mamma_mia66
Jan30-09, 12:04 AM
1. The problem statement, all variables and given/known data
suppose that 35 people are divided in a random manner into two teams in such a way that one team contains 10 people and the other team contains 25 people. What is the probability that two particular people A and B will be on the same team.
2. Relevant equations
I don't have the answer on this problem , I am confused. Please help.
3. The attempt at a solution
35 people team C contains 10 people, so the chance to be in this team is 2/7
team D contains 25 people, so the chance is 5/7
when person A is divided in team C , then there is a chance for person B, to be also in this team, namely 9/34(1 person is already divided)
when person B is divided in team D, then there is a chance for person B, to be also in this team, namely 24/34
so there are 2 cases
case 1 has a chance of 2/7*9/34=18/238
case 2 has a chance of 5/7*24/34=120/238
together this is 138/238=69/119 (=0.58) that they are in the same team (the chance is always bigger than 50%)
suppose that 35 people are divided in a random manner into two teams in such a way that one team contains 10 people and the other team contains 25 people. What is the probability that two particular people A and B will be on the same team.
2. Relevant equations
I don't have the answer on this problem , I am confused. Please help.
3. The attempt at a solution
35 people team C contains 10 people, so the chance to be in this team is 2/7
team D contains 25 people, so the chance is 5/7
when person A is divided in team C , then there is a chance for person B, to be also in this team, namely 9/34(1 person is already divided)
when person B is divided in team D, then there is a chance for person B, to be also in this team, namely 24/34
so there are 2 cases
case 1 has a chance of 2/7*9/34=18/238
case 2 has a chance of 5/7*24/34=120/238
together this is 138/238=69/119 (=0.58) that they are in the same team (the chance is always bigger than 50%)