# Probability of two events in an hour more than 20 mins apart

1. Aug 13, 2012

1. The problem statement, all variables and given/known data

Suppose that two events occur in an hour, and the probability is uniformly distributed. If the time that the first event occurs has the same distribution as the time that the second event occurs, and the two distributions are independent, what is the probability that the second event occurs at least 20 minutes after the first?

2. Relevant equations

The distribution is uniform (i.e. pmf=1/60 for each minute).

3. The attempt at a solution

I'm not sure how to approach this problem. I couldn't find a pattern with this table:
Time of E1, probability of E1 during that time, probability E2 is 20 mins before E1
0..20, 20/60, 0/60
21, 1/60, 1/60
22, 1/60, 2/60
23, 1/60, 3/60,
...

2. Aug 13, 2012

### Ray Vickson

The times of the earlier and later events are _not_ uniform. You need the concept of *order statistics*; see, eg., http://mathworld.wolfram.com/OrderStatistic.html or
http://en.wikipedia.org/wiki/Order_statistic . In fact, if X1 and X2 are the two times, and the order statistics are T1 = min(X1,X2) [first] and T2 = max(X1,X2) [second] you want P{T2 >= T1 + 20}. You will need some information about the bivariate distribution of (T1,T2).

RGV

3. Aug 13, 2012