# A class of 10 students is divided in two teams. Find the probability of students A and B to be in the same team.

My take to this problem is this:

In the case the two teams have the same amount of students. If student A is on team 1 (or 2, it does not matter), then there's 4 slots available for student B. So the probability of B getting that slot is $$\frac{4}{9}$$. So, the probability of A and B to be in the same team is $$\frac{4}{9}$$

How do I deal with finding the probability in the case the teams are not even? (2-8, 3-7, 4-6). How can I do this problem using combinations?

• This is not a meaningful question unless you set down how the teams are formed. If the group is split 50/50 at random, then your analysis is correct. If every student is independently assigned to either team with probability $\frac 12$, then the probability that they are in the same team will be $\frac12$ -- but this includes the degenerate situation of all students being on the same team. – Mees de Vries Nov 14 '18 at 16:12
• You want to use conditional probabilities. So the probability that student A goes to team 1 times the probability that student 2 goes to team 1 given student 1 went to team 1 plus the probability student 1 went to team 2 times the probability student 2 goes to team 2 given student 1 went to team 2. – Jack Moody Nov 14 '18 at 16:15

Suppose that one team of three students is chosen at random from the class. Two given students A and B are in one team if they go to chosen team or if stay at the rest part of $$7$$ students. The total number of possibilities to choose $$3$$ students is $$\binom{3}{10}=120$$, the number of favorable combinations is $$\binom{1}{8}+\binom{3}{8}=8+56=64,$$ and the probability is $$\frac{64}{120}=\frac{8}{15}$$. Here $$\binom{1}{8}$$ is the number of variants to choose three students with A and B and one extra student chosen from the rest $$8$$ students. The summand $$\binom{3}{8}$$ calculates the number of variants to choose $$3$$ students so that A and B stay in the second team.