Question

Starting at 5:00 A.M., every half hour there is a train from city A to city B. Suppose that none of these trains is completely sold out and that they always have room for passengers. A person who wants to travel to city B arrives at the station at a random time between 8:45 A.M. and 9:45 A.M. Find the probability that he waits (a) at most 10 minutes; (b) at least 15 minutes. 

Answer 1

Let $X=$ the number of minutes a person must wait for a train. If a passenger arrives at the stop at a time that is uniformly distributed between 8:45 A.M. and 9:45 A.M, then $X\sim U(0, 60)$

$$f(x)=\dfrac{1}{60-0}=\dfrac{1}{60}, 0\leq x\leq 60$$

(a) The regions ((8:50; 9:00], (9: 20; 9:30 ]) indicate areas that are less than 10 minutes before train arrival. Their spread is $10+10$ mins. By geometric definition of probability, in that case

$$P=\dfrac{10+10}{60}=\dfrac{1}{3}$$

(b) The regions ((9:00; 9:15], (9: 30; 9:45]) indicate areas that are at least 15 minutes before train arrival. Their spread is $15+15$ mins. By geometric definition of probability, in that case

$$P=\dfrac{15+15}{60}=\dfrac{1}{2}$$