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:1/2Step-by-step 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) = 2/9For the second scenario:  (1/3)*(1/2) = 1/6For the third scenario:  (1/3)*(1/3) = 1/9Because all of these scenarios are mutually exclusive the total probability of one of them happen is: 2/9 + 1/6 + 1/9 = 1/2