# Find probability of a situation

John and Mary agreed to meet between 3 and 4 pm to study. Let X be the moment of John's arrival, and Y, the moment of Mary's arrival. These random variables are independent, with density functions, \begin{align} f_{X}(x)=1, \qquad(3<x<4) \end{align} \begin{align} f_{Y}(y)=1, \qquad(3<y<4) \end{align} If the first to arrive wait only 15 minutes for the other, how likely are they to study together that day?

I tried this but I think it is wrong: \begin{align} P(X-Y<0,25) \end{align}

• On this site, you are expected to show what effort you have put in. Please show what you have tried, else you may soon be downvoted and the question closed. – true blue anil Dec 26 '17 at 20:23
• You want to say $P(|X-Y|<0.25)$? This would be better. – user491874 Dec 26 '17 at 20:32
• I tried this: \begin{align} \int_{15}^{15.75}\int_{y}^{y+\frac{1}{4}} 1\,\mathrm dx\,\mathrm dy+\int_{15.75}^{16}\int_{y}^{16}1\,\mathrm dx\,\mathrm dy \end{align} – Andriy Russu Dec 26 '17 at 20:41
• @AndriyRussu You are on the right track, except:  All your times are P.M. ;) ,  You only consider $X>Y$ i.e. you imply John will be later than Mary, so please apply my advice from the comment above. – user491874 Dec 26 '17 at 20:51
• Yes, sorry: \begin{align} \int_{3}^{3.75}\int_{y}^{y+\frac{1}{4}} 1\,\mathrm dx\,\mathrm dy+\int_{3.75}^{4}\int_{y}^{4}1\,\mathrm dx\,\mathrm dy +\int_{3}^{3.75}\int_{x}^{x+\frac{1}{4}} 1\,\mathrm dy\,\mathrm dx+\int_{3.75}^{4}\int_{x}^{4}1\,\mathrm dy\,\mathrm dx \end{align} – Andriy Russu Dec 26 '17 at 20:55

Because $X$ and $Y$ are independent, the joint density is $f_{(X,Y)}(x,y)=1$ in the square $C=[3,4]\times[3,4]$ (and 0 else)

Now, you want calculate $\mathbb{P}(\mid X-Y\mid<0.25)=\mathbb{P}(0<X-Y<0.25)\cdot\mathbb{P}(X>Y)+\mathbb{P}(0<Y-X<0.25)\cdot\mathbb{P}(X\leq Y)$, where the last equality works by total probability law.

Observe the symmetry in the problem, and then $\mathbb{P}(X>Y)=\mathbb{P}(X\leq Y)=\frac{1}{2}$.

Now, for $\mathbb{P}(0<X-Y<0.25)$ use the joint density and:

$\mathbb{P}(0<X-Y<0.25)= \mathbb{P}((X,Y)\in B)=\iint_{B\cap C} 1dxdy$, where $B=\{(x,y)\in \mathbb{R}^2: 0<x-y<0.25\}$.

So, $\mathbb{P}(X-Y<0.25)=area (B\cap C)=\frac{1^2}{2}-\frac{0.75^2}{2}=0,21875$

By symmetry, the solution is $0,21875\cdot 0.5\cdot 2=0,21875$