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.

[TABLE="width: 800"]
[TR]
[TD]a.[/TD]
[TD]Find the probability that on a particular day,[/TD]
[/TR]
[TR]
[TD][/TD]
[TD]i. no cars are requested,[/TD]
[/TR]
[TR]
[TD][/TD]
[TD]ii. at least four requests for cars are received.[/TD]
[/TR]
[TR]
[TD]b.[/TD]
[TD]Calculate the expected daily income received for the rentals.[/TD]
[/TR]
[TR]
[TD]c.[/TD]
[TD]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.
[/TD]
[/TR]
[/TABLE]--------------------
Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[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

[TABLE="width: 1000"]
[TR]
[TD]a.[/TD]
[TD]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. $X\sim P_0(4)$.[/TD]
[/TR]
[TR]
[TD][/TD]
[TD]$ \begin{align*}\text{i.}\,\, P(\text{no cars are requested})&=P(X=0)\\&=\dfrac{e^{-4}(4^0)}{0!}\\&=0.0183\end{align*}$[/TD]
[/TR]
[TR]
[TD][/TD]
[TD]$\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*}$[/TD]
[/TR]
[/TABLE]

b.

[TABLE="class: grid, width: 600"]
[TR]
[TD]$x$[/TD]
[TD]0[/TD]
[TD]1[/TD]
[TD]2[/TD]
[TD]3[/TD]
[TD]$\ge 4$[/TD]
[/TR]
[TR]
[TD]$P(X=x)$[/TD]
[TD]0.0183[/TD]
[TD]0.0733[/TD]
[TD]0.1465[/TD]
[TD]0.1954[/TD]
[TD]0.5665[/TD]
[/TR]
[/TABLE]

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

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

c.

[TABLE="class: grid, width: 600"]
[TR]
[TD]$x$[/TD]
[TD]0[/TD]
[TD]1[/TD]
[TD]2[/TD]
[TD]3[/TD]
[TD]4[/TD]
[TD]$\ge 5$[/TD]
[/TR]
[TR]
[TD]$P(X=x)$[/TD]
[TD]0.0183[/TD]
[TD]0.0733[/TD]
[TD]0.1465[/TD]
[TD]0.1954[/TD]
[TD]0.1954[/TD]
[TD]0.3711[/TD]
[/TR]
[/TABLE]

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

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

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

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.