Probability PDF CDF Help I'm looking at a STEP question and I'm a little confused by the logic of the method, and i'm really hoping someone could clarify what is going on for me. I have a good knowledge (At least I thought), as some STEP II and III questions are accessible but this one , I just can't wrap my head around - there must be a gap in my understanding for sure. 
It would be difficult for me to rewrite the question on here but here is the video of the question AND the solution, (it's not too long I assure you, just the first 2/3 minutes, it's the entry to the question that has got me scratching my head)
https://www.youtube.com/watch?v=YKjQ2RTfltU&frags=pl%2Cwn

An internet tester send $n$ e-mails simultaneously at time $t=0$. Their arrival times at their destinations are independent random variables each having probability density function $ke^{-kt}$ where $t>0$ and $k>0$. The random variable $T$ is the time of arrival of the e-mail that arrives first at its destination. Show that the probability density function of $T$ is $nke^{-nkt}$

I understand the calculation later on in the question for the expectation and even the very last part of the question, I just find it difficult to understand how it is logical to consider what this guy has considered in the first 2/3 minutes of the video. Is there a nicer explanation why he considered $P(T>t)$ and not $P(T<t)$, could it even be done this way? Is this a general method for this 'type' of question? 
I do see what he has done, and understand it, it's just something that I would never have even considered and makes me uncomfortable. 
Thank you for your help, and I do apologise for the awkwardness of the question.
 A: Hint: Let $Y_1, Y_2, \cdots, Y_n$ be independent random variables and let $Z = \min\{Y_1,Y_2,\cdots,Y_n\}$. Then
$$\Pr\{Z > z\} = \Pr\{\min\{Y_1,Y_2,\cdots,Y_n\} > z\} = \Pr\{Y_1>z\}\Pr\{Y_2>z\}\cdots \Pr\{Y_n>z\}.$$
In this type of problems, it's a good idea to consider $\Pr\{Y_i > z\}$.
A: There are two steps:


*

*The probability a particular email arrives by time $t$ is $1-e^{-kt}$ and that it does not arrive by time $t$ is $e^{-kt}$ - use integration of the pdf if necessary

*So the probability none of $n$ independent emails arrive by time $t$ is $\mathbb P(T \gt t)= \left(e^{-kt}\right)^n = e^{-nkt}$ and thus the probability the first arrives before time $t$ is $\mathbb P(T \le t)=1-e^{-nkt}$ - now differentiate this CDF to get the pdf for the first arrival time
You use $\mathbb P(T \gt t)$ at the start of the second step because you can calculate it from the first step  
If you had instead been calculating the pdf for the arrival times $S$ of the last email, you would not have used it, instead saying the probability all of $n$ independent emails arrive by time $t$ is $\mathbb P(S \le t)= \left(1-e^{-kt}\right)^n $ and then differentiating  
