How to benefit from the expected value in this question? The question is :
There are $n$ jobs and $n$ machines. $T_i$ is the random variable for machine $i$ to complete its execution. Let $X= \max\{T_1 , \dots , T_n \}$ and let $Y = \min\{T_1 , \dots , T_n \}$ $T_i - \exp(\lambda_i)$.
Let $n=5$ and let the average time for job $n$ to be executed by $20$(ms). What is the probability that the total time is below $15$ms for $X$ and $Y$ respectively?
I need to know how I can use average time in this question to get to the result. It seems to me that I am missing how expected value can be used in this kind of question. Any advice is highly appreciated.  
 A: In general, if $Z_1,\ldots,Z_n$ are i.i.d. random variables, denote 
\begin{align}
Z_\max &:= \max\{Z_1,\ldots,Z_n\},\\
Z_\min &:= \min\{Z_1,\ldots,Z_n\}.\\
\end{align}
For any $t\in\mathbb R$ we have
$$
\{Z_\max \leqslant t\} = \bigcap_{i=1}^n \{Z_i\leqslant t\}.
$$
It follows from independence that
$$
\mathbb P(Z_\max \leqslant t) = \mathbb P\left(\bigcap_{i=1}^n \{Z_i\leqslant t\}\right)
=\prod_{i=1}^n \mathbb P(Z_i\leqslant t)
$$
and from being identically distributed that
$$
\mathbb P(Z_\max \leqslant t) = \mathbb P(Z_1\leqslant t)^n = F_Z(t)^n,
$$
where $F_Z$ is the distribution function of $Z_1$. Similarly,
$$
\{Z_\min > t\} = \bigcap_{i=1}^n\{Z_i > t\},
$$
so that
\begin{align}
\{Z_\min > t\} &= \mathbb P\left(\bigcap_{i=1}^n\{Z_i > t\}\right)\\
&= \prod_{i=1}^n \mathbb P(Z_i>t)\\
&= \mathbb P(Z_1>t)^n\\
&= (1 -F_Z(t))^n,
\end{align}
so that $\mathbb P(Z_\min \leqslant t) = 1 - (1-F_Z(t))^n$. Now let $T_1,\ldots,T_n$ be independent exponentially distributed random variables with parameter $\lambda$. Then
\begin{align}
\mathbb P(T_\max \leqslant t) &= (1-e^{-\lambda t})^n,\\
\mathbb P(T_\min \leqslant t) &= 1 - e^{-n\lambda t}.
\end{align}
Note that $T_\min$ itself is exponentially distributed, with parameter $n\lambda$. To find the answer for your specific question, plug in $\lambda=1/20$, $n=5$, and $t=15$.
