Amit and Therma meet in a coffee house 5 days in a row. What is the probability that at least once the two will arrive 30 minutes or more apart? 
Amit and Therma meet in a coffee house 5 days in a row, and each day they arrive randomly between 2 and 3 (10 arrivals are independent.  What is the probability that at least once the two will arrive 30 minutes or more apart?

I know I need to break this down into finding the probability that they meat more than 30 mins apart on a daily basis, and only then can I find the probability for 5 days. 
What is the process for finding what the probability is on one day?
 A: From MathWorld we know that the probability density function for a distance d between two randomly picked points on the unit interval is $2(1-d)$. Here the interval is 1 hour and we require the probability of Amit and Therma arriving more than half that time apart, so compute an integral:
$$\int_{\frac12}^1 2(1-x)dx=\frac14$$
This is the one-day probability, and the answer to the original question is straightforward:
$1-(\frac34)^5=\frac{781}{1024}$.
A: There is a nice graphical approach. Using independence, $$P(\ge \text{1 miss}) = 1 - P(\text{meet every day}) = 1 - P(\text{meet on day 1})^5.$$
On each given day, we can represent Amit's and Therma's arrival times as a pair of numbers $(x,y)$ with $0 \le x, y \le 1$ (e.g. (0.5,0.5) means they both arrive at 2:30 pm). Arriving less than 30 minutes apart corresponds to $|x-y| < 0.5$, which is an infinite diagonal band in the $xy$-plane (try graphing it). Its intersection with the square $[0,1]^2$ has an area of $3/4$, so $P(\text{meet on day 1}) = 3/4$.
(I calculated this area by noting that it is the complement in the square of two congruent right-angled triangles. See this quite crude picture.)
A: I would start by giving a name to the quantities we try to find or say something about. So, let
$$ 
A = \text{ arrival time of Amit} \\
T  = \text{arrival time of Thelma}. $$
Not $A,T$ are independent, and "randomly distibuted between 2 and 3." The last part of the statement is somewaht vague. The expression is customarily used in the sense of "uniformly distributed between 2 and 3", so let's go with that.  
We now know the join distribution of the arrival times, since the independence conditions tells us that the join distribution is the product of the individual distributions, thus:  
$
f(x,y) = \text{probability density function of $A, T$ at $(x,y)$ }= \begin{cases}
1 &\text{if $2<x<3$ and $2<y<3$} \\
0 &\text{otherwise.}
\end{cases}
$
Now, $P( |A-T| > 0.5) = \int_{|x-y|>.5} f(x,y) dxdy .$  
And so on.
