CMPSCI 240: Reasoning About Uncertainty

Second Midterm Exam

David Mix Barrington

21 October 2009

Directions:

• Answer the problems on the exam pages.
• There are six problems for 100 total points. Actual scale was A=85, C=55.
• If you need extra space use the back of a page.
• No books, notes, calculators, or collaboration.
• The first four questions are true/false, with five points for the correct boolean answer and up to five for a correct justification.
• When the answer to a question is a number, you may give your answer in the form of an expression using arithmetic operations, powers, falling powers, or the factorial function. If you give your answer using the "choose" notation, also give it using only operations on this list. In addition, if your answer is a non-negative integer less than or equal to 100, you must compute the number for full credit.

  Q1: 10 points
  Q2: 10 points
  Q3: 10 points
  Q4: 10 points
  Q5: 30 points
  Q6: 30 points
  Total: 100 points

• The first two true-false questions involve a random variable X that is equal to 0 with probability 1/2, equal to 1 with probability 1/4, and equal to 2 with probability 1/4. We could produce a value from X by flipping a fair coin, flipping it a second time if and only if the first flip is heads, and returning the total number of times we flip heads.

• Question 1 (10): True or false with justification: The variance of X is exactly three times the expected value of X.

• Question 2 (10): True or false with justification: If we take n values from X independently and add them together, the resulting random variable has a standard deviation of 3n/2.

• Question 3 (10): True or false with justification: If I deal three cards from a standard 52-card deck, with every set of three cards being equally likely, then the probability that the three cards have three different suits is less than or equal to 3/8.

• Question 4 (10): True or false with justification: If I deal three cards from a standard 52-card deck, with every set of three cards being equally likely, then the probability that the three cards contain at least one spade is greater than or equal to 3/4.

• Question 5 (30): Hans is driving several passengers from Dagstuhl to the Frankfurt airport in his van (a Volkswagen, naturlich). He is picking them up at 6:00 a.m., and it is very important that they reach the airport by 8:30. He could take the autobahn, which has an average travel time of 110 minutes and a variance of 400 min², or he could take back roads, which has an average travel time of 130 minutes and a variance of only 25 min².
  • (a,10) Suppose first that each travel time comes from a normal distribution with the given mean and variance. What is the approximate probability that Hans will fail to meet the 150-deadline if he takes the autobahn? Is he more or less likely to meet the deadline by taking back roads? Justify your answer.
  • (b,10) Now we no longer assume that the travel times are normally distributed. Use the Markov Inequality to compute a bound on the probability that Hans fails to meet his deadline in each case, using the given average times and the assumption that the travel time cannot be negative. That is, compute probabilities p_a and p_b such that the Markov Inequality tells you that the chance of being late on the autobahn is at most p_a and that the chance of being late on the back roads is at most p_b.
  • (c,10) Again, we no longer assume that the travel times are normally distributed. Now use the Chebyshev Inequality to get bounds q_a and q_b on the probability of missing the deadline on the autobahn and on the back roads respectively. Remember that this result uses the mean and variance of the given distribution, and no other assumptions.

• Question 6 (30): This multipart question deals with the game of poker dice. A player throws five fair, independent six-sided dice. After the first throw of all five dice, she may pick some or all of the dice to roll a second time, in the hope of making a better combination.
  • (a,5) What is the probability that the player rolls five sixes (the best possible hand) on her first roll?
  • (b,5) What is the probability that she gets three of a kind on her first roll? This means that exactly three dice show one number, and that the other two are different from each other and from the three. (For example, 4-3-4-4-6 is a "three of a kind" combination but 4-4-5-4-4 and 4-2-4-4-2 are not.)
  • (c,5) Now assume that the player has rolled three of a kind, specifically 4-3-4-4-6, and that she has chosen to reroll the two dice that are not 4's. What is the chance that on her second roll she improves her hand to a full house? (A full house is three dice with one number and two dice with a different number, such as 4-2-4-4-2.)
  • (d,5) In the situation of part (c) what is the probability that the player improves her hand on the second roll to four of a kind? What is the probability that she improves it to five of a kind?
  • (e,10) Now assume that the player has three total rolls (as in the commercial game Yahtzee^TM). That is, she first throws five dice, then rerolls some of those dice, then (if she chooses) re-rerolls some of the dice that she rerolled. Again assume that her first roll is 4-3-4-4-6. She will use her second roll, and if necessary her third roll, to maximize her chance of getting five 4's at the end. What is her total probability of succeeding in doing this? (Again, starting with the situation after her first roll.)

Last modified 23 October 2009