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.
(a) You will have to consider the 10 different outcomes. Each combination of $"A-B", "B-Cox", "B-Cramer", ...$ is equally likely to be drawn. There are
 $${5 \choose 2} = 10 $$

10 ways to choose 2 people from a group of 5. Since we are only considering 1 possibility, the answer will be $\frac 1 {10} = 0.1$.

For (b), there are a few possibilities where at least 1 "C" member gets selected. Consider $$"Cox-A", "Cox-B", "Cox-Cramer", "Cox-F"$$ and $$"Cramer-A", "Cramer-B", "Cramer-Cox", "Cramer-F"$$ Notice that $"Cox-Cramer"$ appears twice, so one needs to be eliminated. Turns out there are 7 different possibilities out of 10...so we have $0.7$.

Alternatively, you can consider the different ways no "C" member gets selected. Turns out there are 3 ways  $$"A-B", "A-F", "B-F"$$ resulting in $0.3$ chance of the compliment case happening. Taking  $1 - 0.3$, you will still get $0.7$ as the answer.

For (c), I'd personally recommend taking the "compliment" of events. ie, how many combinations have LESS than 15 years? Turns out there are 4 combinations - $(3,6), (3,7), (3,10) , (6,7)$. You can use the answer from (a) and evaluate this to become $1 - \frac 4 {10} = 0.6.$
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