Distribtution of the maximum of three uniform random variables. How do I get the cumulative density function of $Y$? $X$ is a continuous random variable with pdf
$$f(x) = 1,\quad   0 < x < 1. $$
Three independent observations of $X$ are made. Find the pdf of $Y$ where $$Y = \max\{X_1,X_2,X_3\}.$$
 A: Start by writing the cdf of $Y$ and make the appropriate substitution
$$P(Y\leq y) = P(\max\{X_1,X_2,X_3\} \leq y).$$ 
Notice that
$$\{\max\{X_1,X_2,X_3\}\leq y\}\iff\{X_1\leq y,X_2\leq y,X_3\leq y\}$$
Then
\begin{align*}
P(\max\{X_1,X_2,X_3\} \leq y)&= P(X_1\leq y,X_2\leq y,X_3\leq y)\\
&= P(X_1\leq y)P(X_2\leq y)P(X_3\leq y)\tag 1\\
&=y^3
\end{align*}
where in $(1)$ I used the the fact that the $X_i$ are independent.
This gives that the pdf of $Y$ is
$$f_Y(y) = 3y^2.$$
A: The pdf or cdf you are looking for is part of what is called the "order statistics": https://en.wikipedia.org/wiki/Order_statistic
The easiest path is through the cdf, using the following transformations:
$F_Y(y) \ = \ P(Y\leq y) \ = P(\max{\{X_1,X_2,X_3\}} \leq y)$
$F_Y(y) \ = \ P(X_1 \leq y \ \& \ X_1 \leq y \ \& \ X_1 \leq y).$
Using independence of the $X_k$:
$F_Y(y) \ = \ P(X_1 \leq y) P(X_2 \leq y) P(X_3 \leq y)=(y)^3$
It suffices then to derive wrt $y$ to obtain the pdf:
$$f_Y(y)=3~y^2$$
Edit : Very sorry : I made a mistake that I have corrected now (confusion between $\min$ and $\max$)
