If A is distributed uniformly on [8,10] and B on [9,11], what is the probability that BI was asked this question in an interview, and did not initially answer correctly though I still think my interpretation may have been the correct one. The question was:

There are two delivery trucks, A and B. A makes deliveries between 8am and 10am, and B makes deliveries between 9am and 11am. The deliveries are uniformly distributed for both. What is the probability that any given delivery from B will take place before any delivery from A?

What is your answer, and why?
 A: Since the delivery rates are not specified, lets assume A delivers $a$ packages per hour and B delivers $b$ packages per hour. So there are $2a \cdot 2b$ pairs of delivery times. The window in which A and B overlap in deliver times has only $a \cdot b$ pairs, in half of which A comes before B. So the proportion of pairs in which A comes before B is
$$
\frac{a\cdot b}{2}\frac{1}{2a \cdot 2b} = \frac{1}{8}.
$$
A: It's 1/8. See the figure below, which shows A's delivery time on the x-axis and B's on the y-axis. Since deliveries are uniformly distributed, all points in the square are equally likely to occur. B delivers before A only in the shaded region, which is 1/8 of the total figure.

Another way to think of it is that there's a 50% chance A delivers before B even starts, and 50% chance that B delivers after A is done, meaning there's a 75% chance of one or both of those happening. In the 25% chance they both deliver in the overlapping hour, it's a 50-50 chance of which delivers first.
A: I propose another way of looking at it, only if you had a pc during the interview of course.
We can simulate the process with R, for example.
Let's simulate 1000 values from A and the same from B, we know that both are uniforms, are independent.
a <- runif(1000, 8, 10) # A deliveries
b <- runif(1000, 9, 11) # B deliveries
# [1] 9.485513 8.665070 8.488481 8.840332 8.755384 9.448949 # A deliveries for example

Ok they're not exactly times, but it's the same.
The probability $P(B<A)$ is what we seek. So we just count the number of pairs where $b<a$ in our code.
prob <- sum(b < a)/1000
#[1] 0.112 # almost 1/8

We can also plot the 1000 pairs $(a,b)$, and see the region where B comes first.
plot(a, b)
polygon(c(9, 10, 10, 9),
        c(9, 9, 10, 9), density = 10, angle = 135)


And the prob value above is the proportion of points in the shaded region (looks familiar doesn't it?).
Now we could use the formula for the standard error of a proportion to estimate the standard error of the simulation.
se <- sqrt(prob * (1 - prob) / 1000)
#[1] 0.009972763

And we can build a CI (assuming Normal approximation of the sample distribution of probs).
prob - 1.96*se
#[1] 0.09245338 lower bound
prob + 1.96*se
#[1] 0.1315466 upper bound

A: Stumbled across this and it got in my head.  :-)
The answer seems like it must depend on the relative number of deliveries each truck makes in the hour of possible overlap (9a-10a) -- there's no constant answer.  
For example, suppose each truck makes 2 total deliveries (1 per hour).  They'd each make 1 delivery between 9 and 10 and B wouldn't beat anything from A.  So, the probability is 0 in that case.
Consider a simplified version of the problem where they both only make deliveries between 9-10a (still a uniform distribution).  And, for starters, suppose they make the same number of deliveries, n.


*

*The first delivery for B will beat everything except the first delivery from A (which it ties).  So, with probability $\frac{1}{n}$ (the probability we're the first delivery for B) we beat an event with probability $\frac{n-1}{n}$ (the probability we're not the first delivery of A) 

*The second delivery for B will beat everything except the first two deliveries from A.  So, with probability $\frac{1}{n}$ we beat an event with probablity $\frac{n-2}{n}$ 

*etc.


Putting each of those terms into a summation, we get:
$(\frac{1}{n} \cdot \frac{n-1}{n}) + (\frac{1}{n} \cdot \frac{n-2}{n}) + ...  + (\frac{1}{n} \cdot \frac{n-n}{n})$
Or,
$\sum_{i=1}^{n} \frac{n-i}{n^2}$
Since the probabilities are uniform and half (rounded down) of each occur during the hour of overlap, we only consider half the deliveries of each.  If $n'=\lfloor\frac{n}{2}\rfloor$ and, compared to the whole domain, those events only happen half the time.  So
$\frac{1}{2}\sum_{i=1}^{n'} \frac{n'-i}{n'^2}$
I believe that for $a=b=n$, you get $1/8$.
How to handle the fact A and B do not deliver the same number of packages? Again, to simplify, assume all their deliveries happen between 9-10am.
For every delivery $b$ you consider from earliest to latest, instead of each successive one beating $\frac{1}{a}$ less from truck A, as above (where $a$ is the number of deliveries made by truck A and $b$ the number of deliveries made by $b$), you eliminate $\lfloor \frac{1}{b} \cdot a \rfloor $.  That is, you beat all but a fraction of $a$ proportional to the fraction of $b$ you've thrown out.  So,
$
(\frac{1}{b} \cdot \frac{a - \lfloor 1 \cdot \frac{a}{b} \rfloor }{a}) + (\frac{1}{b} \cdot \frac{n - \lfloor 2 \cdot \frac{a}{b} \rfloor }{a}) + ... + (\frac{1}{b} \cdot \frac{a - \lfloor a \cdot \frac{a}{b} \rfloor }{a})  
$
Or,
$\sum_{i=1}^{b} \frac{a - \lfloor \frac{ia}{b} \rfloor  }{ab}$
Again, accounting for the fact that they only overlap half the time, let $a'=\frac{a}{2}$ and $b'=\frac{b}{2}$:
$\frac{1}{2}\sum_{i=1}^{b'} \frac{a' - \lfloor \frac{ia'}{b'} \rfloor  }{a'b'}$
A: It is zero:
if the B truck has at least one delivery in the period of [9-11] at least one delivery is made after (or equal to ) 10
and that delivery is not before the deliveries of A (which are all before 10)
