If A is uniformly distributed over [-12,16], what is the probability that the roots of the equation

x^2+Ax+A+3=0

are both real?

1. x^2 + ax + a + 48 = 0 Discriminant D = a^2 - 4 * 1 * (a+48) D = a^2 - 4a - 192 For the roots to be real, D > = 0 a^2 - 4a - 192 > = 0 a (a+12) - 16 (a+12) > = 0 (a+12) (a-16) > = 0 a = 16 The required probability P (roots are real) = P (a = 16) The interval is [-19, 22] Therefore P (roots are real) = P (a E [-19, - 12]) + P (a E [16, 22]) = [ (-12) - (-19) ]/[22 - (-19) ] + (22-16) / [22 - (-19) ] = 7/41 + 6/41 = 13/41 Ans: 13/41