Question

A student comes to lecture at a time that is uniformly distributed between 5:09 and 5:14. Independently of the student, the professor begins the lecture at a time that is uniformly distributed between 5:10 and 5:12. What is the chance that the lecture has already begun when the student arrives?

Answer 1

1/2

Explanation:

The lecture has already begun when the student arrives means one of these scenarios happen:

1) the class started at 5:10 and the student arrives at 5:11 or 5:12 or 5:13 or 5:14

2) the class started at 5:11 and the student arrives at 5:12 or 5:13 or 5:14

3) the class started at 5:12 and the student arrives at 5:13 or 5:14

Given student time of arrival is uniformly distributed, then the probability he/she arrives at 5:09 or 5:10 or 5:11 or 5:12 or 5:13 or 5:14 is 1/6.

So, the probability that the student arrives between 5:11 and 5:14 is 1/6 + 1/6 + 1/6 + 1/6 = 2/3.

The probability that the student arrives between 5:12 and 5:14 is 1/6 + 1/6 + 1/6 = 1/2.

The probability that the student arrives at 5:13 or 5:14 is 1/6 + 1/6 = 1/3.

Given class starting time is uniformly distributed, then the probability it starts at 5:10 or 5:11 or 5:12 is 1/3.

Given the two events are independent, the probability of the first scenario is: (1/3)*(2/3) = 2/9

For the second scenario: (1/3)*(1/2) = 1/6

For the third scenario: (1/3)*(1/3) = 1/9

Because all of these scenarios are mutually exclusive the total probability of one of them happen is: 2/9 + 1/6 + 1/9 = 1/2