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.

[Bowser]

Colored balls? That's racist.

[Bowser]

And just off the top off my head, isn't the answer either 10,000 or 10,000,000 possible combinations? Or somewhere inbetween?

I take cashier checks or PayPal for the rep.

[DaFace]

Well, there are only 11 possible combinations, so that part's easy. (0/10, 1/9, 2/8, etc.) In notation, that'd be C(2,10) I think.

As for the probability of each combination being drawn, I'd have to dig back to remember how to do that part.

[Saul Good]

There are 11 combinations. I don't understand the second part of your question.

[Rain Man]

I know there are 11 different combinations if you ignore order (2 red, 8 white, etc). What I'm trying to get at is the number of actual different draws you can have (e.g. 1 red followed by 9 whites// 1 white, then 1 red, then 8 more whites//2 white, then 1 red, then 7 more whites, etc.)

I apologize for not having the terms down right.

I want to build a graph that says if you draw every possible combination in every possible order, how many times does each final combination appear?

For example, if I only draw 4 balls out of the 100, my possible pulls are:

WWWW
RWWW
WRWW
WWRW
WWWR
RRWW
RWRW
RWWR
WRRW
WRWR
WWRR
RRRW
RRWR
RWRR
WRRR
RRRR

So I can count the following

1 instance of 0 reds
4 instances of 1 red
6 instances of 2 red
4 instances of 3 red
1 instance of 4 red

I need to do this again, but with 10. I don't remember the equation to figure it out, and don't remember the nomenclature to even look it up. There's an 82 percent probability that I'm screwed.

[Third Eye]

This is pretty easy, give me a sec.

[Third Eye]

There are 100C10 possible outcomes, which is a big number, roughly 1.73e^13. It will take a minute do the breakdown by type.

[Third Eye]

Actually, you probably need a more exact number, so 17,310,309,456,440 is the total number of combinations. Ok, back to breaking it down by type.

[AustinChief]

The term you are looking for is permutations. (solve using factorial)

[Third Eye]

For 0 W, 10 R: 10,272,278,170
For 1 W, 9 R: 125,271,685,000
For 2 W, 8 R: 657,676,346,250

This is taking a minute longer than I thought it would, having to use a calculator on the PC as my TI-83 rounds at a certain point.

[Third Eye]

Originally Posted by AustinChief

The term you are looking for is permutations. (solve using factorial)

Actually, he's looking for combinations since order doesn't matter. At least I hope he is.

[Red Beans]

The answer is 7 miles per hour. You can thank me later.

[AustinChief]

Originally Posted by Third Eye

Actually, he's looking for combinations since order doesn't matter. At least I hope he is.

ah, i just looked at his example showing order and thought he needed ordered. my bad.
Now that I have gone back and read what he is asking, it's actually even more complicated due to the fixed set he is working with if he is asking for a statistical comparison.

[DanT]

Originally Posted by 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.

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