- Answer the problems on the exam pages.
- There are six problems for 100 total points plus 10 extra credit. 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.
- Parts of Question 6 have numerical answers -- you may express these as fractions or as decimals.

Q1: 10 points Q2: 10 points Q3: 10 points Q4: 10 points Q5: 40 points Q6: 20+10 points Total: 100+10 points

Typo in Question 2 corrected (and recorrected!) 16 April 2009.

**Question 1 (10):***True or false with justification:*Suppose I throw two fair six-sided dice privately and show you the*larger*of the two numbers (we both know that this is what I am going to do). If I show you a six, you can conclude that the sum of the two numbers is equally likely to be 7, 8, 9, 10, 11, or 12.**Question 2 (10):***True or false with justification:*Suppose that A is a boolean method whose output depends on random choices it makes. If I run A 100 times with independent random choices and get an output of`true`

exactly 80 times, I can conclude with about 95% confidence (or more) that the probability of A returning`true`

is between 72% and 88%.**Question 3 (10):***True or false with justification:*Suppose I have four envelopes, one of which contains $100 and the other three of which contain nothing. I have randomly labeled the envelopes as 1, 2, 3, and 4, with every labeling being equally likely. You have already bought the right to open one envelope of your choice and keep the contents. I now offer to open two envelopes, uniformly randomly chosen from the three empty ones, if you give me $20 more -- you may then choose from the remaining two. Then accepting this offer will increase your expected net winnings.**Question 4 (10):***True or false with justification:*In the scenario of Question 3, I offer to open two envelopes uniformly and randomly chosen from all four envelopes, but if I open the envelope with the $100, you get only $25 of it. Then paying me $10 for this option increases your expected net winnings.**Question 5 (40):**A frisbee game consists of a series of points, and substitutions are only allowed between points. A frisbee team captain has collected data on a game in progress -- for each point she has the set of her seven players who played that point, and whether the point was won or lost. She would like to estimate the probability that her team will win a given future point, based on which of her players are playing it. She decides to follow the calculations for a**Naive Bayes Classifier**.- (a,5) What should be her
**prior odds**O(W) for winning the point, before she considers which players are playing? - (b,5) For a particular player p, how should she compute the
**likelihood ratio**L(p|W), for the event of player p being present with respect to the event of winning the point? - (c,5) For the same player p, how should she compute the likelihood ratio L(¬p|W) for the event of player p being absent with respect to the event of winning the point?
- (d,5) How would she then calculate the
**posterior odds**of her team winning the point? - (e,10) Why might she want to use
**smoothing**to adjust the calculations in parts (b) and (c)? Explain one way she might do this. - (f,10) Explain what assumptions she is making for this estimation of the probability to be valid. Give examples of how these assumptions might fail.

- (a,5) What should be her
**Question 6 (20+10):**My friend B is either at home or on campus, and I would like to know which. I call his cel phone and find that his service is not available. This has an 80% chance of happening if he is at home, where the coverage is bad, and a 40% chance of happening if he is on campus, as he still often forgets his phone or forgets to charge it.- (a,10) If my prior estimate of the probability that he is at home was 20%, what should be my new estimate after I learn that B is unavailable?
- (b,10) I recall that the phone network has been unreliable lately, and
that there is a 20% chance that no one is available due to a network outage.
I call my other friend C, who would be available with probability 75% if the
network is not out, and find that she is unavailable. Does this new evidence
make it
*more or less likely*that B is at home, given that I have already failed to reach B? Explain your answer carefully but without numbers. - (c,10 extra credit) Compute my new estimate of the probability that B is at home, given that neither B nor C is available by phone. (Hint: Find the probability that the network is out and use this to revise the probabilities of B being unavailable in either location.)

Last modified 16 April 2009