Perm/Prob (1 Viewer)

[currysauce]

Eight people are to be divided into 2 groups. What is the probability that there wil be 4 in each group?

 

[lucifel]

i am thinking 8C4/(2^8)

because, to pick 2 groups with 4 people in them there are 8C4 ways to do so. And the total number of ways is figured like this. There are 8 people, each with 2 choices. So yeh. Oh wait are we allowed to have empty groups? If not, you gotta subtract 2 from the denominator...

 

[currysauce]

i dont' think thats it

 

[KFunk]

i am thinking 8C4/(2^8)

because, to pick 2 groups with 4 people in them there are 8C4 ways to do so. And the total number of ways is figured like this. There are 8 people, each with 2 choices. So yeh. Oh wait are we allowed to have empty groups? If not, you gotta subtract 2 from the denominator...

⁸C₄/(2⁸ - 2) is what I was thinking at first but I got caught up on something. This situation is different to, say, distributing 8 people amongst two distinct rooms since being in one room is different to being in the other. However, in this situation using (2⁸ -2) establishes the following two situations as different:

(p1) and (p2, p3, p4, ... p8)

(p2, p3, p4, ... p8) and (p1)

For the purposes of grouping people these two situations are actually the same (though it would be different if you had two rooms or if you were distributing 8 objects between 2 people). For this reason I'm going to suggest that the answer might be:

⁸C₄/(⁸C₁ + ⁸C₂ + ⁸C₃ + ⁸C₄)

 

[KFunk]

Oh and sorry, my reasoning for this is:

⁸C₁ = the number of ways to pick 1 person from 8. In pickign 1 from 8 you establish two groups (of 1 and of 7 respectively). This accounts for the number of 1 and 7 group combinations.

⁸C₂ = The number of combinations with a group of 2 and a group of 6 (in taking 2 from 8 you make a group of 2 whilst leaving behind a group of 6)

... etc onto ⁸C₃ and ⁸C₄

Add these together to get the number of ways to arrange 8 people into two groups. Naturally the number of ways for 8 people to be arranged into two groups of 4 is ⁸C₄ so just divide this by the former figure (I hope my reasoning is sound):

⁸C₄/(⁸C₁ + ⁸C₂ + ⁸C₃ + ⁸C₄)

[*also, I assumed that all combinations were equally likely]

 

[currysauce]

nope

the answer is 35/81

any help?

 

[KFunk]

nope

the answer is 35/81

any help?

Tip of the day: Learn to use a calculator

⁸C₄/(⁸C₁ + ⁸C₂ + ⁸C₃ + ⁸C₄) = 0.43209 ... = 35/81

 

[lucifel]

k funk, you're right, i shoulve divided by two, would've got the same answer.

 

[currysauce]

tip of the day, listen to KFunk

 

[KFunk]

Here's something to twist your noodle: technically ⁸C₄ includes the mirror cases since out of the 8 people you can pick:

{p1, p2, p3, p4} or {p5, p6, p7, p8}

If you removed either grouping from the inital 8 you will create the same two groups. Hence it would make sense that ⁸C₄ should be divided by two, or rather, that ⁸C₄/2 is the number of ways to make two groups of 4 out of 8 which would make my answer (and that of the book) incorrect.

My question is: Am I going crazy with pre-HSC stress or is Fitzpatrick a gronk?

 

[rama_v]

Here's something to twist your noodle: technically ⁸C₄ includes the mirror cases since out of the 8 people you can pick:

{p1, p2, p3, p4} or {p5, p6, p7, p8}

If you removed either grouping from the inital 8 you will create the same two groups. Hence it would make sense that ⁸C₄ should be divided by two, or rather, that ⁸C₄/2 is the number of ways to make two groups of 4 out of 8 which would make my answer (and that of the book) incorrect.

My question is: Am I going crazy with pre-HSC stress or is Fitzpatrick a gronk?

I would think the latter