# Intuitive Understanding of exponential distrbution

$$\textbf{Problem}$$: 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.

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*}