In your second post alphatron, I think that I can bring up the analogy of flipping a coin twice. The outcomes are HH, HT, TH and TT. So for our problem, the outcomes are DD, DG, GD, GG where D is defective and G is good. The one thing to keep in mind, is that unlike a coin toss problem, the probabilities change depending on what happened for your first choice. However, they should still not be that difficult to figure out. Once you have figured out the probabilities then you combine the DG and GD into a single category, namely 1 defective.
Danny is basically providing you with a much easier approach, which is to use something called a hypergeometric distribution. We will not cover it formally in class, but you might look online for a discussion. There is a section in the textbook but I would not recommend it. Basically, you have fraction of a product of 2 binomial coefficients over a third binomial coefficient, namely the one you allude to in your first post. In the numerator, you select a certain number of defective mutliplied by a selection of a certain number of good ones: combin(total number of defective, number of defective you are selecting)*combin(total number of good, number of good you are selecting). The denominator is a blind selection: combin(total number of objects, size of selection without dividing into good and defective). How many good and defective you select depends on what your focus is (on defective) and which outcome of the random variable you are looking at (0,1,2). For example, if the outcome is 1, then you are selecting one defective and one good. If you the outcome is 2, then you are selecting 2 defective and 0 good.
