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Here is the question I found. We agree to try to meet between 12 and 1 for lunch at our favorite sandwich shop. Because of our busy schedules, neither of us is sure when we’ll arrive; we assume that for each of us our arrival time is uniformly distributed over the hour. So that neither of us has to wait too long, we agree that we will each wait exactly 15 minutes for the other to arrive, and then leave. What is the probability we actually meet each other for lunch? Let random variables X and Y be independent and uniformly distributed the interval [0, 60]. These variables denote the arrival times of the two people during that hour. So, the two people 1 will meet if |X − Y | ≤ 15. Therefore, we compute enter image description here

Can Any one help me to understand how can I get those three different fx(x)?

Zy tdp
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  • The author forgot a constant and afterwards forgot forgetting it. I suggest you just do it all over again starting from line 2. I think you can just write down what $f_X$ and $f_Y$ are and do the calcs. – Sassatelli Giulio Jul 22 '23 at 14:19

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It is understandable that you don't understand the 3 cases first of all because they should be written with a common unit ; here I have chosen hours :

$$\int_{x=y-1/4}^{x=y+1/4}f(x)dx=\begin{cases}y+1/4& 0<y\le 1/4\\ 1/2& 1/4 <y\le 3/4\\ 5/4-y& 3/4<y \le 1\\\end{cases}$$

But I find it so difficult to understand, because conventions are not settled that I propose you instead a very easy graphical solution based on the following figure :

enter image description here

where a point with coordinates $(X,Y)$ means persons $A,B$ arrive at times $X,Y$ resp. This point is uniformly distributed in $[0,1] \times [0,1]$.

The green elongated hexagon corresponds to dates of arrival complying with the condition.

Its area is the looked-for probability. It is obtained by subtracting to $1$ the complementary probability, i.e., the areas of the two white triangles :

$$1- 2 \tfrac12 (\tfrac34)^2=1-\tfrac{9}{16}=\tfrac{7}{16}$$

Jean Marie
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