Random variable representing distance of first random point I came across the following question in a probability book by A. Papoulis.
We place at random 200 points in the interval (0, 100). The distance from 0 to the first random point is a random variable $z$. Find the cdf $F_z(z)$.
My approach was following.
Probability of a point belonging the $(0, z)$ i.e.
P(A) = $\frac{z}{100}$
Then probability that the first point is $z$ distance away from 0 is
$f_z(z)$ = P({zero points fall in interval (0, $z$)} $\cap$ {99 points fall in interval ($z$, 100)})
i.e. $f_z(z) = \left(1-\frac{z}{100}\right)^{n-1}$
Can you help me understand what I am doing wrong?
 A: You did nothing wrong but you need to continue with your expression. What is missing is just the normalization.
Let more generally
$$f(z,n,k) = \frac{1}{A} (1-\frac{z}{n})^k$$
Here $A$ is a nomalization constant.
In your case we have $n=100$, $k=200$.
In order for $f$ being a probability distribution the sum must be equal to unity, i.e. we must have
$$A = \sum_{z=0}^n  (1-\frac{z}{n})^k$$
The sum can be done in closed form using special functions with the result
$$A = 1+\left(-\frac{1}{n}\right)^k \left(\zeta (-k,1-n)-\zeta (-k)\right)$$
Here $\zeta(x,y)$ the the Hurwitz zeta function and  $\zeta(x)$ the Riemann zeta function.
With $n=100$ and $k=200$ we find numerically $A \simeq 1.15415$, and the first few numerical values of the distribution function (in the format $(z,f)$) are
$$((0, 0.866436), (1, 0.116085), (2, 0.0152388), (3, 0.00195922), (4, 
  0.000246594), (5, 0.0000303709))$$
The expectation value is $E(z)\simeq 0.153604$
Original post
I have misunderstood the Problem of the OP. Hence the following decribes the solution to a different problem which might be of interest as well. 
§1. The discrete case
There are $\binom{n}{k}$ possibilities to place $k$ points in $n$ different positions $1,2,3, ..., n$.
If you fix one point closest to $0$ at position $m$, you can place the remaining $(k-1)$ points in the remaining $n-m$ positions from $m+1$ to n in $\binom{n-m}{k-1}$ manners.
Hence the probability that the position of the point closest to $0$ is $m$ is gven by 
$$p_m = \binom{n-m}{k-1}/\binom{n}{k}\tag{1}$$
The expectation of $m$ is
$$E(m) = \sum_{m=1}^n m p_m = \frac{n+1}{k+1}$$
that of $m^2$
$$E(m^2) = \sum_{m=1}^n m^2 p_m = \frac{n+1}{k+1}=\frac{(n+1) (2 n-k+2)}{(k+1) (k+2)}$$
And the central second moment is
$$\sigma^2 = E(m^2)-E(m)^2 = \frac{k (n+1) (n-k)}{(k+1)^2 (k+2)}$$
2. The continuous case
This case consists in finding the distribution function (pdf) of the minimum of $k$ independent random varables $x_{i}, i=1,...,k$ distributed according to $U(0,1)$.
Due to a lack of time at the moment I provide just the result
$$f(k,w)  = k (1 - w)^{k - 1}\tag{2}$$
The first two moments of the minimum $w$ are
$$E_c(w) = \int_{0}^1 w f(k,w)\,dw = \frac{1}{k+1}$$
$$E_c(w^2) = \int_{0}^1 w^2 f(k,w)\,dw = \frac{2}{(k+1) (k+2)}$$
And the central second moment is
$$\sigma_{c}^2=E(w^2)-E(w)^2 = \frac{k}{(k+1)^2 (k+2)}$$
§3. Discussion
The moments for the continous case are the limits for $n\to \infty$ of those of the discrete case:
$$ E_c(w) = \lim_{n\to \infty} \frac{1}{n} E(m) $$
$$ E_c(w^2) = \lim_{n\to \infty} \frac{1}{n^2} E(m^2) $$
$$ \sigma_{c}^2 = \lim_{n\to \infty} \frac{1}{n^2} \sigma^2 $$
A: The distance to the first point is less than z if and only if there is at least one point in the interval (0,z].
If x is the distance to the first point x < z => there is at least one point in the interval (0,z]
Let n(x,z) be the number of points in the interval (x,z], then
$F_{Z}(z) =P(x \leq z) = P(n(0,z) > 0) = 1 - P(n(0,z) = 0)$
$= 1-(1-\frac{z}{100})^{200}$
