Every evening John either visits his girlfriend who lives downtown, or visits his mother who lives uptown, but not both. In order to be completely fair, he goes to the bus stop every evening at a random time between 6pm and 7pm, and takes either an uptown or a downtown bus (from the same bus stop), whichever comes first. Each of the two kinds of buses arrives at the bus stop every 30 minutes with a fixed regular schedule: the arrival times for downtown buses are a or (30 + a) minutes past each hour, and that for uptown buses are b or (30 + b) minutes past each hour, where 1 ≤ a < b < 30 are two integers.
How many times is John expected to visit his mother every month (i.e. 30 days)?
My approach yields 10 days. But I would like to know if there's a formal approach to solving this! Thanks!
Well the answer depends on $a$ and $b$.
Let $X$ be the variable denoting his arrival in minutes past 6pm. That is $X\sim Uniform(0,60)$
There are 3 events where the downtown bus is taken. $\{X\leq a\}$, $\{b<X\leq 30+a\}$ and $\{X>30+b\}$. Therefore $$P(\text{John visits his girlfriend})=P(X\leq a)+P(b<X\leq 30+a)+P(X>30+b)$$ Using that the CDF is given by $P(X\leq x)=\frac{x}{60}$, we have
$$P(\text{John visits his girlfriend}) = \frac{a}{60} + \frac{30+a-b}{60} + 1- \frac{30+b}{60}= 1 + \frac{a-b}{30} $$ Likewise we can compute $$P(\text{John visits his mom})=1-P(\text{John visits his girlfriend})=\frac{b-a}{30}$$ In a total of 30 days, the expected number of visits to his mom will be $$30P(\text{John visits his mom})=b-a$$