currysauce
Actuary in the making
Eight people are to be divided into 2 groups. What is the probability that there wil be 4 in each group?
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Perm/Prob (1 Viewer)
- Thread starter currysauce
- Start date
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...
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
Actuary in the making
i dont' think thats it
<sup>8</sup>C<sub>4</sub>/(2<sup>8</sup> - 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 (2<sup>8</sup> -2) establishes the following two situations as different:lucifel said:
(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:
<sup>8</sup>C<sub>4</sub>/(<sup>8</sup>C<sub>1</sub> + <sup>8</sup>C<sub>2</sub> + <sup>8</sup>C<sub>3</sub> + <sup>8</sup>C<sub>4</sub>)
Oh and sorry, my reasoning for this is:
<sup>8</sup>C<sub>1</sub> = 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.
<sup>8</sup>C<sub>2</sub> = 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 <sup>8</sup>C<sub>3</sub> and <sup>8</sup>C<sub>4</sub>
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 <sup>8</sup>C<sub>4</sub> so just divide this by the former figure (I hope my reasoning is sound):
<sup>8</sup>C<sub>4</sub>/(<sup>8</sup>C<sub>1</sub> + <sup>8</sup>C<sub>2</sub> + <sup>8</sup>C<sub>3</sub> + <sup>8</sup>C<sub>4</sub>)
[*also, I assumed that all combinations were equally likely]
<sup>8</sup>C<sub>1</sub> = 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.
<sup>8</sup>C<sub>2</sub> = 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 <sup>8</sup>C<sub>3</sub> and <sup>8</sup>C<sub>4</sub>
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 <sup>8</sup>C<sub>4</sub> so just divide this by the former figure (I hope my reasoning is sound):
<sup>8</sup>C<sub>4</sub>/(<sup>8</sup>C<sub>1</sub> + <sup>8</sup>C<sub>2</sub> + <sup>8</sup>C<sub>3</sub> + <sup>8</sup>C<sub>4</sub>)
[*also, I assumed that all combinations were equally likely]
currysauce
Actuary in the making
nope
the answer is 35/81
any help?
the answer is 35/81
any help?
Tip of the day: Learn to use a calculatorcurrysauce said:
<sup>8</sup>C<sub>4</sub>/(<sup>8</sup>C<sub>1</sub> + <sup>8</sup>C<sub>2</sub> + <sup>8</sup>C<sub>3</sub> + <sup>8</sup>C<sub>4</sub>) = 0.43209 ... = 35/81
currysauce
Actuary in the making
tip of the day, listen to KFunk
Here's something to twist your noodle: technically <sup>8</sup>C<sub>4</sub> 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 <sup>8</sup>C<sub>4</sub> should be divided by two, or rather, that <sup>8</sup>C<sub>4</sub>/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?
{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 <sup>8</sup>C<sub>4</sub> should be divided by two, or rather, that <sup>8</sup>C<sub>4</sub>/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
Active Member
I would think the latterKFunk said: