probability distribution of the maximum 
Let T be a random variable giving the time to failure of led lights that follow exponential distribution with a mean value of 15 000 hours.
We put three new lights at the same time. Find the cumulative distribution of $T_{max}$ and then find its probability density function.

I assume the cumulative distribution will be a product of $(1-e^{(-1/15000)t_1}) (1-e^{(-1/15000)t_2})(1-e^{(-1/15000)t_3})$ but I am not sure how to get the probability density function. Any hints?
 A: It's important to start off with properly defined variables and events, and to develop the calculations carefully, so you can see where the mistakes are (don't jump steps, every time you did, you had mistakes that were then hard to spot). Something like this:
Let $T_1$ be the time until failure for light 1, and similarly define $T_2$ and $T_3$.
Let $W$ be the time until all three lights fail.
The event $W> w$ is equivalent to $(T_1 > w)\cap(T_2 > w)\cap(T_3 > w)$. Hence
$\:P(W > w)=P((T_1 > w)\cap(T_2 > w)\cap(T_3 > w))$
$\qquad\qquad\quad=P(T_1 > w)\,\cdot\,P(T_2 > w)\,\cdot\,P(T_3 > w)$ (independence).
Now $P(T_1>w) = 1-P(T_1\leq w)=...$ et c.
Then work out $P(W\leq w)$ from that.
When you can do it like this without any risk of an error, there are some shortcuts you can get away with ... but to be honest, I tend to use few shortcuts on problems of this sort. In the end, it's faster to do it properly the first time than to redo it three (or more) times.
(Nevertheless, I managed to make a mistake here; I originally had $\leq$ where I should have had $>$. It was, however, easier to correct because what was there was more explicit.)
