Question

What is the possibility such that the inequality x^2 + b > ax is true, when a=32.4 and b=76.5 and x∈[0,30].

(a) 1.91

(b) 4.3

(c) 2.94

(d) 6.1

The question was asked during an interview for a job.

I need to ask this question from Geometric Probability topic in chapter Discrete Probability of Discrete Mathematics

Answer 1

Correct choice is (a) 1.91

To elaborate: x2+76.5>32.4x is equivalent to x^2−32.4x+76.5 > 0. By completing the square, x^2 − 32.4x + 266.44 – 266.44 + 76.5 = (x−16.2)^2 − 189.94>0, which is the same as (x−16.2)^2 > 189.94,  which implies that either x−16.2 > 13.78 ⇒ x > 29.98, or x−16.2 < −2.42 ⇒ x  < 13.78. Assume that it is a uniform distribution. So, the probability that x > 29.98 is 30 − 29.98 = 0.02 and the probability that x < 189.94 is 1.89. The desired probability is 0.02 + 1.89 = 1.91.