Brent Hanneson wrote:Extension question:
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

Brent Hanneson wrote:You're right, DD; the answer is E
Here's how I would tackle this.
First place player A on any team.
We want the probability that B and C are placed on the same team as A.
The probability that B is placed on A's team is 1/4 (determined earlier).
Now, if A and B are already on the same team, then what is the probability that C is also placed on this same team?

_ _ _ | _ _ _ | A B _
There are 7 possible spaces remaining, but only one space will put C on the same team as A and B. So, the probablity that C joins A and B is 1/7.

So, P(B AND C are placed on A's team) = 1/4 x 1/7 = 1/28

Brent Hanneson wrote:

dtweah wrote:If 9 students are to be seperated into three teams, what is the probability that any two students are on the same team?
A. 1/4
B. 1/3
C. 1/2
D. 1/6
E. 2/7

I believe the answer is A.
Let's look at it this way. Here are our three teams, each with three spaces for students:
_ _ _ | _ _ _ | _ _ _

We want the probability of two players (say players A and B) ending up on the same team.

First assign player A to one team. For example, we might place him/her on the third team as follows:
_ _ _ | _ _ _ | A _ _

Now we place player B. In how many ways can we place player B? There are 8 spots remaining, so there are 8 ways to place player B.
In how many ways can we place player B such that players A and B are on the same team? There are only 2 spots remaining on player A's team, so there are 2 ways.
So, the probaility is 2/8 = 1/4