If p and q are chosen randomly from the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} with replacement, then the probability that the roots of the equation x2 + px + q = 0
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
are real is 33/50
b
are imaginary is 19/50
c
are real and equal is 3/50
d
are real and distinct is 3/5
answer is B.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
Roots of x2 + px + q = 0 will be real if p2≥4q.The possible selections are as follows:Therefore, number of favourable ways is 62 and total number of ways is 102 = 100. Hence, the required probability is62/100 = 31/50.The probability that the roots are imaginary is1 - 31/50 = 19/50.Roots are equal when (p, q) ≡ (2, 1), (4, 4), (6,9). The probability that the roots are real and equal is 3/50. Hence, probability that the roots are real and distinct is 3/5.