Waiting Bus 2

A bus arrives every 10 minutes on a fixed schedule. The driver independently refills on gas with a 10% probability for each appearance, adding 1 hour to the bus's travel time if it occurs. For instance, the bus can arrive at times 0, 10, 20, 30, etc. In this example, the bus takes 70 minutes to arrive after the fourth arrival time as it refills gas here. If you arrive at a uniformly random time throughout the day, what is the expected time until the next bus appears?

Answer

In this scenario, 90% of the intervals are normal 10-minute intervals and 10% are extended to 70 minutes; over 10 intervals this averages to 9 normal intervals (9×10 = 90 minutes) and 1 extended interval (70 minutes), for a total cycle of 160 minutes. Since arrival is uniformly random, the probability of arriving during a normal interval is $\tfrac{90}{160} = \tfrac{9}{16}$ (with an expected wait of $$10/2=5$$ minutes) and during an extended interval is $\tfrac{70}{160} = \tfrac{7}{16}$ (with an expected wait of $$70/2=35$$ minutes). Thus, the overall expected waiting time is

\frac{9}{16}\times 5 + \frac{7}{16}\times 35 = \frac{45 + 245}{16} = \frac{290}{16} = \frac{145}{8} \text{ minutes},

so the expected time until the next bus arrives is $\boxed{\frac{145}{8}\text{ minutes}}$ .