Suppose that you roll one fair die. \begin{itemize} \item Let $A$ be the event that the number shown is even. \item Let $B$ be the event that the number shown is a multiple of three. \end{itemize} \begin{problemparts} \problempart Suppose that the die has six sides numbered $1, 2, \ldots, 6$. Are events $A$ and $B$ independent? Justify your answer. \vspace{2in} \problempart Suppose that the die has four sides numbered 1, 2, 3, and 4. Are events $A$ and $B$ independent in this case? Justify your answer. \vspace{2in} \problempart Suppose that the die has eight sides numbered $1, 2, \ldots, 8$. Are events $A$ and $B$ independent now? Justify your answer. \end{problemparts} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newpage \problem Eight people go out for dinner at a restaurant and sit down at a round table. Everyone orders a different meal, and soon the waiter returns with those eight dishes. Unfortunately, he has largely forgotten who ordered what. He manages only to serve {\em three} of the eight people what they ordered; each of the other five diners gets someone else's meal. One diner suggests that they spin the table and see if chance can do better than the befuddled waiter. If they spin the table to an orientation \textbf{different from the original one} selected uniformly at random, what is the expected number of people that end up with the correct meal in front of them? %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newpage \problem {\bf [15 pts]\ } In a certain course, a student's exam is graded according to the following, rigorous procedure: \begin{itemize} \item With probability $\frac{1}{10}$, the exam is accidentally dropped behind the radiator and the student is arbitrarily given an 84. \item If the exam is not lost, then with probability $\frac{6}{7}$ it is graded by a tutor, and with probability $\frac{1}{7}$ it is graded by a lecturer. \item Tutors grade by scoring each problem and taking the sum. \begin{itemize} \item There are ten true/false questions worth 2 points each. Each question is marked wrong with probability $\frac{1}{3}$. \item There are four problems worth 15 points each. The score on each of these problems is determined by taking the sum of two rolls of a fair, six-sided die and adding 3. \item The single 20 point question is awared either 9 points or 17 points with equal probability. \end{itemize} \item Lecturers grade based on overall impression: \begin{itemize} \item The exam is initially scored 40 with probability $\frac{3}{10}$. \item The exam is initially scored 50 with probability $\frac{5}{10}$. \item The exam is initially scored 60 with probability $\frac{2}{10}$. \item The lecturer then rolls two fair dice, takes the product of the results, and adds this to the initial score to obtain a final score. \end{itemize} \end{itemize} Assume that all random choices during the grading process are mutually independent. What is the student's expected score on the exam? %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newpage \problem {\bf [15 pts]\ } There is a dinner party for $n \geq 2$ couples. From among the set of all possible seating arrangements, the hostess chooses one uniformly at random. What is the expected number of couples such that the husband and wife are sitting next to each other? \solution{ Let $I_k$ be an indicator variable for the event that the $k$th couple is sitting together. Then the number of couples sitting together is $I_1 + I_2 + \cdots + I_n$. The probability that a couple is seated together is $2 / (2n-1)$, since a wife is equally likely to occupy each of the $2n-1$ positions relative to her husband, and there are 2 positions right next to him. Consequently, $\pr{I_k = 1} = 2 / (2n-1)$, and we can reason as follows: \begin{eqnarray*} \expect{\text{couples sitting together}} & = & \expect{\sum_{k=1}^n I_k} \\ & = & \sum_{k=1}^n \expect{I_k} \\ & = & \sum_{k=1}^n \pr{I_k = 1} \\ & = & \sum_{k=1}^n \frac{2}{2n-1} \\ & = & \frac{2n}{2n-1}. \end{eqnarray*} } \problem In a variant of Carnival Dice, a player picks a number between 1 and 6. Then she rolls three six-sided dice that are fair and mutually independent. \begin{itemize} \item If her number never comes up, she must pay one dollar. \item If her number comes up once, she wins one dollar. \item If her number comes up twice, she wins two dollars. \item If her number comes up three times, she wins $k$ dollars. \end{itemize} For what value of $k$ is this a fair game? Six couples arrive late to a movie. There are four rows, each containing three consecutive empty seats. From among all possible seating arrangments, the twelve people blunder into one uniformly at random in the darkness. What is the expected number of couples such that the husband and wife end up sitting beside one another in the same row?