Question

If p and q are chosen from the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, with replacement, the probability that the roots of the

equation x² + 2px + q = 0 are real is-

• A. 0.84
• B. 0.16
• C. 0.62
• D. 0.38

Answer 1

Answer: (A)

0.84

Answer 2

Required probability

= 1 – P (the roots of x² + 2px + q = 0 are non-real) The roots of x² + 2px + q = 0 will be non-real if (2p)² – 4q < 0 i.e. if p² < q.

We enumerate the possible values of p and q. When q = 1, there is no value of p

When q = 2, 3, 4, possible value of p is 1 When q = 5, 6, 7, 8, 9 possible values of p are 1 and 2

When q = 10, possible values of p are 1, 2 and 3. Thus, the number of pairs for which 2x² + 2px + q = 0 have non-real roots is

0 + 3 × 1 + 5 × 2 + 1 × 3 = 16 Also, total number of possible pairs is 10 × 10 = 100

Thus, probability of the required event

= 1 – $\frac { 16 } { 100 }$ = 0.84