Question text is in black, solutions are in blue.
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.
FALSE. For example, the chance that the total is 12, given that the larger number is 6, is 1/11 rather than 1/6. We compute Pr(max is 6 and total is 12) divided by Pr(max is 6) which is (1/36)/(11/36).
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%.
TRUE. The number of true outputs is a binomial random variable with n = 100 and with p equal to the unknown probability. If p were 0.8, the variance of this random variable would be npq = 16 and thus the standard deviation would be 4. There would be about a 95% chance that the variable (with p = 0.8) would be within two standard deviations of its mean, so there is about a 95% chance that the real observed number is within two standard deviations of np, and thus that np is between 72 and 88.
TRUE. Your expected winnings with four envelopes are $25, and with two envelopes are $50, so you pay $20 and gain $25 in expected value for a net expected profit of $5. Unlike the Monty Hall scenario, the method of choosing envelopes to open leaves the remaining two envelopes equally likely, since your choice is not made before the envelopes are opened.
TRUE. There are two cases, each of probability 1/2. If the good envelope is opened you get $25, and if two bad envelopes are opened you are left with a choice between one good and one bad envelope, which has expected value of $50. So your expected winnings are now (1/2)($25 + $50) = $37.50, and subtracting the $10 you pay your expected net winnings are $27.50, more than your previous expected winnings of $25.
Her best estimate is the ratio between total points won so far and total points lost so far.
This should be Pr(p|W) divided by Pr(p|¬W). We can estimate Pr(p|W) by the percentage of winning points in which p was on the field, and Pr(p|¬W) by the percentage of losing points in which p was on the field.
This is Pr(¬p|W) divided by Pr(¬p|¬W), which we can estimate by the fraction of winning points where p was not on the field divided by the fraction of losing points where p was not on the field.
She takes the prior odds, multiplies them by the product of L(p|W) for every player p who is going to be on the field, and then multiplies that by the product of L(¬p|W) for every player p who is not going to be on the field.
If p is never on the field for a winning point, or never off the field for a
winning point, or similarly for a losing point, one of the likelihood ratios
for p will be zero or infinite, which will force the posterior odds to be zero
or infinite. This is unlikely to be what we want because it would destroy all
the information about the other players, and it is probably not true that the
presence or absence of a single player guarantees the winning or losing of the
next point.
Smoothing consists of adding one to each of the four counts for p (on
field/winning, off field/winning, on field/losing, and off field/losing) before
computing the percentages. This has a small effect on large counts but tends
to move ratios of small counts toward being even, thus lowering the impact of
these ratios on the posterior odds.
She is ignoring the effect of which opponents are on the field, or assuming that these effects will average out. It's possible that the opponents' choice of players might be affected by her choices, which would mean that they would not average out. Also, like the Naive Bayes Classifier, this method assumes that the effects of the players are independent. This would be wrong if two players on her team worked particularly well or particularly badly together. The model also ignores changes in a player's ability over the course of the game, such as might be due to fatigue or injury.
Let UB be the event that B is unavailable, and h the hypothesis that B is at home. We are only considering one other hypothesis, so ¬h means that B is on campus. To get posterior odds for h, we multiply the prior odds O(h), which are Pr(h)/Pr(¬h) = 0.2/0.8 = 1/4, by the likelihood ratio L(UB|h). The latter is Pr(UB|h)/Pr(UB|¬h) = 0.8/0.4 = 2. Thus my posterior odds are (1/4)(2) = 1/2 and my posterior estimate of the probability is (1/2)/(1 + (1/2)) = 1/3.
It makes it less likely that B is at home. B's unavailability was evidence tending to show that he is at home, but C's unavailability increases our estimated likelihood of a network outage, which would be an alternate explanation of B's unavailability. So the evidence of B not being home becomes weaker than when we didn't consider the possibility of the outage.
First let's revise our estimate of the probability of an outage NO given the evidence about C. Our prior odds of the outage were 1/4, and we multiply by the likelihood ratio L(UC|NO) = Pr(UC|NO)/PR(UC|¬NO) = 1/(1-0.75) = 4. So our posterior odds are 1, and we think the chance that there is a network outage is now 50%. Thus we can revise Pr(UB|h) to 90% (1/2 times 100% for the outage, plus 1/2 times 80% for his being at home with no outage) and Pr(UB|¬h) to 70% (1/2 of 100% plus 1/2 of 40%), and L(UB|h) becomes 0.9/0.7 = 9/7. The new posterior odds are (1/4)(9/7) = 9/28 and the posterior estimate of the probability is (9/28)/(1 + 9/28) = 9/37 = 24.3%, higher than the 20% before the evidence of UB but lower than the 33.3% calculated in (a).
Last modified 16 April 2009