Combinatorics Statistics Question The problem I am working on is:
An academic department with five faculty members—Anderson, Box, Cox, Cramer, and Fisher—must select two of its members to serve on a personnel review committee.
Because the work will be time-consuming, no one is anx-ious to serve, so it is decided that the representative will be selected by putting the names on identical pieces of paper
and then randomly selecting two.
a.What is the probability that both Anderson and Box will
be selected? [Hint:List the equally likely outcomes.]
b.What is the probability that at least one of the two members whose name begins with C is selected?
c. If the five faculty members have taught for 3, 6, 7, 10,
and 14 years, respectively, at the university, what is the
probability that the two chosen representatives have a
total of at least 15 years’ teaching experience there?

For a), I figured that since probability of Anderson being chosen is $1/5$ and Box being chosen is $1/5$ the answer would simply be $2/5$. It isn't, though. It is $0.1$ How did they get that answer? I might need help with parts b) and c) as well.

most simple way to understand this problem.
(i just did this problem just now just now) lol
A, B, Co, Cr, and F exist.
pick 1 of 5 at random. (1/5)
pick another at random, but this time around there are only 4 choices. so (1/4)
multiply the two values. (1/20)
BUT! that's considering that A is picked at first try then B on the second.
you also have to consider the possibility that B would be picked first and etc.
so you add (1/20) + (1/20) = (2/20) = (1/10) = 0.1
pce
