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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?

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    $\begingroup$ 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. $\endgroup$ – Mees de Vries Nov 14 '18 at 16:12
  • $\begingroup$ 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. $\endgroup$ – Jack Moody Nov 14 '18 at 16:15
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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.

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