Maximum of Three Uniform Random Variables Here is the question, I am studying for exam P, and am using a study guide with a solution guide. I am stumped on this problem, and the solution in the back was very confusing. Any clarity I can get would be tremendously appreciated. Here is the problem:
Three individuals are running a one kilometer race. The completion time for each individual is  a random variable. $X_i$ is the completion time, in minutes, for person $i$.
$X_1$: uniform distribution on the interval [2.9, 3.1]
$X_2$: uniform distribution on the interval [2.7, 3.1]
$X_3$: uniform distribution on the interval [2.9, 3.3]
The three completion times are independent of one another.
Find the expected latest completion time (nearest .1)
I let $Y = \max\{X_1, X_2, X_3\}$
$F_Y(y) = P[Y \leq y] = P[\max\{X_1, X_2, X_3\} \leq y]$
So, in short, $5(y-2.9)(2.5)(y-2.7)(2.5)(y-2.9)$, 
However, the author uses a piecewise function, with what I have above, but he has two different intervals which I don't understand where he got. This is what he has:
$$F_Y(y) = \begin{cases} 5(y-2.9)(2.5)(y-2.7)(2.5)(y-2.9), &  2.9 \leq y \leq 3.1 \\ 2.5(y-2.9), & 3.1 \leq y \leq 3.3 \\
\end{cases}$$
Sorry, for the jumbled piecewise function, I'm not sure how to space it out more.
My questions, is how did he get the second part where the interval is $3.1 \leq y \leq 3.3$

If $y > 3.1$, then you know that the first two runners are already done.  So the probability that the last runner finishes in at most $y$ minutes is the probability that $X_3 \le y$; i.e., $$\Pr[Y \le y] = \Pr[X_3 \le y] = \frac{y-2.9}{3.3-2.9}, \quad 3.1 < y \le 3.3.$$
