Two people agree to meet each other at the corner of two city streets between 1pm and 2pm. However, neither will wait for the other for more than 30 minutes. If each person is equally likely to arrive at any time during the one hour period, determine the probability that they will in fact meet.

This question is very similar to Probability of two people meeting during a certain time. But I'm afraid I'll have to ask again, because it's not quite what my query is about.

I've tried letting X and Y be the two random variables, and they are independent of each other. I tried phrasing it in terms of $P(|X-Y|<30) $ but I don't know how to solve this, as i don't know the individual probability of either X and Y? Any advice would be greatly appreciated. Sorry in advance if I seem to be repeating the question (in link) and for any wrong title or tag labelling.

I have also thought of another method:

If A arrives before B, then the probability that he arrives during first half an hour is 0.5 . He'll then wait for 0.5 hours. If He arrives during the last half an hour hours, with proability 0.5 . He then waits for 0.5 hours on average. Thus his total wait time is 0.5*0.5*2=0.5. For A and B to meet, B must arrive when A is waiting. Thus the probability that B arrived when A was waiting is 0.5 . Similarly if B arrives before A, the probability that they meet is 0.5 . Thus, Total probability that they meet is 1. Where did I go wrong? As the answer suggested is 0.75.

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    $\begingroup$ Aren't $X$ and $Y$ both uniformly distributed between $0$ and $60$? (And I assume it is intended that they arrive independently, though it's not stated explicitly in the problem description.) $\endgroup$ – Brian Tung Apr 3 '16 at 22:24
  • $\begingroup$ @BrianTung, However, if using uniform distribution, if I do invnorm on GDC, it doesn't mean that I can solve it? (as i don't know the mean and the variance?) Pls advise. $\endgroup$ – CCC Apr 3 '16 at 22:35
  • $\begingroup$ Maybe I'm misunderstanding the question. Why isn't this question like the one you link to? $\endgroup$ – Brian Tung Apr 3 '16 at 22:37
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    $\begingroup$ The variables are not normally distributed, so you cannot use mean and variance. You can integrate the probability density function over the region where the two people meet; since that function is constant and the region is more or less as depicted in that other question, the methods are essentially identical. $\endgroup$ – Brian Tung Apr 3 '16 at 22:48
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    $\begingroup$ Probability is 75% according the following simulation (in R). "n <- 10^5 # number of trials a <- runif(n, min = 0, max = 60) b <- runif(n, min = 0, max = 60) sum(abs(a-b) < 30)/n" $\endgroup$ – snoram Apr 3 '16 at 23:06

The R code below essentially repeats @snoram's nice simulation, but using only 50,000 points. Then I go on to plot the points in the square with vertices at $(0,0)$ and $(60,60)$, with the points corresponding to the condition that the two people meet plotted in light blue.

From there, it is obvious that the two excluded regions (black) each have $1/8$th of the area. Because the joint distribution is uniform on the square this means that the probability of the 'condition' is $3/4.$

Of course, you can draw the boundaries of the condition, and thus solve the problem, without simulation. (It is $not$ necessary to know the distribution of $|X - Y|$.) In integral notation, you need $\int\int_C 1/60^2\,dx\,dy,$ where $C$ is the region corresponding to the condition. You'd have to break $C$ up into two parts in order to set numerical limits on the integral signs. But I think the geometrical argument suffices.

 m = 50000
 x = runif(m, 0, 60);  y = runif(m, 0, 60)
 cond = (abs(x-y) < 30);  mean(cond)
 ## 0.74938  # aprx answer; actual answer 3/4
 plot(x, y, pch=".")
 points(x[cond], y[cond], pch=".", col="skyblue2")

enter image description here


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