What are the possibilities of $2$ people being in the same group (group of $5$ people) among $30$ people?
I thought of using conditional probability: given that a group has been chosen the probability that a person falls into it is $1/36$. I am wrong?
To organize the comments above:
We may approach directly via definitions and counting principles. There are $\binom{30}{5,5,5,5,5,5}$ different ways in which we assign the people to groups. Of these, there are $6\cdot \binom{28}{3}\cdot \binom{25}{5,5,5,5,5}$ ways in which we assign people to groups such that Mr. A and Mr. B are both in the same group. (Six ways to choose the group, $\binom{28}{3}$ ways to choose who else is in their group, and $\binom{25}{5,5,5,5,5}$ ways to choose how to distribute the remaining people). Taking the ratio then, the probability is: $\frac{6\cdot \binom{28}{3}\binom{25}{5,5,5,5,5}}{\binom{30}{5,5,5,5,5,5}}$
Alternatively, we can break apart by disjoint events. The probability that Mr. A and Mr. B are in the same group is equal to the sum of the probabilities of Mr. A and Mr. B both being in the first group plus the probability that Mr. A and Mr. B are both in the second group, etc... noting that it is impossible for them to both be in more than one group at a time. We get for this an answer of $6\times \frac{\binom{28}{3}}{\binom{30}{5}}$ as found in the comments above (Noting again the nuance of the question that we don't care /which/ group it was that they were in, just that they were together, something that the OP originally forgot).
My preferred solution:
Let Mr. A force his way to the front of the line and get assigned a group first. Now, lay out the available remaining places in a line. Mr. B is equally likely to have been placed in any of the remaining $29$ places, only $4$ of which would be in the same group as Mr. A. The probability is then $\frac{4}{29}$. Very minimal calculation and arithmetic needed here.
All of the above approaches result in the same answer once simplified.