Probability that the sum of roots is negative

Question

Three distinct integers are randomly chosen between -5 and 5 (both included, and 0 is excluded), and a polynomial is formed whose roots are those three numbers. What is the probability that:
i) The absolute value of the constant term is less than 20
ii) The coefficient of $x^2$ lies between $-8$ and $-12$ (both inclusive)
iii) The coefficient of $x^2$ is positive.

I have solved the first two parts by considering the cases, getting the answers $\frac12$ and $\frac3{40}$.

The problem lies with the third part. I do not know how to approach it. If the roots of the equation are $a,b,c$ then the coefficient of $x^2$ would be given by $-(a+b+c)$ and if that is positive, then it means that $(a+b+c)$ is negative. I know that, in total, there are $\binom{10}3=120$ possibilities. So how would I find how many required possibilities would be there? Considering the cases seem to be tiresome and difficult as there are many such possibilities, and I might miss some.

Answer 1

Let $s$ be the sum of roots. As mentioned in the comments:

• The numbers of selections with $s>0$ and $s<0$ are the same, due to a bijection between the two sets: if $\{a,b,c\}$ is a selection with $a+b+c=s>0$, then $\{-a,-b,-c\}$ has $s<0$ and vice versa.
• There are 8 selections with $s=0$.
• Exactly one of $s>0$, $s=0$ and $s<0$ is true for each selection.

Thus the number of selections with $s<0$ is $\frac{120-8}2=56$. The probability that $s<0$, hence the $x^2$ coefficient being positive, is $\frac{56}{120}=\frac7{15}$.