There are 3 math teams in the area, with 5, 7, and 8 students respectively. Each team has two co-captains. If I randomly select a team, and then randomly select two members of that team to give a copy of $\emph{Introduction to Geometry}$, what is the probability that both of the people who receive books are co-captains?
There's a $\dfrac{1}{3}$ chance that I will select each team. Once I have selected a team, let $n$ be the number of students on that team. There are $\dbinom{n}{2}$ ways to choose a pair of those students to give books to, but only one of those pairs will be the two co-captains, which means that once I have selected that team, the probability that I give books to the co-captains is $$\dfrac{1}{\dfrac{n(n-1)}{2}}=\dfrac{2}{n(n-1)}.$$Since the teams have $5,$ $7,$ and $8$ students, this means that the total probability is $$\dfrac{1}{3}\left(\dfrac{2}{5(5-1)}+\dfrac{2}{7(7-1)}+\dfrac{2}{8(8-1)}\right)$$which after a bit of arithmetic simplifies to $\boxed{\dfrac{11}{180}}$.
\dfrac{11}{180}
Level 5
Counting & Probability
Problem: There are 3 math teams in the area, with 5, 7, and 8 students respectively. Each team has two co-captains. If I randomly select a team, and then randomly select two members of that team to give a copy of $\emph{Introduction to Geometry}$, what is the probability that both of the people who receive books are co-captains?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the probability that both selected members are co-captains, we first need to determine the total number of students across all teams. The teams have 5, 7, and 8 students, respectively, leading to a total of 5 + 7 + 8 = 20 students.

Next, we calculate the total number of ways to select a team and then select 2 members from that team. The probability will depend on the team selected. 

For the first team with 5 students, the number of ways to choose 2 members is given by the combination formula C(n, k), where n is the total number of students in the team and k is the number of students to choose. Thus, for the first team, C(5, 2) = 10.

For the second team with 7 students, C(7, 2) = 21.

For the third team with 8 students, C(8, 2) = 28.

Now, we calculate the total number of ways to choose 2 members from any team:
Total ways = C(5, 2) + C(7, 2) + C(8, 2) = 10 + 21 + 28 = 59.

Next, we need to find the number of favorable outcomes where both selected members are co-captains. Each team has 2 co-captains.

For the first team, the number of ways to choose 2 co-captains is C(2, 2) = 1.

For the second team, C(2, 2) = 1.

For the third team, C(2, 2) = 1.

Thus, the total number of favorable outcomes is 1 + 1 + 1 = 3.

Now, we calculate the probability of selecting 2 co-captains. The probability for each team is the number of favorable outcomes divided by the total outcomes for that team, multiplied by the probability of selecting that team.

The probability of selecting the first team is 1/3, and the probability of selecting 2 co-captains from that team is 1/10. Therefore, the contribution to the overall probability from the first team is (1/3) * (1/10) = 1/30.

For the second team, the contribution is (1/3) * (1/21) = 1/63.

For the third team, the contribution is (1/3) * (1/28) = 1/84.

Now, we need to find a common denominator to sum these probabilities. The least common multiple of 30, 63, and 84 is 1260.

Converting each fraction:
1/30 = 42/1260,
1/63 = 20/1260,
1/84 = 15/1260.

Adding these gives:
42/1260 + 20/1260 + 15/1260 = 77/1260.

Now, we simplify 77/1260. The greatest common divisor of 77 and 1260 is 7, so we divide both the numerator and the denominator by 7, yielding:
77 ÷ 7 = 11,
1260 ÷ 7 = 180.

Thus, the final probability that both selected members are co-captains is 11/180.<|END_OF_LATENT|> The final answer is $\dfrac{11}{180}$.