Question

If p and q are chosen randomly from the set {1, 2, 3, 4. 5, 6, 7, 8, 9, 10}, with replacement, determine the probability that the roots of the equation x² + px + q = 0 are real.

Answer 1

The required probability = 1 - (probability of the event that the roots of x² + px + 4 = 0 are non-real).

The roots of x² + px + 4 = 0 will be non-real, if and only if

p² - 4q < 0 ie,if p² < 4q

The-possible values of p and q can be possible according to the following table.

Value of q	Value of q	no of pair of p,q
1 2 3 4 5 6 7 8 9 10	1 1, 2 1, 2, 3 1, 2,3 1, 2, 3, 4 1, 2, 3, 4 1, 2, 3, 4, 5 1, 2, 3, 4, 5 1, 2, 3, 4, 5 1, 2, 3, 4, 5, 6	1 2 3 3 4 4 5 5 5 6

Therefore, the number of possible pairs = 38

Also, the total number of possible pairs is 10 x 10 = 100. 

.'. The required probability = 1 - 38/100 = 1 - 0.38 = 0.62