Class, 10/3

We passed over the Fermi problem bullet after I re-explained the geometric mean method.

We reminded everyone of the basic equation of conditional probability that underlies everything we are doing: P(A,B)=P(A|B)P(B)=P(B|A)P(A). We talked about the three equivalent facts about whether a distribution is independent or not. It is independent if and only if P(A|B)=P(A) for every A and B, it is independent if P(A,B)=P(A)P(B) for every A and B, and if P(A|B)=P(A) for every A and B, then necessarily P(B|A)=P(A) for every A and B.

We then showed how to construct the unique table of joint probabilities when we are given the marginal probabilities: Just multiply the marginal in a row with the marginal in a column and put the result in the corresponding cell.

We then took a table of independent probabilities and, by changing four cells in a square, just adding an arbitrary number to two of the cells on a diagonal and subtracting the same number from the cells on the other diagonal, and got a table where the probabilities are not independent.

We discussed the Monty Hall problem and variants. We found that if there are four doors, and Regular Monty opens two of them, each showing a goat, then the probability of getting the prize goes from 1/4 to 3/4 if we switch. We found that it goes from 1/4 to 3/8 if Monty opens one door and we switch to one of the others. We did this by a spreadsheet calculation. We then thought of a simpler way: Since your probability of initially picking the right door is 1/4, the probability that one of the other doors has the prize is 3/4. That doesn't change when Monty opens one of them and shows you a goat. So, since there are two doors left, the probability that you'll get the right one if you switch is 1/2 times 3/4, or 3/8.

I left you with another related problem to think about: There are three cards which have been made by pasting together two cards so that the backs are visible. One has two red backs, one has a red back and a blue back, and one has two blue backs. The cards are put in a hat and shaken, and you pick one out, looking at only one back. It is red. What is the probability that the other side is red?

We'll discuss that next time.

We went onto the cancer problem. We took a population of 10000 individuals. The problem statement says that 1% of the population has the gene, so that's 100 who have the gene and 9900 who don't. Of those who have the gene, 98 will be detected by the test and 2 missed (false negative). Of those who don't have the gene, the test will falsely identify 5% as having the gene, or 495 in all (false positives) and will correctly say that the remaining 9405 do not have the gene. Looking at just the positives, there are 98 true positives and 495 false positives, so that the probabilty that a person has the gene if they test positive is only 98/593 or about 1/6; the remaining 495/593, or about 5/6, do not have the gene.

We ran out of time here and will continue on Monday, finishing this problem and then going on in the study guide.

In answer to a question, I stated that if there is an item that we don't get to in our review, then similar items will not appear on the test. I also stated that I expected there to be five questions on the test, and that there should be enough time for everyone to do all of them. I pointed out that usual test-taking strategy says to go for the easy ones first and save the bulk of the time for the harder ones. I also said that it is very important to at least try to answer every question, since I cannot give credit if an item goes completely unanswered. We agreed that people who come early (not earlier thatn 10 AM, please) could start early, and that you may be able to stay an extra 5 minutes (but not more, because of the class that meets next in this room) to finish.

Be sure to bring your calculators. I do not have a loaner calculator!