probability qs (1 Viewer)

[lum]

10 people are divided into 5 groups of two
what's the probability of 2 particular ppl in same group

this is what i thought

no. ways for 2 particular ppl in a group is 2C2, and the 8 others can be arranged in 8C2 x 6C2 x 4C2 x 2C2

total no. ways for 5 groups of 2 ppl is 10C2 x 8C2 x 6C2 x 4C2 x 2C2

therefore probability is 1/ (10C2)

is that right?

 

[KFunk]

I'm never sure whether I'm using the correct logic when I'm doing probability but my first guess would be:

Let the two people be A and B. Let A be in a specific group. There are 9 people remaining which can be paired with A, one of which is B.

This yields a 1/9 probability that A and B will be together.

 

[KFunk]

10 people are divided into 5 groups of two
what's the probability of 2 particular ppl in same group

this is what i thought

no. ways for 2 particular ppl in a group is 2C2, and the 8 others can be arranged in 8C2 x 6C2 x 4C2 x 2C2

total no. ways for 5 groups of 2 ppl is 10C2 x 8C2 x 6C2 x 4C2 x 2C2

therefore probability is 1/ (10C2)

is that right?

10C2 x 8C2 x 6C2 x 4C2 x 2C2 treats the groups like permutations, e.g. if the people are pi ( i = 1, 2, ... 10) then you're treating:

p1,p2 ... p3,p4 ... p5,p6 .... p7,p8.... p9,p10

as though it were different to

p3,p4 ... p5,p6 .... p7,p8.... p9,p10...p1,p2

(for the purpose of making 'groups' you can see that these are actually the same). However, if you're going to do that then there are 5 ways A and B can be together (they can be in positions 1, 2, 3, 4 or 5. So that gives you:

5/¹⁰C₂ = 1/9

 

[who_loves_maths]

lum, your method would be correct if the groups in the question are distinct, able to be labelled A, B, C, D, E. {but you need to consider that the group with the two ppl can be in any of those five sets, i.e. times by 5.}
if not, then you have double counted.

the (8C2 x 6C2 x 4C2 x 2C2) would then need to be divided by 4!
and (10C2 x 8C2 x 6C2 x 4C2 x 2C2) by 5!
before dividing them.

 

[Stefano]

KFunk is 'teh masterer combinatrix'

 

[lum]

whoa, so the answer's just 1/9? thx dudes , i knew i was missing the "getting rid of doubled up combinations" part

 

[Sirius Black]

help on a similar q

8 members of a tennis club meet to play two simultaneous sets of double tennis on two separate by otherwise identical courts. In how many different ways can 8 members of the club be selected?

Is it same to the q if they ask how many different ways to arrange 8 ppl into 4 groups of two?

Thanx

 

[KFunk]

Because you have doubles matches I think you have to split them into groups of 4 first so the way I'm looking at it it's like:

⁸C₄/2 = the number of ways to make two unique groups of 4 from 8.

For group one there are ⁴C₂/2 ways to make two unique groups of 2 from 4 and likewise for the second group.

Combined this gives ⁸C₂(⁴C₂)²/8 = 315



*Edit: I just thought of another way to do it. Put a single player (A) in place in one of the courts. The are 7 people who can be A's partner, leaving ⁶C₂ pairs that can verse A and their partner. In the next court leave one person in place, then there are 3 people who can partner them leaving the other two left over to play against them. This gives:

(3)(7)(⁶C₂) = 315

 

[Stefano]

Because you have doubles matches I think you have to split them into groups of 4 first so the way I'm looking at it it's like:

⁸C₄/2 = the number of ways to make two unique groups of 4 from 8.

For group one there are ⁴C₂/2 ways to make two unique groups of 2 from 4 and likewise for the second group.

Combined this gives ⁸C₂(⁴C₂)²/8 = 315



*Edit: I just thought of another way to do it. Put a single player (A) in place in one of the courts. The are 7 people who can be A's partner, leaving ⁶C₂ pairs that can verse A and their partner. In the next court leave one person in place, then there are 3 people who can partner them leaving the other two left over to play against them. This gives:

(3)(7)(⁶C₂) = 315

I told you!