Probability of achieving Maximum Value in a Probability Density Function As part of my (incomplete) lecture notes I've been provided with this example but, alas, with no supplemented solution. As this is the first (& only) example following on from Theorem 1.4 (provided below) I have little idea of achieving the solution. Hopefully you can help me and explain to me your methodology as I'm a little lost currently and would really appreciate any extra help!
Firstly, the theorem: 

Theorem 1.4: Let $X_1, X_2, \dots , X_n$ be a random sample from a continuous distribution with cumulative distribution function $F(x)$ and probability density function $f(x)$. The probability density function of the maximum and minimum of $X_1, X_2, \dots , X_n$ are given by $g(z) = nf(z)[F(z)]^{n−1}$ and $h(w) = nf(w)[1−F(w)]^{n−1}$, where $z = \max(x_1,x_2,...,x_n)$ and $w = \min(x_1,x_2,...,x_n)$.

Example Question:

The random variable $X$ has probability density function $f(x)=12x^2(1−x)$ for $0 \leq x\leq 1$.
  Obtain the probability density function of the sample maximum, when a random sample of size $n$ is taken from $X$. Hence, or otherwise, find the probability that the largest maximum is $1$.

Thank you very much,
Mckenzie.
 A: I will not use the quoted theorem, but instead derive the distribution of the maximum $Y$ of the $X_i$. The same argument works more generally, and can be used to prove the max part of the theorem you quoted. 
The density function of the $X_i$ is $12x^2(1-x)$ on $(0,1)$ and $0$ elsewhere. We first find the cdf $F_Y(y)$ of $Y$.
The only interesting values of $y$ are $0\lt y\lt 1$, so we assume $y$ is in this interval. We have $Y\le y$ if and only if all the $X_i$ are $\le y$. The probability any individual $X_i$ is $\le y$ is
$$\int_0^y 12x^2(1-x)\,dx.$$
This is $4y^3-3y^4$.  So by independence the probability all the $X_i$ are $\le y$ is $(4y^3-3y^4)^n$.  Now we know $F_Y(y)$ and can differentiate to find the density function of $Y$. We get $12ny^2(1-y)(4y^3-3y^4)^{n-1}$.
Remarks: 1) The proof of the min part of the theorem you quoted goes along similar lines. For the probability the min of the $X_i$ is $\gt z$ is the probability all the $X_i$ are $\gt z$.
2) The probability that the maximum is exactly $1$ is $0$. That does not require the above machinery. For any random variable $W$ with continuous distribution, and any real $a$, we have $\Pr(W=a)=0$.
