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Stats Question

Registered User regular
I am in now way trying to get someone to do my homework. If there are four management posisitions, and there are 5 womeon and 7 men, what is the chance that the trainee class is all men? We're supposed to do a (n,C,r) for probability, but shouldn't the probability be 20%, because we know the outcomes can only be 0 women, 4 men; 1 woman, 3 men; et cetera, giving us 5 possible outcomes. I have the combination answer, but I still don't know why I would need to do the combination in the first place.

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Posts

• Registered User regular
Because there are multiple possible ways that you can choose candidates and end up with say 3 men and 1 women, or 4 men, or 2 and 2. Each candidate is distinct, so woman1...woman5 and man1...man7. For all 4 men you could have:
man1, man2, man3, man4
man1, man2, man3, man5
...
man4, man5, man6, man7
etc.

There are only 5 possible outcome bins with respect to the number of men and women, but there are many possible combinations and they don't result in an even split between bins, so the probability is not just 1/5.

• Registered User
(0 women, 4 men) is not as likely as (1 women, 3 men) or (2 women, 2 men), etc.

If you do what Daenris did and list out every single possible combination, then count the portion that are entirely men, you won't get 20%:

m1m2m3m4
m1m2m3m5
...
w1w2m1m2
w1w2m1m3
...
etc.

• Registered User regular
Because their are alot more ways that it can be 4 men (7C4) instead of just one.

Second example: There is going to by 7C2*5C2 different ways it could be two of each. This is because their are 7C2 ways of picking two out of 7 and 5C2 ways of picking 2 from 5.

Where you say there are 5 possible outcomes. That is true, but each of these outcomes have different probabilities. In the same way that rolling a 7 on two dice isn't 1/11 even though there are 11 different outcomes.

• Registered User regular
The answer would be 100% if you are cynical and believe in the glass ceiling.

The simplest way to refute your statement is that there are only 11 possible outcomes when you roll and sum up 2 six-sided dice (the standard 2d6 roll in most RPGs), but that doesn't mean you have a 1/11 chance to get a 2 or 12. You actually have a 1/36 chance to get a 2, and a 1/6 chance to get a 7. In other words, the outcomes are not equivalent to the probability. This is a common misconception that leads to things like The Birthday Paradox and the like.

Steam ID: Hahnsoo, Steam Name currently: Hahnsopolis | PSN: Hahnsoo | Monster Hunter Tri: Hahnsoo, E8HJCA
• Registered User regular
I suspected as much. So for stats am I supposed to be selectively pedantic?

• Registered User regular
Yeah, I see it now. I have to account for the population, the distinct variables in the population, and their relavance to each other. Hmmm. Stats is weird.

• Registered User regular
Also, can you assume that all of the trainees are going to be equally qualified and thus it comes down to random chance? I'm not sure what the context of this problem would be, since it's worded in a way that it can totally be a trick question. Is there any assumption that all of the candidates are equally likely to get one of the slots?

Steam ID: Hahnsoo, Steam Name currently: Hahnsopolis | PSN: Hahnsoo | Monster Hunter Tri: Hahnsoo, E8HJCA
• Registered User regular
Yeah, I see it now. I have to account for the population, the distinct variables in the population, and their relavance to each other. Hmmm. Stats is weird.

Yes, stats is weird. Also, don't trust an answer just because it "sounds" right. That's part of the fun of stats. I use the term "fun" loosely.

Also, I get either 7.0% or 7.3%, depending which way I go. I think it's 7.0%.
Also, can you assume that all of the trainees are going to be equally qualified and thus it comes down to random chance? I'm not sure what the context of this problem would be, since it's worded in a way that it can totally be a trick question. Is there any assumption that all of the candidates are equally likely to get one of the slots?

It's a stats problem, so I doubt it's a trick question. It sounds like a basic combination problem.

• Registered User regular
uhhh 7C4 / 12C4

7C4 is the number of combinations that can be made with your 7 men, and there are 12C4 possible groups.

7*6*5*4 / 4 * 3 * 2 * 1 = 35
12 * 11 * 10 * 9 / 4 * 3 * 2 * 1 = 495

35 / 495

• Registered User regular
My calculator says 7C4 is 35, not 21.

But yeah.

• Registered User regular
oh yeah I skipped 5 in my counting, so fix the numbers

• Registered User regular
mcdermott wrote: »
Also, can you assume that all of the trainees are going to be equally qualified and thus it comes down to random chance? I'm not sure what the context of this problem would be, since it's worded in a way that it can totally be a trick question. Is there any assumption that all of the candidates are equally likely to get one of the slots?

It's a stats problem, so I doubt it's a trick question. It sounds like a basic combination problem.

This. Statistics isn't really interested in asking you a question and then saying 'ololz, all menz r janitors'*. Now, you could very well be asked to describe flaws in an approach, as understanding the limitations and assumptions of a methodology is important, but it should be obvious when that happens.

Equal probability of each item is actually a normal assumption in statistics. If it doesn't hold, you'll probably be told.
When used without qualification, the word "random" usually means "random with a uniform distribution."

* Or, in fact, the reverse of this, which is a real concern in some fields. But menz is a fun word.

What if this weren't a rhetorical question?
• Decorative Monocle Registered User regular
Yeah, I see it now. I have to account for the population, the distinct variables in the population, and their relavance to each other. Hmmm. Stats is weird.

Take the advice of someone who's taken several classes in statistics. When it comes to calculating probabilities, throw your "common sense" out the window. It never works. Do the process.

• J.2C When's KoFRegistered User regular
Stats professor and phd student here! I go through a process like this whenever I do counting problems:

probability = (# of ways something happens)/(# of ways anything happens)
= (# of ways to choose 4 men from 7)/(# of ways to chose any 4 from total)
= (7C4)/(12C4) = .... = 7/99

aka what musanman wrote.

If Blanka receives any nerfs in 2013, and I mean any, I will eat a box of cocks and change characters forever.
@Folken_fgc
• Registered User regular
Terrendos wrote: »
Yeah, I see it now. I have to account for the population, the distinct variables in the population, and their relavance to each other. Hmmm. Stats is weird.

Take the advice of someone who's taken several classes in statistics. When it comes to calculating probabilities, throw your "common sense" out the window. It never works. Do the process.

NEVER EVER EVER

Seriously, the words "common sense" and "sounds about right" are your worst fucking enemies in a basic stats class, because the entire point of the class is often to throw problems at you were the answer is either not intuitive or actively counter-intuitive, thus forcing you to learn the process.

• Registered User regular
Hahnsoo1 wrote: »
The answer would be 100% if you are cynical and have experienced the glass ceiling