
Bradley T. answered 03/18/23
Tutor for Python and High School and Middle School Math
After reading the problem, the first thing to notice is that these events are happening in a cycle. Bus arrives, waits 2 minute, then leaves. We want to find the probability in one cycle, and because all cycles are identical, it will be the same probability for all cycles. For now, we go through one cycle. We begin at time 0.
- Let's say that at time 0 minutes, the old bus leaves, you just miss it.
- You then have to wait 9 minutes.
- Remember, the two minutes the bus waited are counted in the 11 minutes until the next bus arrives.
- A bus arrives at time 9 minutes.
- You wait 2 minutes.
- The bus leaves at time 11.
It might be helpful to look at this visually.
Arriving at time 0, the person has to wait 11 minutes total, greater than the 9 min we want to know about. At time 1, the person waits 10 minutes. It is when the person arrives at time 2 that the person waits exactly 9 minutes. Arriving any later would mean waiting less, up until the bus leaves, starting a new cycle. So when time (t) < 2 min, the person waits more than 9 minutes. There are a total of 11 minutes. In other words, in this cycle of 11 minutes, only during the first 2 minutes, would the person wait more than 9 minutes for the bus to leave. That is 2/11 is the fraction of time, when if I show up at a random time, I will have to wait more than 9 minutes. All the cycles are the same, and do not affect each other. That fraction is in fact the probability.
Another way to look at it, if I choose any random minute within those 11 minutes, only during 2 of them would I have to wait more than 9 minutes for the bus. So the probability is 2/11

Mark M.
Illustrative graphic!03/18/23