A car and a bus arrive at a railroad crossing at times independently and uniformly distributed between 7:15 and 7:30. A train arrives at the crossing at 7:20 and halts traffic at the crossing for five minutes. Calculate the probability that the waiting time of the car or the bus at the crossing exceeds three minutes.

• 0.25
• 0.27
• 0.36
• 0.40
• 0.56

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Solution: A

To be delayed over three minutes, either the car or the bus must arrive between 7:20 and 7:22. The probability for each is 2/15. The probability they both arrive in that interval is (2/15)(2/15). Thus, the probability of at least one being delayed is 2/15 + 2/15 – (2/15)(2/15) = 56/225 = 0.25.

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.