Car Rental Shop: Probability, Income, and Expansion Analysis

[anemone]

A car rental shop has four cars to be rented out on a daily basis at $50 per car. The average daily demand for cars is four.

a.	Find the probability that on a particular day,
	i. no cars are requested,
	ii. at least four requests for cars are received.
b.	Calculate the expected daily income received for the rentals.
c.	If the shop wishes to have one more car, the additional cost incurred is $20 per day. Determine whether the shop should buy another car for rental.

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[anemone]

Unfortunately no one submitted the full correct answer to last week POTW. However, an honorable mention goes to both I like Serena and lfdahl for getting the first part of the problem correct.:)

Solution:

Spoiler

a.	Let $X$ be the number of demands/requests of car on a particular day. We then noticed $X$ follows the Poisson distribution with mean $4$, i.e. $X\sim P_0(4)$.
	$ \begin{align*}\text{i.}\,\, P(\text{no cars are requested})&=P(X=0)\\&=\dfrac{e^{-4}(4^0)}{0!}\\&=0.0183\end{align*}$
	$\begin{align*}\text{ii.}\,\, P(\text{at least four requests for cars are received})&=P(X\ge 4)\\&=1-P(X\le3)\\&=1-e^{-4}\left(\dfrac{(4^0)}{0!}+\dfrac{(4^1)}{1!}+\dfrac{(4^2)}{2!}+\dfrac{(4^3)}{3!} \right) \\&=1-0.0183-0.0733-0.1465-0.1954\\&=0.5665\end{align*}$

b.

$x$	0	1	2	3	$\ge 4$
$P(X=x)$	0.0183	0.0733	0.1465	0.1954	0.5665

$\displaystyle E(x)=\sum_{\text{all x}} x\cdot P(X=x)=0(0.0183)+1(0.0733)+2(0.1465)+3(0.1954)+4(0.5665)=3.2185$

$\text{Hence the expected daily income}=(3.2185)(50)=\$160.925$

c.

$x$	0	1	2	3	4	$\ge 5$
$P(X=x)$	0.0183	0.0733	0.1465	0.1954	0.1954	0.3711

$\displaystyle E(x)=\sum_{\text{all x}} x\cdot P(X=x)=0(0.0183)+1(0.0733)+2(0.1465)+3(0.1954)+4(0.1954)+5(0.3711)=3.5896$

$\text{Hence the expected daily income}=(3.5896)(50)=\$179.48$

$\text{additional daily income}=\$179.48-\$160.925=\$18.555$

Since the added daily income $\$18.555$ is less than the added daily cost $\$20$, the shop should make the wise decision by not buying another car for rental.