What is the Probability That Two Specific People Will Be on the Same Team?

  • Thread starter Thread starter mamma_mia66
  • Start date Start date
  • Tags Tags
    Probability
mamma_mia66
Messages
51
Reaction score
0

Homework Statement



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.


Homework Equations



I don't have the answer on this problem , I am confused. Please help.

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%)
 
Physics news on Phys.org


I get the same answer.
 


Then it is okay. Thanks.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top