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I was just wondering if someone could maybe check my logic of working this question out. As a disclaimer I realise that there is questions of similar nature on the forum but, I have not looked due then if I see how it been worked out it doesn't help me if there fault in my logic.

Question:

Bob and Jane agree to meet at a known location between 2pm and 3pm. The problem is that neither of them know what time the other will arrive.

We may assume that both of them are equally likely to arrive at any time between 1pm and 2pm.

However Bob is prepared to wait 10 minutes for Jane to arrive before leaving. And Jane is prepared to wait only 5 minutes for Bob to arrive before she leave.

a) If Bob arrived at 1 pm what would be the probability that they meet?

b)If Bob arrived at 1.05 pm, what would be the probability that they meet?

c) If Bob arrived at 1.30 pm, what would be the probability that they meet

d) Draw a sketch of the probability of meeting as a function of Bob arrival time where the arrival time is the time elapsed in hours from 1pm.

So for part a) my thinking is this. If Bob arrives at 1pm then for them to meet, Jane could arrive at the following times my assumption is that it on the minute and not the second from looking at other questions.

Jane can arrive at: {1,1:01,1:02........1:10}pm so then the probability of them metting is $P=\frac{10}{60}=\frac{1}{6}$

for b) Jane can arrive at : {12:55,12:56........12:59} and {1,1:01......1:10} so the probability of them meeting is $P=\frac{15}{60}=\frac{1}{4}$

for c) because Jane could arrive at {1:25....1:29} and {1:30......1:40} so the probability of them meeting is $P=\frac{15}{60}=\frac{1}{4}$

for d) I remember reading in a old text book which iv lost at the moment that this is can be solved geometric probability and I think the diagram I am looking for is a line of y=x-10 where $x\geq0$ as this would included all possibilities from the seconds.

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I'll assume the agreement was to meet between 1 and 2 pm, not 2 and 3, because otherwise the rest of the question makes no sense.

I think the usual idea is not to assume Jane arrives only on at one of the on-the-minute times, because the arithmetic for on-the-minute times is more complicated than if you assume that Jane might arrive at any time during the interval, giving an infinite number of possibilities for what time she might arrive.

Notice that if only on-the-minute times are possible, then there are $61$ possible times to arrive, not $60$. That is, one can arrive at $0, 1, 2, 3, \ldots, 58, 59,$ or $60$ minutes past the hour. There are $61$ numbers in that list: the sixty numbers $1, 2, 3, \ldots, 59,$ and $60$, and the number $0.$

Similarly, there are $11$ numbers, not $10,$ in the list of numbers $0, 1, \ldots, 10.$ So if you let Jane arrive at 1:00, 1:01, 1:02, 1:03, 1:04, 1:05, 1:06, 1:07, 1:08, 1:09, or 1:10, she has $11$ possible arrival times. (Count them!)

Working only in whole minutes, therefore, the answer to part a) is $\frac{11}{61}.$ Or maybe it's $\frac{10}{61},$ because if Jane arrives at 1:10 then Bob is already leaving by a different path and doesn't see her.

But if we allow Jane to arrive at any time between 1 and 2 pm, with a uniform continuous probability distribution, then she has a $\frac{1}{60}$ chance to arrive between 1:00 and 1:01 and $\frac{10}{60}$ chance to arrive between 1:00 and 1:10. While it is possible for her to arrive at 1:10 exactly, the probability of that exact time is zero, so whether Bob is still waiting or already leaving at that exact instant makes no difference.

The continuous-distribution assumption also helps in part d), because when we graph Bob's arrival times at every point along one axis and Jane's arrival times at every point along the other axis, not just at the $61$ whole-minute times, the graph of the set of events in which they successfully meet is a polygon whose area we can easily compute. If we dealt only in whole-minute times then we'd have to count the individual lattice points, for which we can also derive formulas but not ones quite as simple. For example, the area of a right triangle with legs $n$ and $n$ is $\frac12 n^2,$ but it has $\frac12(n^2 + n)$ lattice points (if $n$ is an integer).

In short, under your assumptions, your answer for part a) is incorrect, but if you make the same assumptions that I would, your answer is correct.

For part b), you say Jane might arrive at 12:56 (among other times), but that is not correct: it is assumed that Jane arrives between 1 and 2 pm, and 12:56 is not between 1 and 2 pm. On the other hand, even if Jane is later than 1:10, as long as she arrives by 1:15 Bob will still be waiting. You did not consider that possibility, but when you do consider it then the probability of meeting becomes $\frac14.$ So again you have the same answer I would give, but the wrong reasons for it.

For part c) you have the right range of times, but miscounted them: there are $16$ distinct whole-minute times from 1:25 to 1:40 inclusive. And again if you allow Jane to arrive at any time in that interval the formula is simplified and your result turns out to be correct.

For part d), the line $y = x - 10$ is useful if $x$ is Jane's arrival time in minutes (and possibly fractions thereof) after 1 pm and $y$ is Bob's arrival time. That line separates the "good" events from the events where $y < x - 10,$ in which Bob arrives more than $10$ minutes before Jane and therefore leaves before she arrives. Another line to look at is the one that separates the "good" events from the ones where Bob arrives more than 5 minutes after Jane and she has already left.

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  • $\begingroup$ Firstly thank you for the detailed answer, which has helped me allot. I did original think by the seconds but thought because of the other questions maybe it minutes but I see now my mistakes. Once again thank you. $\endgroup$ – james2018 Sep 30 '18 at 17:29
  • $\begingroup$ Also after a quick read on uniform distribution I think I can see now what, is going. So for a) if I draw a graph the region I am looking in is $P(1 \leq x \leg1:10)=\frac{1}{3600} \times 600= \frac{1}{6}$ I have followed the same idea for b) and c) ,and see that I do get $\frac{1}{4}$ for both. Not sure why latex error but it a less than equal to for that interval $\endgroup$ – james2018 Sep 30 '18 at 18:06
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    $\begingroup$ You don't have to change units to seconds. The idea is to assume any fractional amount of time is possible, so no matter what units you use you can subdivide a time period as finely as you need to. The calculations will then work out to the same probabilities in any units, exactly as you computed them. $\endgroup$ – David K Oct 1 '18 at 12:02

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