I have a question that goes like this. Two people, X and Y, decide to meet at a particular time. The probability of them both being late is uniformly distributed between 0 and 60. Person X is always the first to reach the meeting point. What is the probability that X will have to wait less than 10 minutes. I tried to approach it in this manner: $$\Pr(Y - X\leq 10) = \int_{10}^{60}\int_{0}^{50} p_X(x)\cdot p_Y(y) \operatorname dx \operatorname dy$$ where $p_X(x) = 1/60$ and $p_Y(y) = 1/60$

Is this a correct way to approach? Could anyone please help.


Geometric probability simplifies the problem.

On a $60\times60\;\; X-Y$ grid, plot the arrival times of $X$ and $Y$.

Since $X$ always comes before $Y$, the feasible region is the area above the main diagonal $y = x$,

and the "meeting" region is the area between the main diagonal and $y = x+10$

Thus $Pr = \dfrac{0.5(60^2 - 50^2)}{0.5(60^2)}$

  • $\begingroup$ Thanks so much! I just drew the region of interest and understood the concept perfectly. I slightly can't figure out how to get the area of the region of interest, it doesn't seem to be a parallelogram. Can you please explain in more detail how you got the expression? $\endgroup$ – QPTR Oct 16 '15 at 1:56
  • $\begingroup$ From the area of the triangle above the main diagonal, subtract the area of "non-meeting" triangle above y = x+10. $\endgroup$ – true blue anil Oct 16 '15 at 3:35
  • $\begingroup$ Hmm, I am finding it hard to visualize this. Isn't their a main parallelogram in between and a triangle with base and height = 50 above and a triangle with base and height of 60 below. So, if I do 60^2 - (0.5*50*50) - 0.5*60*60 / 60^2 (the numerator representing the region between y=x and y=x+10) won't this be it? $\endgroup$ – QPTR Oct 17 '15 at 3:14
  • 1
    $\begingroup$ Have a look at the diagram at math.stackexchange.com/questions/103015/… . The main difference is that there either could arrive first, so the whole square was considered, here we consider only the area above the main diagonal. $\endgroup$ – true blue anil Oct 17 '15 at 4:09

We know that the arrival times $X$ and $Y$ are both uniformly distributed on the same interval, yet somehow $X < Y$ almost surely.   This behaviour means they cannot be independent; so the joint density is not the product of their marginal densities.

Instead, let's say then that $X,Y$ are the first and second least order statistics of $\{X_1,X_2\}$, which are iid random variables, uniformly distributed.   (Which ever is least, we call $X$, the other $Y$. )   This gives us a model of their behaviour.

We then want to find: $\Pr(\lvert X_1-X_2\rvert < 10)$ where $(X_1, X_2) \mathop{\sim}\mathcal{U}[0;60]^2$

Hint: the easiest way to find this is to sketch the area of interest on the $[0;60]^2$ square, and consider that:   $\Pr(\lvert X_1-X_2\rvert < 10)=1-\Pr(\lvert X_1-X_2\rvert \geq 10) \\ = 1 - 2\int\limits_0^{50}\int\limits_{(x_1+10)}^{60} f_{X_1,X_2}(x_1,x_2)\operatorname d x_2\operatorname d x_1$.

  • $\begingroup$ Thank you so much for the answer, and explaining the dependent part as well. Would the joint density function be the area of the region of interest that is the parallelogram? Also, could please explain how you obtained the limits on the integrals ? If X1 is the one who is late, then for dx2 shouldn't the limits go from 0 to 50 and for dx1 x2+10 to 60 ? $\endgroup$ – QPTR Oct 16 '15 at 1:49
  • $\begingroup$ @QPTR $\mathsf P(X_2\geq X_1+10) = \int_0^{50}\int_{x_1+10}^{60}...\operatorname d x_2\operatorname d x_1$. And $\mathsf P(X_1\geq X_2+10) = \mathsf P(X_2\geq X_1+10)$, by symmetry, since $f_{X_1,X_2}(x_1, x_2) = f_{X_1,X_2}(x_2, x_1)$ $\endgroup$ – Graham Kemp Oct 16 '15 at 3:30
  • $\begingroup$ Thank you! I would like to accept both answers since they both shed light in a different way, but I can only accept one, so tonnes of thanks for the explanation. $\endgroup$ – QPTR Oct 17 '15 at 2:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.