From Algorithms to Z-Scores - matloff - University of California, Davis

2.11. EXAMPLE: BUS RIDERSHIP 19
Again, this is a very common approach. But be sure to take care that we are in an “if and only if”
situation. Yes, R+B = 4, B > 0 implies R = 3, B = 1, but we must make sure that the converse
is true as well. In other words, we must also confirm that R = 3, B = 1 implies R+B = 4, B > 0.
That’s trivial in this case, but one can make a subtle error in some problems if one is not careful;
otherwise we will have replaced a higher-probability event by a lower-probability one.
2.11 Example: Bus Ridership
Consider the following analysis of bus ridership. (In order to keep things easy, it will be quite
oversimplified, but the principles will be clear.) Here is the model:
• At each stop, each passsenger alights from the bus, independently, with probability 0.2 each.
• Either 0, 1 or 2 new passengers get on the bus, with probabilities 0.5, 0.4 and 0.1, respectively.
• Assume the bus is so large that it never becomes full, so the new passengers can always get
on.
• Suppose the bus is empty when it arrives at its first stop.
Let Li denote the number of passengers on the bus as it leaves its i th stop, i = 1,2,3,... Let’s find
some probabilities, say P (L2 = 0).
For convenience, let Bi denote the number of new passengers who board the bus at the i th stop.
Then
P (L2 = 0) = P (B1 = 0 and L2 = 0 or B1 = 1 and L2 = 0 or B1 = 2 and L2 = 0) (2.47)
2
= P (B1 = i and L2 = 0) (2.48)
=
i=0
2
P (B1 = i)P (L2 = 0|B1 = i) (2.49)
=
i=0
0.5 2 + (0.4)(0.2)(0.5) + (0.1)(0.2 2 )(0.5) (2.50)
= 0.292 (2.51)
For instance, where did that first term, 0.5 2 , come from? Well, P (B1 = 0) = 0.5, and what about
P (L2 = 0|B1 = 0? If B1 = 0, then the bus approaches the second stop empty. For it to then leave
that second stop empty, it must be the case that B2 = 0, which has probability 0.5.