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Probability

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by scoobydooby » Tue May 26, 2009 7:12 am
probability of selecting any 1 (say A) member: 1/9
this member can be on any one of the 3 teams. probability of putting A on any one team =3*1/9=1/2

probability of selecting one particular member (say B) given that A has been selected: 1/3*1=1/3

would go for B.
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by hk » Tue May 26, 2009 7:44 am
My answer is B

There are 9 students, say A, B, C, D, E, F, G, H, I

lets say the question is asking for the probability that A and B are in the same team.

Prob of A to be in any of the 3 teams is 1/3
Now B has be in the same team and so the probability is 1/3

Now probability that A and B are in the same team = 1/3 * 1 = 1/3
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Re: Probability

by Brent@GMATPrepNow » Tue May 26, 2009 8:39 am
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
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by Brent@GMATPrepNow » Tue May 26, 2009 8:44 am
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
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by VP_Jim » Tue May 26, 2009 8:44 am
I'm with Brent on this one. Player A can go to any team, so he is out of the picture.

Essentially, this question is asking: what is the probability that Player B will end up on Player A's team? That makes it a lot easier!
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by dumb.doofus » Tue May 26, 2009 10:24 am
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
I think the answer is E.. i.e. 1/28

I have still not mastered the quick way to do things like you showed above Brent, so my solution is pretty rudimentary. I look forward to a quicker way to solve this from you..

Solution:

Possible events i.e. denominator: (9C3*6C3)/3! (Divided by 3! coz the order doesn't matter) ----------- (1)

Given A, B and C are in one team..

now the number of possible combinations = 6C3/2 (divided by 2 coz order doesnt matter) ------------ (2)

So probability = (2)/(1)
= 1/28
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by Brent@GMATPrepNow » Tue May 26, 2009 11:03 am
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
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by dumb.doofus » Tue May 26, 2009 11:22 am
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
Yes.. this is much simpler.. Many thanks for sharing..
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Re: Probability

by dtweah » Tue May 26, 2009 11:40 am
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
Nice approach Brent. OA is A.

Identify 2 two persons. That leaves 7. There are 7C1 ways one of the 7 can join the two. That leaves 6 people. 6C3 or 3C3. But order doesn't matter so have divide by 2 giving 7x 10.

3 groups of 3 each out of 9 is 280. 9C3 x 6C3 3C3)/3!
70/280=1/4.


Regarding Brent's extension question.

Fix 3 students. The remaining 2 groups come from 6 students. (6C3, 3C3)/2
10/280=1/28
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by Mani JVS » Sat May 30, 2009 11:05 pm
total number of cases is C(9,2)

Total Number of ways = choosing 2 from 9 =C(9,2).

out of 3 people, 2 people can be chosen in C(3,2) ways, and we can have 3 teams of 3 people, and hence 3 * C(3,2) is the total number of ways in which 2 people can be in the same team.

So, Prob = (Favorable Number of Cases)/(Total Number of Cases)
= 3 * C(3,2) / C(9,2) = 3 * 3 * 2 / (9 * 8) = 1/4

In a similar way, if we need 3 people to be in the same team,

Prob = 3 * C(3,3) / C(9,3) = 3 * 3 * 2 / (9 * 8 * 7) = 1/28
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