Verifying answer of probability question This is the question: Eight teams are competing in basketball game. Three out of them are European. If the teams are paired randomly, what is the probability that none of the European teams have to play against eachother? And this is how I have solved it: (5C1*3C1 + 5C2)/8C2 Is this correct? Edit : Explanation This is a combination problem because the order doesn't matter in this case. Now, I have to get two teams at a time from total 8. So, the total number of combinations are 8C2. Either I can choose two non-european teams or one non-european and other european. So, when for both non-european I'll take two teams from 5 teams (since there are 5 non-european teams). Moreover, I can only choose one team from 5 non-european and 3 from european (And these because both will happen at the same time).

Let's number the teams and let $E_i$ denote the event that European team $i$ does not have to play against a European team.

The probability equals: $$P(E_1\cap E_2\cap E_3)=P(E_1\cap E_2)=P(E_1)P(E_2\mid E_1)=\frac57\frac45=\frac47$$The first factor is the probability that European team1 will not have to play against a European team.

(there are $5$ suitable choices out of $7$ choices in total)

The second is the probability that European team2 will not have to play against a European team under the condition that European team1 will not have to play against a European team.

(European team1 plays against a non-European team, so in this situation there are $4$ remaining suitable choices out of $5$ remaining choices in total)

If European team1 and European team2 do not have to play against a European team then it is for sure that also European team3 does not have to play against a European team.