You come to a bus stop served by 3 bus lines: A, B, C. Line A comes every 2 minutes, B every 3 minutes, C every 5 minutes. Each bus line is independent to each other (i.e. they don't have to be in-phase or synced). You board the SECOND bus you see. What's the expected time of waiting?

Solution

First consider line A, regardless of the other lines. Probability that we have to wait $t$ minutes or less is $t/2 (0 \leq t \leq 2)$. Likewise, probability that the first bus from line A doesn't come in $t$ minutes is $1 - t/2$. For line B and C, a similar formula applies.

Now, we consider nine cases of the first two buses we see: AA, AB, AC, ..., CC.

Case 1: AA

This means that bus A comes at time $t$, then at time $t+2$, and in the mean time, bus B and C never come. Obviously $0 \leq t \leq 1$ for otherwise B would have come already. The probability that B doesn't come in $t+2$ minutes is $1 - (t+2)/3$, and probablity that C doesn't come in $t+2$ minutes is $1 - (t+2)/5$. Also, keep in mind that your waiting time is not $t$ but $t+2$. The expected waiting time is thus:

$$\int_0^1 (t+2) (1-\frac{t+2}{3})(1-\frac{t+2}{5})dt = \frac{37}{180}$$

Case 2: AB

This means that bus B comes at time $t$, bus A comes before $t$ and bus C comes after $t$. Note that for $0 \leq t \leq 2$ that means bus A comes in the time between $(0,t)$, but if $2 \leq t \leq 3$, that means bus A comes in the time between $(t-2, 2)$.

$$\int_0^2 (t) (t/2)\frac{2}{3}(1-\frac{t}{5})dt + \int_2^3 (t) \frac{1}{3} (\frac{2-t}{2})(1-\frac{t}{5})dt = \frac{28}{45} + \frac{37}{120}$$

Case 3: AC

$$\int_0^2 (t) (t/2) \frac{2}{5}(1-\frac{t}{3})dt + \int_2^3 (t) \frac{1}{5}(\frac{2-t}{2})(1-\frac{t}{3})dt= \frac{4}{15} + \frac{23}{360}$$

Case 4: BA

$$\int_0^2 (t) (t/3)(1-\frac{t}{5})dt = \frac{28}{45} $$

Case 5: BB

Impossible, because bus A must come at least once in the period between two Bs.

Case 6: BC

$$\int_0^2 (t) (t/3)(1-\frac{t}{2})dt = \frac{2}{9}$$

Case 7: CA

$$\int_0^2 (t) (t/5)(1-\frac{t}{3})dt = \frac{2}{15}$$

Case 8: CB

$$\int_0^2 (t) (t/5)(1-\frac{t}{2})dt = \frac{1}{15}$$

Case 9: CC

Impossible, because bus A must come at least once in the period between two Cs.

The sum of those values is $\frac{83}{36} = 2.3055$ minutes

## Wednesday, April 4, 2012

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