# 50 States w/2 senators each. Probablity of various committee compositions

From Carol Ash's, "Probability Tutoring Book". pg. 28 section 1-4, question 10.

There are 50 states and 2 senators from each state. Find the prob that a committee of 15 senators contains:

a. at least 1 from each of Hawaii, Mass. and Penn.

b. at least 1 from the 3 state region composed of Hawaii, Mass. and Penn.

c. 1 from Hawaii and at least 1 from Mass.

-
Usually, the best way to see if you argument breaks down is to do it for some smaller example. e.g. try it with 3 states and committee sizes of 3. – Thomas Andrews Jan 1 '13 at 14:32

Your answers all overcount, I’m afraid. For (a) consider a committee that contains both senators from Hawaii and one senator each from Massachusetts and Pennsylvania. If the senators from Hawaii are A and B, you count this committee once with A as the Hawaiian senator counted in the $2^3$ factor and B as one of the $12$ counted in the $\binom{97}{12}$ factor, and once with B as the Hawaiian senator counted in the $2^3$ factor and A as one of the $12$ counted in the $\binom{97}{12}$ factor. Every committee that contains two senators from at least one of the three named states is overcounted.

The other calculations overcount in similar fashion.

-
I should have read the section a little more carefully. There's a note and example: > Don't try to count at least n thingos by presetting n thingos to be sure and then going on from there. It just won't work. – Robert S. Barnes Jan 1 '13 at 20:19
@Robert: I’m not terribly surprised that there’s such a note: it’s a very common ‘second-level’ mistake, meaning the sort of mistake that you don’t make until you’ve actually started to get a grip on the ideas. – Brian M. Scott Jan 1 '13 at 20:25

She answered all these questions using inclusion-exclusion, however I answered with direct calculation. Are my answers below correct? If not, why?

My thought is that when we say "exactly" or "at least" I pick a member of a group to meet the condition. If it says exactly I eliminate the remaining members of that group before picking the remainder, or if it says at least I leave them in and then pick the remainder.

a. $\huge\frac{2^3\binom{97}{12}}{\binom{100}{15}}$

b. $\huge\frac{6\binom{99}{14}}{\binom{100}{15}}$

c. $\huge\frac{2^2\binom{97}{13}}{\binom{100}{15}}$

-