Finite Math

Question

One cage contains 4 white mice and 2 black ones, and another cage contains 3 white mice and 2 black ones. A cage is chosen at random and three mice are selected. Find the expected number of white mice in the sample.

Answer 1

Cody,

In problems of this sort, you need to start by considering how you calculate probabilities and how you combine them. For each individual cage, the probability distribution is determined by the combinations you could pick and their individual probabilities of being selected. Then, the average numbers of white mice expected for choosing each cage, may be averaged (since each of these results is independent of the other!).

So, how to calculate for the first cage? I like to approach problems like this as permutations all the way, unless I can immediately see my way through using combinations only. So for cage 1, for example, you know you will get at least 1 white mouse -- you must run out of blacks. Consider each of the cases separately -- one white mouse, two white mice, three white mice (if blindness is linked to white color, they may have a problem...;). For the one white mouse case, there would be 4 ways of picking it, 2 ways of picking the first black mouse, and one way of picking the 3rd mouse. Or would there? No; you must pick *both* the black mice, so only the 4 ways for the white mouse. OK, now we must mix them up (to make the permutations): the white mouse picked could be chosen as #1, #2, or #3 -- that's a factor of three on possible choices, and the black mice could be mixed in either order in the two remaining positions (do you see why I deferred mixing the black mice until this step? It's a comparable operation, mixing all the chosen mice up. You might also consider this step as finding the permutations of the combination of specific mice, i.e. 3x2x1 or 3!). So that's 4x3x2=24 permutations (ordered choices of specific mice, such that get only one white). Next, consider the two white case (still first cage): here, 4 choices for 1st white, 3 choices for 2nd white, 2 choices for the black. Here, the black choices ARE lumped into the initial pick. Then, mixing these up by 6 permutations, and we have 4x3x2x3x2=144 permutations in all. Lastly, the case for 3 white mice: only one white mouse isn't used, there are 4 ways of doing that. And still the 6 ways of permuting, for a set of 24 specific permutations. To form the grand total, you need to include the number of white mice contributed by each of the permutation subtotals, so: The grand total: 24(1) + 144(2) + 24(3) = 384 mice expected, out of 6x5x4 sets ordered x6 ways or 720 specific orderings. So the average probability from cage 1 is 384/720 = 0.5333... mouse/ mouse chosen; since there are 3 mice chosen the average number of white mice outcome is 3x this, or 1.6 exactly.

I'll leave it up to you to do cage 2, and take the average.

You might want to note, that the average mice expected from cage one is NOT 4/6 * 3 =2 !!! Moral: don't take shortcuts (or the cat might get you, so I hear!).

Hope this helps you to think this (and such problems) through.