HCOL 195 10/19/09

We discussed the test. No problems with #1. On #2 the most effective way to do it is to list the possibilities and count up those that have the first child a boy (hence the king) and then count the individual cases: two brothers, two sisters, one brother and one sister. The four cases that are relevant are


Note that BBG and BGB are not the same. Therefore there is a 1/4 probability of two brothers, the same for two sisters, and a 1/2 probability of one brother and one sister. These add up to 1.

In problem #3 there are 10 SON (1,2,…,10); The likelihood for each SON is (SON/SON)*((SON-1)/SON)*(2/SON)*(2/SON). The denominator is always the SON since that’s how many fish there are in the lake each time we catch one. The first two numerators represent the number of untagged fish left in the lake, and the second two the number of tagged fish in the lake, for the four fish we caught. One student started with the smallest SON=5, but that’s not what the statement of the problem says. Also, tagged vs. untagged are not states of nature, they are data.

Problem #4 is easiest done by using natural frequencies: If we have 2000 patients (may as well use that number as it is directly useful for the last question), then 1%, or 20 of patients will have the disease and 1980 will not. Of the 20 that have the disease, 19, or 95%, will test positive. The remaining one will test negative. Of the 1980 patients who don’t have the disease, 4%, or 79 will test positive (it’s really 79.2, but we can round here without sensible error). That’s the answer to the number of false positives in the group of 2000 patients. The probability of having the disease, given that you test positive, is 19/(19+79)=19/98, or a little over 0.19.

The fifth problem has a table of independence. The marginals are .25 and .75 in the horizontal direction and .5, .2 and .3 in the vertical direction. Each entry in the joint table is the product of the corresponding marginals, which proves the result. To make it independent, you can add a fixed number to two rows and subtract the same number to two columns; this would involve four numbers changed in the table.

For the last problem, pick to use either gains or losses and stick to it. Losses is easiest; then there is a loss of $800×10 million or $8 billion if we require installation; if we do not require it, then there will be a loss of 10,000x$5 million, or $50 million due to lives lost that might have been saved. The second loss is greater, so we should reject that branch and require installation of the safety device. Note that you use each number exactly once: Some students tried to use the numbers on both branches, once as a gain and once as a loss, but that doesn’t work.

I asked whether people would prefer $100,000 as a sure thing or a 50% chance at $1 million and a 50% chance of nothing. About half the class preferred the sure thing, and half the gamble. We then said, what if the probability of getting the $1 million were 0.1, 0.2,…,0.9, 1.0. As the probability ramped up, more people were willing to take the gamble, but two students would only go for the $1 million if it were a “sure thing”, that is the probability were 1.

Then we talked about being on a jury. We decided that the four possibilities are: AI (acquit someone who is innocent), CI (convict someone who is innocent), AG (acquit someone who is guilty) and CG (convict someone who is guilty). We discussed which of these were the best and the worst outcomes. While it is clear that making a right decision (AI or CG) is good, and making a wrong decision (CI or AG) is bad, we didn’t come to agreement as to how to order the two good ones and the two bad ones. We’ll bring this up again later.


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