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Xavier and Yolanda plan to meet for lunch between noon and 1 p.m. They arrive independently with uniform distribution on [0, 1]. Yolanda will wait 30 min. for Xavier, but Xavier will only wait 15 min. for Yolanda. What is the probability that they meet?

I solved a problem similar to this by finding $P(\vert X-Y\vert\leq 30)$, but in that example they were both waiting for 30 minutes. How would I set up the bounds for the distribution function in this scenario, when they are waiting for different times?

BCLC
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Matt G
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2 Answers2

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See here for various ways of solving such problems, the easiest being using geometric probability, which is the first of the approaches given, and explained fully.

The only twist here is the lack of symmetry, easily resolved.
Draw a $60 \times 60$ square representing the $1$ hour period, with Xavier's arrival time on the X-axis, and Yolanda's on the Y-axis.

Draw $2$ parallel lines to the main diagonal from $(15,0)$ to $(60,45)$
and from $(0,30)$ to $(30,60)$.

We need to exclude the areas of the $2$ triangles forming the "non-meeting" area, so

$Pr = 1 -$ $\dfrac{(0.5*30*30) + (0.5*45*45)}{60*60}$

true blue anil
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Take $Y$ to be the time Yolanda arrives and $X$ the time Xavier arrives. We have two cases. $X \leq Y$ or $Y \leq X$.

  • If $X \leq Y$ in order for them to meet we need that $Y \leq X + 15$. Since $X$ and $Y$ are independent this happens with probability $P(X \leq Y) \cdot P(Y \leq X + 15)$.
  • If $Y \leq X$ in order for them to meet we need that $X \leq Y + 30$. Since $X$ and $Y$ are independent this happens with probability $P(Y \leq X) \cdot P(X \leq Y + 30)$.

The final probability that they meet is the sum of the above probabilities $$ P_{meet} = P(X \leq Y) \cdot P(Y \leq X + 15) + P(X \leq Y) \cdot P(X \leq Y + 30)$$

Ana Echavarria
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