We started out discussing the problem set. The first problem (King and Brother) can be diagrammed with a probability tree. In three of the four branches (all equal probability 1/4), there is a king, but in only one of them does the king have a brother. Thus, the probability is 1/3 that the king has a brother. The result can also be obtained, as in the right side of the picture, by simply listing all the possibilities and counting: 1 BB, but 3 cases in all where there is a boy. Again, 1/3.

Yet another way to do it is to use the formula for conditional probability, P(A,B)=P(A|B)P(B), written in the form P(A|B)=P(A,B)/P(B). With A=”king has a brother” and B=”there is a king), we find that P(A,B)=1/4 and P(B)=3/4 (again, looking at the tree or the listing of cases), so by division we find P(A|B)=1/3.

Similar methods can be used when there are three children. In particular, in only one case are there two brothers, and in three cases the king has one brother. There are seven cases in all where there is a king (only GGG does not have a king). So the probability that the king has two brothers is 1/7, and the probability that there is just one brother is 3/7, using any of the three methods.

I noted that the most systematic way to list all the possibilities is to think of them in terms of binary numbers. Listing all the binary numbers from 0 to 7, as on the left side of the chart, and then letting 0=B and 1=G, gives us all the cases without the possibility of making a mistake by leaving one of them out.

For the dice problem, it’s a good idea to think of the two dice as having different colors, e.g., a red one and a green one. This means that a red ‘5’ and a green ‘3’, for example, is a different case from a red ‘3’ and a green ‘5’. The dice are distinguishable. The easiest way to list all the cases is to put one die on the left side of a grid and the other die at the top of the grid. Then we can fill in the grid with the totals for the two dice.

We see that there are 5 different cases where the total is 8 (circled in red), and 11 different cases where a ‘5’ is showing on at least one of the dice. There are two cases where a ‘5’ is showing and the total is ‘8’ (will circled both in red and green). By just counting cases and knowing that there are 36 equally probable cases we find that the probability that the total is 8 is 5/36, the probability that a ‘5’ is showing, given that the total is 8 is 2/5, and the probability that the total is 8, given that a ‘5’ is showing, is 2/11. This can also be calculated using the conditional probability formula:

We then started on the drug test problem (which is similar to polling and other similar problems). The cure rate of the drug is r, and the states of nature are all the possible values of r, of which there are infinitely many. Obviously we cannot list them all. After some discussion we ended up with ten (on the whiteboard) states of nature, the probability of being between (0.0 and 0.1), between (0.1 and 0.2), etc., for ten states of nature. We represent each interval by the probability at the midpoint of the interval, that is, 0.05, 0.15, etc.

For a prior we chose a uniform one, with a probability 1/10 on each point. We noted that this is reasonable if we really have no prior notion about the cure rate. On the other hand, if we are looking at a political poll, we can be pretty sure that it’s not going to be hugely lopsided, so something like a bell-shaped curve for the prior might be more reasonable, putting higher prior probability for rates near the middle and lower probability near the extremes.

The next step is the likelihood. In our little test, 3 people were cured and 7 were not cured. That’s the data. I started out by noting that Lindsay was in the test, and she was cured. What’s the probability of that? It’s r. Kelsey also was in the test and was cured. The probability of that is also r. Since the two events are independent (that is, the probability that any individual is cured is independent of what happened to the other individuals), we can simplify the product law of probability (as in the problem set due on Monday) to r*r for the probability of both events.

I left everyone with the problem of figuring out how to write down the likelihood for all the data, that is, the 3 cures and the 7 who were not cured. We’ll discuss this on Monday.

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