Is the answer for this question from the Fitzpatrick textbook wrong? (1 Viewer)

[bleh1234]

8 people are to be into 2 groups. What is the probability that there will be 4 in each group?
I'm getting 35/127 but the answer says 35/81.
This is my working: (8C4) ÷ 2 / (8C1 + 8C2 + 8C3 + (8C4 ÷ 2)
Thank you very much

 

[braintic]

8 people are to be into 2 groups. What is the probability that there will be 4 in each group?
I'm getting 35/127 but the answer says 35/81.
This is my working: (8C4) ÷ 2 / (8C1 + 8C2 + 8C3 + (8C4 ÷ 2)
Thank you very much

I agree with your answer. But the working is much simpler:
7C3 / (2^7 - 1)

 

[bleh1234]

Thank you
Can you please explain the 2^7-1 part?

 

[braintic]

Thank you
Can you please explain the 2^7-1 part?

Pick one person (say 'person A') to be in one group. Each of the other 7 people have 2 choices - they are in A's group or they are in the other group. That's 2^7 possibilities. Then subtract the case where they all end up in A's group.

 

[bleh1234]

OH!
Awesome, thanks again