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Two bankers each arrive at the station at some random time between 5PM and 6PM (arrival time for each of them is uniformly distributed). They stay exactly five minutes and then leave. What is the probability that they will meet on a given day?

I am not sure how to go about modelling this problem as uniform distribution and solving it. Appreciate any help.

Here is how I start with it: Assume banker A arrives X minutes after 5PM and B arrives Y minutes after 5PM. Both X and Y are uniformly distributed between 5PM and 6PM. So pdf of X, Y is $\frac{1}{60}$. Now A and B will meet if $|X - Y| < 5$.

So required probability is $P(|X - Y| < 5)$ = Integral of joint distribution function of $|X - Y|$ from $0$ to $5$?

Now not sure how to write the equation from this point onwards and solve it.

Answer: $\frac {23}{144}$

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  • $\begingroup$ Do you know how to solve multivariate probabilities in general. That is, to find the region you have to integrate your multivariate distribution over and then to calculate the integral? $\endgroup$
    – Marc
    Commented May 30, 2016 at 12:21
  • $\begingroup$ @Marc I am trying to get my head around how to model multivariate equations. I believe integral is next step which should be ok. $\endgroup$
    – Gerry
    Commented May 30, 2016 at 12:24

2 Answers 2

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Hint: If $f_{X,Y}$ is the multivariate pdf then you want to solve the following integral $$ \int_{0}^{60}\int_{\max\{0,x-5\}}^{\min\{60,x+5\}} f_{X,Y}(x,y)dydx. $$

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    $\begingroup$ So $f_{X,Y} = \frac{1}{60} \frac{1}{60}$ . How did you arrive at limits of inner integral. Pls explain $\endgroup$
    – Gerry
    Commented May 30, 2016 at 12:41
  • $\begingroup$ Well $x$ has to be 5 minutes from $y$. However they also both have to be between 5 and 5 PM. Thats why the min and max are there, to make sure $x$ and $y$ satisfy both these requirements. $\endgroup$
    – Marc
    Commented May 30, 2016 at 12:52
  • $\begingroup$ sorry still not clear to me. $\endgroup$
    – Gerry
    Commented May 30, 2016 at 13:44
  • $\begingroup$ We have two equations: $x-5 < y < x+5$ and $0<y<60$. We want $y$ to satisfy both, so we have to take the minimum and maximum. $\endgroup$
    – Marc
    Commented May 30, 2016 at 14:11
  • $\begingroup$ Thanks for your help. Can you also help with the outer integral. The inner one will give $$ \int_{0}^{60}\frac{[min(60, x+5) - max(0, x-5)]}{3600}dx. $$ $\endgroup$
    – Gerry
    Commented May 30, 2016 at 15:16
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Hint:

We will consider the situation in terms of minutes. We notice that that they will meet only if $|X-Y|\leq 5/60$, where $X,Y$ is the time of arrival as a fraction of the hour. We notice that $X,Y\overset{iid}{\sim} \text{unif(0,1)}$. Then they problem becomes $$P(|X-Y| \leq 5/60).$$ It will help to draw a picture. No integration required.

You can do it considering the $X,Y\overset{iid}\sim\text{unif}(1,60)$ too. enter image description here

Notice that we want the blue part. Also, notice that we can instead consider the complement \begin{align*} P(|X-Y|<5) &= P(-5<X-Y<5)\\ &=P(\text{blue})\\ &= 1-P(\text{Not blue})\\ &= 1-2\bigg[\underbrace{(1/2)\cdot55\cdot55\cdot 60}_{(1)}/\underbrace{60^3}_{(2)}\bigg]\\ &= 1-\left(\frac{55}{60}\right)^2 \\ &= \frac{23}{144} \end{align*}

where in

  1. That is the volume of one wedge
  2. That is the entire volume.
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  • $\begingroup$ Can you show how to do it using Integral. Thanks $\endgroup$
    – Gerry
    Commented May 30, 2016 at 13:37
  • $\begingroup$ No, I specifically did not want to integrate. You are simply going to be finding the volume above the blue strip. You're going to have to break up the blue strip into 3 regions. $\endgroup$
    – Em.
    Commented May 30, 2016 at 13:41
  • $\begingroup$ @Em. : Suppose a bomb will explode at a random moment from 5 PM to 6.05 PM . What will be the probability that both bankers will be there at that moment ? $\endgroup$
    – mrwong
    Commented Feb 23, 2019 at 7:57
  • $\begingroup$ @mrwong Please use the Ask Question button to ask a new question and include your own work and understanding of the problem. Otherwise, your question will be closed. $\endgroup$
    – Em.
    Commented Feb 23, 2019 at 20:43
  • $\begingroup$ @Em. Thanks for your reply. But there is a reason why I don't ask by myself. I believe not so many people can get a correct answer to my question . If you are interested you are welcome to give an answer here at any time . $\endgroup$
    – mrwong
    Commented Feb 24, 2019 at 3:50

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