\problemdata                                                             
        {Exponential Distribution} % latex-friendly label for the prob.
        {Recurrence} % The topic(s) of the problem content.  
        {} % Source (if known)                                          
        {S01 Final, 14} % Usage (list ones you are aware of). 
        {Tina Wang, S02} % Last revision info (author, date).   
                                                                         
                                                                         
\begin{problem}           
The 042 staff is taking a post-class vacation trip to California in
Professor Karger's car.  The tires on this car have an expected
lifetime of 20,000 miles before going flat.  Model the distances the
tires can go before going flat as independent, exponentially
distributed random variables.

\begin{problemparts}
\problempart A skilled driver, Professor Karger can ignore the first
flat tire and keep going until he gets a second flat.  What is the
expected distance Professor Karger can expect to drive before
stopping?

\solution{ 
For one tire, $\lambda=\frac1{20,000}$. For 4 tires, $4\lambda=\frac{1}{5,000}$. The expected distance before the \emph{first} tire goes flat is $\frac{1}{4\lambda}=5,000$ miles. After that, the expected distance for the second tire to go flat will be $\frac{1}{3\lambda}=\frac{20,000}{3}$. Therefore, Professor Karger can expect to drive for \[
5,000+\frac{20,000}{3}=11,667\]
miles before stopping.}



\problempart What is the probability that Professor Karger will have to
stop before the end of the 3000 mile trip to California?

\solution{ 
Let $p$ denote the probability that a particular tire will last for $\geq 3,000$ miles. Then:
\[
p=e^{-\frac{3,000}{20,000}}=e^{-\frac{3}{20}}=0.861
\]
The probability that Prof. Karger will \emph{not} have to stop is equal to the probability that he gets \emph{at most 1} flat tire during the trip, which is Pr(no flat tire)+Pr (exactly 1 flat tire). Therefore, the probability that he \emph{will} have to stop is 
\[
1-p^4-\binom{4}{3}p^3(1-p)=0.0959\]
Of course we get the same answer if we calculate the probability that \emph{at least} 2 tires go flat, which is: Pr(exactly 2 flat tires)+Pr(exactly 3 flat tires)+Pr(exactly 4 flat tires), or in other words
\[
(1-p)^4+\binom{4}{3}(1-p)^3p+\binom{4}{2}(1-p)^2p^2=0.0959
\]
 
} 
\problempart Thanks to his intensive training on the icy slopes of
Swedish mountains, Dr. Engebretsen can even handle two flat tires, so
long as they occur on the same side of the car.  How long can Dr.
Engebretsen expect to drive before two flat tires \emph{on opposite
  sides of the car} force him to pull over?
\solution{ 
The expected distance before the \emph{first} tire goes flat is $\frac{1}{4\lambda}=5,000$ miles. After that, the expected distance before \emph{one} of the two tires on the opposite side of the car goes flat will be $10,000$ miles. So the total is $15,000$ miles.
}


\end{problemparts}
\end{problem}
                                 