UW-Madison 431 - Introduction to Probability Theory Spring 2009
First Midterm Solutions
1. In a chemistry class there are 16 boys and 16 girls. For a certain lab assignment they have to
work in pairs. Suppose that the lab partners are assigned randomly (each outcome is equally
likely).
(a) (12 points) Set up an appropriate sample space for this experiment and determine its size.
A possible outcome for this experiment is 16 pairs made out of the numbers 1,2,3, . . . , 32.
If you want an algebraic description, you can write
©©{a1, a2},{a3, a4}, . . . , {a31, a32}ª: 1 ≤ai≤32, ai6=ajif i6=jª
If we also number the pairs (i.e. there is a first pair, a second pair...) then we could count
the number of possible outcomes with the multinomial ¡32
2,2,...,2¢=32!
216 . If we do not care
about the ordering of the pairs then we have to divide this with 16! (the number of ways
we can order 16 pairs). So the size of the state space is 32!
216(16!) . (You can also use the state
space with the ordered pairs, but then you have to modify the solutions for part (b) and
(c) accordingly.)
(b) (12 points) What is the probability that each boy will be the lab partner of a girl?
Suppose that we number the boys from 1 to 16. The first can choose out of the 16 girls,
after that the second one can choose one out of the remaining 15 and so on. This means
that there are 16! possible boy-girl pairings. (Another solution: if we number the boys
from 1 to 16 then if we assign the girls then this will produce a permutation of the girls.)
Since each outcome is equally likely, the probability is
16!
32!
216(16!)
=(16!)2216
32!
(c) (12 points) What is the probability that each girl will be the lab partner of a girl?
Copying the solution of part (a) there are 16!
28(8!) ways we can make 8 girl-girl pairs. Similarly,
there are 16!
28(8!) ways we can pair the remaining boys which means that the probability is
(16!
28(8!) )2
32!
216(16!)
=(16!)3
(32!)(8!)2
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