小A与小B约好早晨在咖啡店见面.两人到达咖啡店的时间均为10am-11am.

且每人最多只等对方20分钟，求两人成功碰面的概率是多少？







记得好像用面积法做，具体忘了。。

求助

这道题就成这个问题

x1 and x2 follows uniform distribution U[0,1]
what is the probability that |x1-x2|>1/3?

case a:
0<=x1<1/3
then as long as x2<=x1+(1/3), |x1-x2|>1/3
so the probability is x1+(1/3)

case b:
1/3<=x1<2/3
then as long as x1-(1/3)<=x2<=x1+(1/3)
so the probability is 2/3

case c:
2/3<=x1<=1
similar to case a
probability is (4/3)-x1

then integrate all the probability, we get 5/9

if we use area method, consider (x1, x2) as a point in square [0，1]^2
then |x1-x2|<1/3 is bounded by two lines, x1-x2<1/3 and x2-x1<1/3

by subtracting the two triangles, the area between the lines is 5/9

done.

then the answer is a double integral on fa and fb

\int\int|ta-tb|<1/3 {f(ta)*f(tb)} dta dtb


use two step integral to do it

Suppose change the waiting time to 40 min and two person's arrival time follows uniform distribution, e.g.
f(x) = 1 (10 < x < 11)
f(y) = 1 (10 < y < 11)
Then you answer will be
\int_{10}^{11}\int_{y-2/3}^{y+2/3) 1 dx dy
= 4/3
??

\int_{0}^{1}\int_{max(0, x-1/3)}^{min(1, x+1/3)}f(x)*f(y)dydx

华新不支持latex encoding，大家写出来看，_{}表示subscript，^{}表示superscript

小A在第n分钟到达

n<=20
p=(n+20)/60

20<n<40
p=2/3

n>=40
p=(60-n+20)/60

画图是这样的
p
│
│ /￣￣\
│/　 \
│
└─────min
0 20 40 60
　

左上，右下两个40x40的直角三角形是不能见面的时间。
不能见面的概率 = (40x40) / (60x60) = 4/9

谢谢各位

可惜答案忘了

线性规划？

不会面积法,分三个二十分钟,从一个人的角度考虑.不过不知道对不对.
头二十分钟是1/3*1/3
第二个是1/3*2/3
最后也是1/3*2/3

假设两人到的时间X, Y是uniform distribution and independent. 可对让人不能相遇的概率进行分析。
可分为三种情况进行讨论：
当X <= 20, Y > x + 20, 让人不可相遇。P(Y > x + 20, X = x) = 1/60(40-x).
当20 < X <= 40,Y > x + 20 or Y < x-20 良人不可相遇。P(...) = 1/3;
当40 < X <= 60,Y < x-20 良人不可相遇。 P(...)=1/60(x-20);
最后可以得出两人不相遇的概率是5/9。答案为4/9