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Question

Find the probability that a person will catch a bus from 8 to 9 am, given that bus arrives at a random time between 8 and 9 am and waits for 11 minutes, and the person also arrives at a random time between 8 and 9 am and waits for 11 minutes (all arrival times are equally likely).

Attempt

I have tried to mark all the times and intersections of bus and person on a number line from 8 to 9, and tried to find the probability that they'll meet in the first 11 minutes, and the last 11 minutes. I have added them together and got 0.366. However that's not the right answer, and adding 11/38 (prob of first 11 mins, prob of last 11 mins and prob that they'll meet somewhere in between alll added together).

Any hints to where I'm going wrong will be greatly appreciated.

FD_bfa
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John Doe
  • 502
  • Rather than a line, try a square as in https://math.stackexchange.com/questions/103015/chance-of-meeting-in-a-bar – Henry Apr 20 '22 at 16:58

1 Answers1

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The Mistake in your Attempt

Your answer is close and so first I'll explain the mistake that you made, and then I'll address how we can fix that.

The probability that you catch the bus, assuming that the bus can leave no later than 9am is $0.366$ (which is what you found). The issue with this, is that you haven't considered the fact that the bus could arrive at 8:58 (for example), and so even though it will stay for $11$ minutes, we still need to consider the fact that the person could catch the bus here too (since the bus will leave at 9:09am here).

The Solution

The way we fix this is by noting that both the person and the bus could arrive at 9am (and so they would both stay for 11 minutes in this situation, which would take us up to 9:11am). If you re-compute your probabilities with this consideration, you will arrive at the correct answer.

Your strategy of marking the intersections will still work, however, the interval we need to consider is over $71$ minutes (from 8-9:11am rather than $60$ minutes which is what you have done).

Note: if you are looking to solve this more rigorously, then the way to solve this is by looking at the CDF of a continuous uniform random variable X over the closed interval $[0,71]$ and considering the probability that $X$ $\in$$[a-11,a+11]$ for $11\le a\le60$. We then add to this to the probability that $X$ $\in$$[0,a+11]$ for $0\le a\le 11$ where $a$ is the arrival time of the bus.

FD_bfa
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  • Thank you! I have added 1/71 + 1/71 to get 0.3099, however that's not the right answer. Did I misunderstand your explanantion? And as for the CDF - how do i find $\mu$ and $\sigma$ ? – John Doe Apr 20 '22 at 13:41
  • We don't need to calculate the mean and standard deviation for the CDF as we are looking at the continuous uniform distribution on [0,71] which just has CDF $\frac{X}{71}$. The formulae for the mean and standard deviation aren't too complicated but they aren't needed for this particular question @JohnDoe – FD_bfa Apr 20 '22 at 14:58