约会概率

令 $S$ 为学生在 4:00 后的分钟数，$S\sim\mathrm{Unif}(0,60)$；令 $G$ 为 Gabe 在 4:00 后的分钟数，$G\sim\mathrm{Unif}(30,60)$。

能见面等价于 $|G-S|\le 10$。

在 $(G,S)$ 平面上的可行区域面积占比计算可得概率为

$$\frac{11}{36}.$$

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Original Explanation

First, let's model this mathematically. Let $G$ and $S$ be the number of minutes after 4 PM that Gabe and the student show up, respectively. Then we have that $G \sim \text{Unif}(30,60)$ and $S \sim \text{Unif}(0,60)$. The event that the meeting occurs is the just the event that $|G-S| \leq 10$, as they must differ by up to $10$ minutes.

We should draw out the region in the plane to see what we are working with. If you put $G$ on the $x$-axis and $S$ on the $y$-axis, we have a tall rectangle. When drawing the two lines corresponding to $|G-S| \leq 10$, you will notice that it is easier to calculate the probability of the complement, as those are easier regions to work with. You will see that there is one trapezoid and one triangle. It is easy enough to use some basic algebra to note that the slope is $1$, so the line will go as far vertically as it does horizontally. You will find that the base of the trapezoid is $30$, and the two heights are $20$ and $50$, so its area is $1050$. The two sides of the triangle are $20$ each, so the area is $200$. Thus, the probability of the complement is $\dfrac{25}{36}$. Thus, the probability in the question is $\dfrac{11}{36}$.