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