Counting (Combinations) Suppose there are 10 people apart of a club: A, B, C, D, E, F, G, H, I, and J. They decide to go to a restaurant for a club outing, but there isn't one table to seat all of them, so they decide to take one table that seats four people and two tables that seat three people. Based on this, what is the probability that Person G and Person J sit at the same table?


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*Ways for both Person G and Person J to both sit at the 4-person table: 6 = (4 choose 2)

*Ways for both Person G and Person J to both sit at the 3-person table: 3 = (3 choose 2)

*Ways for both Person G and Person J to both sit at the other 3-person table: 3 = (3 choose 2)


Total of $6+3+3=12$ ways for Person G and Person J to both sit at the same table.
I've determined the number of ways that Person G and Person J can sit together at each table, but I don't know how to properly count the total different ways (i.e., the denominator) to determine the probability.

By conditional probability, we have 
$$P(J\text{ is at $G$’s table}) =
P(G\text{ at $4$-table})P(J\text{ at $G$’s table} | G\text{ at $4$-table}) + P(G\text{ at a $3$-table})P(J\text{ at $G$’s table}| G\text{ at a $3$-table})$$
$$\left(\frac4{10}\right)\left(\frac39\right)+\left(\frac6{10}\right)\left(\frac29\right) = \frac2{15} + \frac2{15} = \frac4{15}$$
