Probability of waiting before meeting in case of two uniformly distributed random variables

Question

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.

Answer 1

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

Answer 2

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:   $\Pr(\lvert X_1-X_2\rvert < 10)=1-\Pr(\lvert X_1-X_2\rvert \geq 10) \\ = 1 - 2\int\limits_0^{50}\int\limits_{(x_1+10)}^{60} f_{X_1,X_2}(x_1,x_2)\operatorname d x_2\operatorname d x_1$.