A little help in probability and stats?

[Rain Man]

Man, I'm out of touch on this stuff.

I have 100 orbs*, 50 of which are red and 50 of which are white. I draw 10 of them out randomly without replacement.

How many total combinations can be drawn, and how many of them occur in each of the 10 potential outcomes (e.g., 0 red and 10 white, 1 red and 9 white, etc.)

I need to do this calculation to save the earth from being hit by a large asteroid, so quick help would be appreciated. If we survive, I'll give you rep.





* - I'm not going to say balls because it'll divert the thread.

[Saul Good]

You guys are making this too hard. The fact that there are 100 balls doesn't factor into the equation because we're looking for possibility, not probability. If I understand the question correctly, it's two to the tenth power.

[Third Eye]

For 3 W, 7 R: 1,957,734,240,000
For 4 W, 6 R: 3,659,628,210,000
For 5 W, 5 R: 4,489,143,937,600

From here, you have everything you need. There are the same number of 4 W and 6 R as there are 6 W and 4 R.

[Third Eye]

Originally Posted by DanT

Hey Rain Man,

That's the hypergeometric distribution. Excel has a function hypergeomdist() for that.

Here's the Wolfram link on the hypergeometric distribution:

http://mathworld.wolfram.com/Hyperge...tribution.html


--Dan

This is absolutely right if all he needs is the probability of drawing any particular number. Since he asked for the number of combinations, I gave them to him.

[AustinChief]

Originally Posted by DanT

Hey Rain Man,

That's the hypergeometric distribution. Excel has a function hypergeomdist() for that.

I was just looking online for a calc for it.. this is the best I could find http://stattrek.com/online-calculato...geometric.aspx

Should be able to run it for each possible outcome and get your statistics.

(unless I'm not understanding what he is asking for)

[AustinChief]

Ok here should be the solution if I understand what you need.


obviously, 11 combinations

red,white - prob.

0,10 - 0.000593419672585829
1,9 - 0.00723682527543693
2,8 - 0.0379933326960439
3,7 - 0.11309643221148
4,6 - 0.211413217031686
5,5 - 0.259333546225535
6,4 - 0.211413217031686
7,3 - 0.11309643221148
8,2 - 0.0379933326960439
9,1 - 0.00723682527543693
10,0 - 0.000593419672585828

[Pitt Gorilla]

Originally Posted by Saul Badguy

You guys are making this too hard. The fact that there are 100 balls doesn't factor into the equation because we're looking for possibility, not probability. If I understand the question correctly, it's two to the tenth power.

I thought this initially, as he simply asked for the possible outcomes. HOWEVER, he also noted that he's sampling 10 (ostensibly pulling one at a time) without replacement. The without replacement part wouldn't matter if he was only interested in possible outcomes.

We needed a clearer description of the actual problem, IMO.

[AustinChief]

Originally Posted by Pitt Gorilla

I thought this initially, as he simply asked for the possible outcomes. HOWEVER, he also noted that he's sampling 10 (ostensibly pulling one at a time) without replacement. The without replacement part wouldn't matter if he was only interested in possible outcomes.

We needed a clearer description of the actual problem, IMO.

yeah, I'm kinda guessing at what is wanted in regards to the probabilities. If he even wants probabilities.

[Third Eye]

Originally Posted by Pitt Gorilla

I thought this initially, as he simply asked for the possible outcomes. HOWEVER, he also noted that he's sampling 10 (ostensibly pulling one at a time) without replacement. The without replacement part wouldn't matter if he was only interested in possible outcomes.

We needed a clearer description of the actual problem, IMO.

The problem is plenty clear, and has been answered both in number of combinations and probability. Both answers are correct as well.

[DanT]

You good, RainMan?

There are 100*99*98*...*91 = 100! / 90! = 100! / (100-10)! ways to draw 10 balls from 100, denoted P(100,10), if we could distinguish the balls apart. That's the total number of elementary outcomes. Given that we can't distinguish among the white balls or among the red balls, then we can group all these elementary outcomes into 2 ** 10 distinct outcomes where the order would matter.

Here for example, is one of those

WRWRR RRRRW

The above permutation of repeated objects has 3 white orbs and 7 red orbs. Given that I could get 3 white orbs in any of P(50,3) ways from the urn of 50 white balls and I could get 7 red orbs in any of P(50,7) ways, there are

P(50, 3) * P(50, 7 ) ways that I could generate that particular permutation from the elementary outcomes.

Now suppose that I want to group all of the permutations together according to how many unique white balls it has. In other words, I no longer care about the order in which those white ball appear.

Well, each permutation has the same number of elementary possiblities, P(50, 3) * P(50, 7). Note that I could identify a permutation by the location of the white balls. So, for 3 white balls out of 10, there are P(10,3) / 3! distinguishable locations, which we can denote as C(10,3) combinations. So that means that there are C(10, 3) * P(50, 3) * P(50, 7) elementary outcomes that result in a total of 3 white balls out of 10.

If we want to compute the probability of getting 3 white balls out of 10, you just divide the total number of elementary possibilities that yield 3 white balls,

C(10,3) * P(50,3) * P(50, 7) by the total number of elementary possiblities,

P(100, 10 ).

That's how you get the probability mass function for the hypergeometric. It simplifies very nicely to the formula that is expressed simply using combinations, the formula you'll see in authoritative sources.

I hope you kept that asteroid from hitting us. If you break that rock down a little bit and send it toward the Chiefs front office, I'd appreciate it.

[Baby Lee]

http://www.r-project.org/

Came late to the discussion, but between this bit of open source and the discussions on the site, assuming a baseline of knowledge in P&S, you solve not only this question but any others that come up for you.

[DanT]

Originally Posted by Baby Lee

http://www.r-project.org/

Came late to the discussion, but between this bit of open source and the discussions on the site, assuming a baseline of knowledge in P&S, you solve not only this question but any others that come up for you.

That's a great product. Lots of cool stuff you can do with that freeware!

[Third Eye]

Originally Posted by DanT

If we want to compute the probability of getting 3 white balls out of 10, you just divide the total number of elementary possibilities that yield 3 white balls,

C(10,3) * P(50,3) * P(50, 7) by the total number of elementary possiblities,

P(100, 10 ).

If you don't mind me asking, why wouldn't you just do 50C3*50C7/100C10?

Sorry for the different notation, that's just the way I've always seen it.

[AustinChief]

Originally Posted by Baby Lee

http://www.r-project.org/

Came late to the discussion, but between this bit of open source and the discussions on the site, assuming a baseline of knowledge in P&S, you solve not only this question but any others that come up for you.

Nice, had never seen that!

[DanT]

Originally Posted by Third Eye

If you don't mind me asking, why wouldn't you just do 50C3*50C7/100C10?

Sorry for the different notation, that's just the way I've always seen it.

I don't mind the question. Starting from the number of elementary outcomes and then grouping them together leads to the same formula, but laying out each step and and using standard terms along the way could be helpful to a reader who wonders where the formula comes from or why there are different terms for related concepts, like permutations and combinations, possibilities and probabilities.

[Black Bob]

It's kind of funny that "Rain Man" is asking for help with numbers and stats....