Order Statistics Problem: Wackerly/Mendenhall/Scheaffer, 5th Ed., Problem 6.58 Problem Statement:
Suppose that the number of occurrences of a certain event in time
interval $(0,t)$ has a Poisson distribution. If we know that $n$ such events
have occurred in $(0,t),$ then the actual times, measured from $0,$ for the
occurrences of the event in question form an ordered set of random variables,
which we denote by $W_{(1)}\le W_{(2)}\le\cdots\le W_{(n)}.$ [$W_{(i)}$ actually
is the waiting time from $0$ until the occurrence of the $i$th event.] It can
be shown that the joint density function for $W_{(1)}, W_{(2)},\dots,W_{(n)}$
is given by
$$f(w_1, w_2,\dots,w_n)=
\begin{cases}
\dfrac{n!}{t^n},&w_1\le w_2\le\cdots\le w_n\\
0,&\text{elsewhere.}
\end{cases}
$$
[This is the density function for an ordered sample of size $n$ from a uniform
distribution on the interval $(0,t).$] Suppose that telephone calls coming
into a switchboard follow a Poisson distribution with a mean of ten calls per
minute. A slow period of $2$ minutes' duration had only four calls.

*

*Find the probability that all four calls came in during the first
minute; that is, find $P(W_{(4)}\le 1).$

*Find the expected waiting time, from the start of the $2$-minute period,
until the fourth call.

My Work So Far:
What's extremely confusing to me in this problem is how to interpret all the numbers I'm given. So I'm told the underlying Poisson distribution has $\lambda=10.$ Where does that figure into solving this problem, if at all? Then we're examining a slow period of $2$ minutes: where does that figure into solving this problem? Should $n=4$ in the joint density function above? Or should $n=10?$
I think if I could please have a nudge in the right direction for the first part, I imagine I could easily perform the integral to get the second.
Many thanks for your time!
 A: [EDIT] Only keeping this answer to make sense of the comments from whuber, which have the solution implicit in them.
Well, I've got an answer to Part 1, I hope:
We start with the underlying Poisson distribution:
$$p(y)=\frac{10^y\,e^{-10}}{y!},\; y=0, 1, 2,\dots$$
We will need the cumulative distribution function
$$F(y)
=P(Y\le y)
=\sum_{x=0}^y \frac{10^x\,e^{-10}}{x!}
=\frac{\Gamma(y+1, 10)}{\Gamma(y+1)}.$$
The function in the numerator is the upper incomplete gamma function.
This is a discrete distribution, so computing the order statistics is
completely different from continuous distributions. Following the wikipedia
article, we define
\begin{align*}
p_1(y)&:=P(Y<y)
=F(y)-p(y)
=\frac{\Gamma(y+1, 10)}{\Gamma(y+1)}-\frac{10^y\,e^{-10}}{y!}
=\frac{\Gamma(y+1, 10)-10^y\,e^{-10}}{y!}\\
p_2(y)&:=P(Y=y)
=p(y)
=\frac{10^y\,e^{-10}}{y!}\\
p_3(y)&:=P(Y>y)
=1-F(y)
=1-\frac{\Gamma(y+1, 10)}{\Gamma(y+1)}.
\end{align*}
Now then, for the order statistics, we have that
$$P(W_{(k)}\le y)=\sum_{j=0}^{n-k}\binom{n}{j}(p_3(y))^j 
(p_1(y)+p_2(y))^{n-j}$$
in general, so that
\begin{align*}
P(W_{(4)}\le 1)
&=\sum_{j=0}^{10-4}\binom{10}{j}(p_3(1))^j (p_1(1)+p_2(1))^{10-j}\\
&=\sum_{j=0}^{6}\binom{10}{j}(p_3(1))^j (p_1(1)+p_2(1))^{10-j}.
\end{align*}
Note that
\begin{align*}
p_1(1)&=e^{-10}\\
p_2(1)&=10\,e^{-10}\\
p_3(1)&=1-11\,e^{-10}.
\end{align*}
Then we simplify
\begin{align*}
P(W_{(4)}\le 1)
&=\sum_{j=0}^{6}\binom{10}{j}\left(1-11\,e^{-10}\right)^j \left(11\,e^{-10}\right)^{10-j}\\
&\approx 1.30308\times 10^{-11}.
\end{align*}
[Correct Answer]

*

*As whuber mentions in the comments, the probability of a call occurring in the first minute is $1/2$ with a uniform distribution. Hence the probability of all four calls occurring in the first minute is simply $1/16.$


*We compute the expected value of $W_{(4)}.$ To do so, we need the density and distribution for the uniform distribution:
\begin{align*}
f(t)&=
\begin{cases}
\dfrac12, &t\in[0,2]\\
0,&\text{elsewhere,}
\end{cases}\\
F(t)&=
\begin{cases}
0,&t<0\\
\dfrac{t}{2},&t\in[0,2]\\
1,&t>2.
\end{cases}
\end{align*}
Then, according to the work done in Section 6.6 of the book, the density function for the maximum order statistic is given by
\begin{align*}
g_{(4)}(t)
&=4\,[F(t)]^3 f(t)\\
&=
\begin{cases}
4(t/2)^3(1/2),&t\in[0,2]\\
0,&\text{elsewhere}
\end{cases}\\
&=
\begin{cases}
t^3/4,&t\in[0,2]\\
0,&\text{elsewhere.}
\end{cases}
\end{align*}
It follows that the expected value of $W_{(4)}$ is
$$\int_0^2 \frac{t^4}{4}\,dt=\frac85.$$
This makes sense: we would expect the value to be greater than the midpoint, but certainly not greater than $2.$
