What is the probability that Andrea, Melissa and Carol end up being on the same team?

Question

I am stuck on a word problem about picking teams. I thought it would be very simple, but to my surprise, I could not solve it. So here's the problem..

Andrea, Melissa, and Carol are in a class of 27 girls. The teacher chooses students at random to make up teams of three. What is the probability that Andrea, Melissa and Carol end up being on the same team?

Answer 1

Consider that there are $27$ slots available.

$\Large\boxed{.}\Large\boxed{.}\Large\boxed{.}\;\;\Large\boxed{.}\Large\boxed{.}\Large\boxed{.}\;\;\Large\boxed{.}\Large\boxed{.}\Large\boxed{.}\;\;\Large\boxed{.}\Large\boxed{.}\Large\boxed{.}\;\;\Large\boxed{.}\Large\boxed{.}\Large\boxed{.}\;\;\Large\boxed{.}\Large\boxed{.}\Large\boxed{.}\;\;\Large\boxed{.}\Large\boxed{.}\Large\boxed{.}\;\;\Large\boxed{.}\Large\boxed{.}\Large\boxed{.}\;\;\Large\boxed{.}\Large\boxed{.}\Large\boxed{.}\;\;$

Andrea can occupy any slot.
To be in the same group, Melissa can choose any $2$ of the $26$ remaining,
and Carol is left with only $1$ choice out of the $25$ left.

$Pr = \dfrac2{26}\dfrac1{25}$

Answer 2

There are $\frac{27!}{3!^9}$ ways to make up $9$ teams of $3$ numbered $1$-$9$.

There are $9\frac{24!}{3!^8}$ ways to make up $9$ teams of $3$ numbered $1$-$9$ with Andrea, Melissa, and Carol on the same team. This counts $9$ teams that Andrea, Melissa, and Carol could be on, times the number of ways to make up $8$ teams of $3$ numbered $1$-$8$, for the other $24$.

Thus, the probability is is $\frac9{\binom{27}{3}}=\frac1{325}$.