CMPSCI 291b: Reasoning About Uncertainty

Practice Exam for Midterm Exam

David Mix Barrington

8 March 2008

Directions:

• Answer the problems on the exam pages.
• There are six problems for 100 total points. Probable scale is A=90, C=63.
• 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.
• The other questions have numerical answers -- 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.

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

• Question 1 (10): True or false with justification: A professor is making up a test with four true/false questions, and chooses the answer "true" or "false" with equal probability, independently for each question. Then the probability that he makes exactly two true and two false is less than 1/3.

• Question 2 (10): True or false with justification: If I choose 2n elements independently, with repetition, from a set of n elements, then the probability that I choose each element at least once approaches 1 as n approaches infinity.

• Question 3 (10): True or false with justification: If I choose n elements independently, with repetition, from a set of n elements, then the probability that I choose each element one approaches 0 as n approaches infinity.

• Question 4 (10): True or false with justification: The number of length-5 strings over the alphabet {a,b,c}, where the letters come in alphabetical order, is greater than 30.

• Question 5 (30): In college basketball, there are two different ways a player can be awarded two free throws. In a one-and-one, the player attempts her first throw and is allowed a second throw only if the first throw succeeds. In a two-shot foul, the player is allowed the second throw whether the first is successful of not. Assume that Julia's free throws are independent events, each with a 3/5 (or 0.6) probability of success.

  • (a,5) If Julia shoots a one-and-one, what is the probability that she gets two points? That she gets one point? That she gets no points? (Each successful free throw counts for one point.)

  • (b,5) What is her expected number of points in a one-and-one?

  • (c,5) What is the variance of the number of points she gets in a one-and-one? (You may find the formula Var(X) = E(X²) - E(X)² to be useful.)

  • (d,5) What are the probabilities of 0, 1, or two points respectively if Julia shoots a two-shot foul?

  • (e,5) What is her expected number of points for a two-shot foul?

  • (f,5) What is the variance of the number of points she gets in a two-shot foul? (Hint: This question is substantially easier than part (c) if you see how to do it.)

• Question 6 (30): Angelina, Brad, and Celine are playing a poker-like card game where each gets one card and the player whose card has the highest rank wins. (It's possible for two or three players to tie if they have cards of the same rank.)

  • (a,5) How many possibilities are there for the distribution of one card to the each players? (Here we consider cards of different suits as being different possibilities.)

  • (b,5) What is the probability that each player gets a card of a different rank?

  • (c,5) What is the probability that Angelina wins and the three cards have different ranks?

  • (d,10) What is the overall probability that Angelina wins the game? (This is the answer to (c) plus the probability that Brad and Celine have cards of the same rank and Angelina's card has a higher rank.)

  • (e,5) I forget to mention that when the players bet, each can see the other players' cards but not their own. How can Brad compute the probability that he wins the game, based on seeing Angelina's and Celine's cards but not his own?

Last modified 8 March 2008