The coefficients of the quadratic equation x^2+bx+c=0 are determined by tossing a fair die twice, the first outcome to determine b and second to determine

c. what is the probability that the equation has real root (s), i. e., what is the probability that the discriminant b^2-4c ≥0?

This problem can be solved by a 36x36 contingency table.

The sample space of the sum of two throws of a die is 6*6=36.

We have the same sample space for the value of "b", and that of "c".

Let the rows of the 36x36 table represent b, and columns, c.

Calculate, if b^2 ≥ 4c, mark a 1 in the square crossing the values of b & c.

At the end, count the number of "ones" in the table.

I get 986 ones out of 36^2=1296.

So the probability is 986/1296.

Alternatively, a simpler way is to calculate the number of occurrences of each outcome for two throws.

E. g.

outcome freq

2 1

3 2

4 3

5 4

6 5

7 6

8 5

9 4

10 3

11 2

12 1

Then we reduce the size of the table to 11x11, but we have to multiply the results by the weight (freq) to get the total probability.