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# 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 equation x2 + px + q = 0 are real.

6 years ago
Hello Student,
The required probability = 1 – (Probability of the event that the roots of x2 + px + q = 0 are non – real if and only if
p2 – 4q < 0 i.e. if p2 < 4q.
We enumerate the possible values of p and q, for which this can happen in the following table.
 q P Number of pairs p, q 1 1 1 2 1, 2, 2 3 1, 2, 3 3 4 1, 2, 3 3 5 1, 2, 3, 4 4 6 1, 2, 3, 4 4 7 1, 2, 3, 4, 5 5 8 1, 2, 3, 4, 5 5 9 1, 2, 3, 4, 5 5 10 1, 2, 3, 4, 5, 6 6
Thus, 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

Thanks