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 x 2 + px + q = 0 are real.

Question

Aditi Chauhan · Answer

Hello Student,

Please find the answer to your question

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
Aditi Chauhan
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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 x 2 + px + q = 0 are real.

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.

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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 x 2 + px + q = 0 are real.

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.

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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.