Question: > 
> $\textbf{Problem}$: An athletic facility has $5$ tennis courts. Pairs of players arrive at the
> courts and use a court for an exponentially distributed time with mean $40$ minutes.
> Suppose a pair of players arrives and finds all courts busy and k other pairs waiting in
> queue. What is the expected waiting time to get a court?
> 
> 
> 

> 
> $\textbf{Answer}$: As long as the pair of players are waiting, all five courts are occupied by other players. When all five courts are occupied, the time until a court is freed up is exponentially distributed with mean $40/5=8 $ minutes. For our pair of players to get a court, a court must be freed up $k+1$ times. Thus, the expected waiting time is $8(k + 1)$.
> 
> 
> 

The issue that I'm having in understanding is why does "The time until a court is freed up is exponentially distributed with mean $40/5=8$ minutes" rather than simply $40$ minutes. Since every court is independent and using the memoryless property.

On the contrary, if I look at $$Z=min(X\_{1}, X\_{2}, X\_{3}, X\_{4}, X\_{5})$$. where $X\_{i}$ is time taken by a pair to play in court $i$ and is exponentially distributed time with mean $40$ minutes.  

Then $Z$ itself becomes and exponentially distributed with mean $40\*5 = 200$ minutes.

I am unable to understand it intuitively. Any help is appreciated.

Answer: 
Each $X\_i$ has rate $\lambda=\frac1{40}$ (with minutes as the time unit), so $Z$ has $\lambda=\frac5{40}=\frac18$ for a mean of $8$ minutes. You have applied the formula incorrectly: the rates add, not the means.
Let $Z=\min(X\_1,...,X\_5)$.

We have $X\_1=Z$ with probability ${\large{\frac{1}{5}}}$, and by the memoryless property we have 
$$
E[X\_1-Z{\,\mid\,}X\_1 > Z]=40
$$
hence
\begin{align\*}
&
E[X\_1-Z]
=
\left({\small{\frac{1}{5}}}\right)\!(0)
+
\left({\small{\frac{4}{5}}}\right)\!(40)
\\[4pt]
\implies\;&
E[X\_1-Z]=32
\\[4pt]
\implies\;&
E[X\_1]-E[Z]=32
\\[4pt]
\implies\;&
40-E[Z]=32
\\[4pt]
\implies\;&
E[Z]=8
\\[4pt]
\end{align\*}