sana.noor wrote:If the students are A, B, C, D, E, F, G, H, and I, what is the probability that students A, B and C will be on the same team?
(A) 1/3
(B) 1/9
(C) 1/16
(D) 1/27
(E) 1/28

This question is incomplete.
How many teams are we creating? How many on each team?

Cheers,
Brent

I see.

So, the original question divided the 9 people into 3 teams and asked for the probability that 2 specific people are on the same team. The extension question still divides the 9 people into 3 teams but asks for the probability that 3 specific people are on the same team. I guess I didn't reword the question since the original question was part of the thread.

Given this wording, your solution is perfect.

Cheers,
Brent

Hi sana.noor,

Assuming that you're picking a team with just 3 people on it, here's one way to get to the solution:

We want A, B and C on the same team; there are 9 total people to choose from and since it's a "team", it doesn't matter what order they're in (ABC is the same team as CBA).

So, the first person could be any of the 3; THAT person is a given - now it's just a matter of figuring out IF the other two people are there too.

Let's say A is first....

The probability that B or C is next is 2/8. Assuming one of the two is picked, then there's 1 left....
The probability of the final person is 1/7.

(2/8)(1/7) = 2/56 = 1/28

Final Answer: E

GMAT assassins aren't born, they're made,
Rich