\begin{problem}
Fifteen telephones have just been received at an authorized service center.  Five of these telephones are cellular, five are cordless, and the other five are corded phones.  Suppose that these components are randomly allocated the numbers 1, 2, … , 15 to establish the order in which they will be serviced.
\item
What is the probability that all the cordless phones are among the first ten to be serviced?\\

$\frac{(\binum{5}{5})(\binum{10}{5})}{(\binum{15}{10})}=\frac{\frac{5!}{5!}\cdot \frac{10!}{5!\cdot 5!}}{\frac{15!}{5!}{10!}}=\frac{\frac{10\cdot 9\cdot 8 \cdot 7 \cdot 6}{5\cdot 4\cdot 3\cdot 2}}{\frac{15\cdot 14\cdot 13\cdot 12\cdot 11}{5\cdot 4\cdot 3\cdot 2}}=.0839=8.39%$

\item
What is the probability that after serciving ten of these phones, phones of only two of the three types remain to be serviced?\\

The answer from above is for one type of phone.  Multipy it by 3 for each individual type of phone.  The result is 25.17%.

\item
What is the probability that two phones of each type are among the first six served?\\

$\frac{15\cdot 14\cdot 11\cdot 10\cdot 5\cdot 4}{15\cdot 14\cdot 13\cdot 12\cdot 11\cdot 10}=.1282=12.82%$

\end{problem}

\begin{problem}
Show that if one event A is contained in another even B (i.e. A is a subset of B), then $P(A)\leq P(B)$.  For general events A and B, what does this imply about the realtionship among $P(A\bigcup B), P(A) /text{and} P(AUB)$?\\
\item
Find $P(A\bigcup B)$.\\
$1-P([AUB]')=P(A)+P(B)-P(A\bigcup B) \hspace{2em} .9=.7+.5-P(A\bigcup B)\\
P(A\bigcup B)=.3\\$

\item
Give $P(A|B)$.\\
$\frac{P(A\bigcup B)}{P(B)}=\frac{.3}{.5}=.6$

\item
Give $P(B|A)$.\\
$\frac{P(A\bigcup B)}{P(A)}=\frac{.3}{.7}\approx .43$

\end{problem}


\begin{problem}
On any given evening, a raccoon shows up in Selma's yard with probability 25%.  Assume that appearances of the raccoon are independent from day to day.\\
\item
What is the probability that the raccoon does not appear at all over a ten-day period?\\
$.75^{10}=.0563=5.6$


\item
What is the probability that the raccoon first appears on the tenth day?\\
$.75^9\cdot .25^1=.0188=1.88%$

\item
What is the probability that the raccoon appears exactly once in 10 days?\\
The same as above because it doesn't matter which day it shows up on.\\

\item
Given that the raccoon appeared on the first day, what is the probability that the raccoon will appear again during the ten-day period?\\
$1-.75^9=.9249=92.49%$\\
The first day doesn't matter because the appearances are independent.