You are not logged in.
Pages: 1
2 people decided to meet each other on a certain day between 5 and 6 pm. They arrive (independently) at a uniform time between 5 and 6 and wait for 15 minutes. What is the probability that they meet each other?
y
_ _ _ _
| / |
| / /|
| / / |
|/ / |
|_ /_ _ | x
The rectangle (actually square) represent the distribution region of the combination of random variables X and Y, where X=the arrival time (in hour) of the first person and Y=the arrival time of the other person. Since they are independent of each other, combined distribution f(x,y)=f(x)f(y)=1/1(1/1) on Rectangle (5,5)-(6,5)-(6,6)-(5,6).
Now please notice the diagonal section (5,5)-(5.25,5)-(6,5.75)-(6,6)-(5.75,6)-(5,5.25) which is the middle part of the rectangle cut by two lines y-x=1/4 and x-y=1/4. And this concludes all the possiblities that x and y arrive within 1/4 hour to each other - y either late or earlier than x for no more 1/4 hour.
By integrating the distribution f(x,y)=1 over the region you will get the probability.
(Note: They will never arrive exactly at the same time (y=x) since the probability will be zero- rediculuous? Only because you are applying the spooky continuous model)
X'(y-Xβ)=0
Offline
Pages: 1