; a= 1..3$ if $a=4; c=1,2$ if $a=5,6;c =1$. There are $6+4+3+2+1+1 = 17$ ways to do this.

If $b = 5$ then $ac \le 6$. If $a=1$ then $c = 1...6$ if $a= 2$ then $b=1..3$ and if $a = 3; c=1,2$ and if $a \ge 4; c=1$. So $6+3+2+3 = 14$ ways to do this.

If $b = 4$ then $ac \le 4$ So if $a= 1,2,3,4$ then $c \le 4,2,1,1$ so there are $4+2+1+1 = 8$ ways to do that

If $b =3$ then $ac \le 2$ So if $a= 1,2$ then $c \le 2,1$ so there are $2+1=3$ wways to do that.

If $b = 2$ then $ac \le 1$ so $a=c=1$. $1$ way to do that.

$b$ can not be $1$ because $4ac \ge 4 > 1^2$.

So there are $17+14+8+3+1 = 43$ ways to do this.

So the probability is $\frac {43}{6^3} \approx 20\%$ (very slightly less.)

Calculating the probability of a function being real when it has random variables

2 Answers2