Any math wiz here? Need help with a problem.

[Lager]

Solved. Thank you aux and Furyline.

 

[Lager]

I thought there were many smart students here with 1400+ SAT scores and stuff. Please help.

 

[bleeb]

Originally posted by: lager
I thought there were many smart students here with 1400+ SAT scores and stuff. Please help.

The simplicity of this problem is way too low to be looked at from the extraordinary minds of ATOT. Please give us problems with regards to physics, quantum physics, combinatorics, Optimization. =P

 

[Lager]

Originally posted by: bleeb

Originally posted by: lager
I thought there were many smart students here with 1400+ SAT scores and stuff. Please help.

The simplicity of this problem is way too low to be looked at from the extraordinary minds of ATOT. Please give us problems with regards to physics, quantum physics, combinatorics, Optimization. =P

Then solve it so that I know you're not stupid.

 

[CubicZirconia]

I must be missing something here. When you say "5 women voted together and 3 men voted together," and "If there was no bias between 5 women voting for the same thing and 3 men voting for the same thing," what exactly do you mean? The men voted in one room and the women voted in another? Or that all the men voted for one thing and all the women voted for another? How many options were there?

I guess I'm reading into this too much. It's worded too funny.

 

[zer0burn]

do you own hw I took this stuff in 1st year highschool

 

[DrPizza]

It's so simple, I don't want to risk that it's a trick question and destroy my credibility as a mathematician and physicist.

hint:
what are all the possibilities?
women Yes men No
women No men yes
women yes men yes
women no men no

Figure it out.

 

[Lager]

Originally posted by: CubicZirconia
I must be missing something here. When you say "5 women voted together and 3 men voted together," and "If there was no bias between 5 women voting for the same thing and 3 men voting for the same thing," what exactly do you mean? The men voted in one room and the women voted in another?

Basically, what is the probability or (odds/chance) of the women and men voting to the same thing, the same genders voting for the same thing, as 5 women voted together and three men voted together.

The odd that women and men voted together for the same thing.

 

[Lager]

 

[CubicZirconia]

Originally posted by: lager

Originally posted by: CubicZirconia
I must be missing something here. When you say "5 women voted together and 3 men voted together," and "If there was no bias between 5 women voting for the same thing and 3 men voting for the same thing," what exactly do you mean? The men voted in one room and the women voted in another?

Basically, what is the probability or (odds/chance) of the women and men voting to the same thing, the same genders voting for the same thing, as 5 women voted together and three men voted together.

The odd that women and men voted together for the same thing.

Well then Dr. Pizza explained it for you. The fact that there are 5 women and 3 men is irrelevant.

 

[OulOat]

Originally posted by: DrPizza
It's so simple, I don't want to risk that it's a trick question and destroy my credibility as a mathematician and physicist.

hint:
what are all the possibilities?
women Yes men No
women No men yes
women yes men yes
women no men no

Figure it out.

He never said there is only two options. I don't understand the question, if there can be more than 2 options, does that mean that all the women have to vote for the same option x and all the men has to vote for the same option y where x<>y, or that just the sets of men and women votes has cannot intersect?

 

[sharkeeper]

Women vote democrat.

Cheers!

 

[Dacalo]

This isn't too difficult, think about it a little more

 

[CubicZirconia]

I think you should copy the problem word for word out of the book. It might just be me, but I really don't understand what you are asking.

Or it could just be that I suck at math.

 

[Lager]

Updated the question more clearly.

 

[royaldank]

Not 1 in 256 chance.

 

[Lager]

Is the question more clear now? I hope it is more understandable.

Edit, no, it is not 1/256.

 

[royaldank]

I don't think I completly understand the question. What is the answer?

second guess is 1 in 56.

 

[aux]

Count the ways in which the vote can be 5-3; this is "8 choose 3". The probability is 1/"8 choose 3", because only in one of the cases it is split 5 women for, 3 men against.

 

[Lager]

Originally posted by: royaldank
I don't think I completly understand the question. What is the answer?

The question is..

What is the odd of the vote turning out 5 women and 3 men if it were not biased.

 

[Lager]

Originally posted by: aux
Count the ways in which the vote can be 5-3; this is "8 choose 3". The probability is 1/"8 choose 3", because only in one of the cases it is split 5 women for, 3 men against.

Excellent work, that's makes whole lot more sense.

 

[CubicZirconia]

Originally posted by: aux
Count the ways in which the vote can be 5-3; this is "8 choose 3". The probability is 1/"8 choose 3", because only in one of the cases it is split 5 women for, 3 men against.

We have a winner. If you have an "nCr" function on your calculator it will confirm this.

You can even type it into google. Nice

 

[Furyline]

I say 1/56

With 8 people, 3 voting against, there are 56 combinations of this happening, (ex. woman #1, w#4, man#2; w#2, m#1, m#3; etc.) In only one of those combinations do all the men vote no. So 1/56.

I hope that explains it. (and I hope that's right)
edit: well I'm about 5 minutes too late

 

[Lager]

Originally posted by: Furyline
I say 1/56

With 8 people, 3 voting against, there are 56 combinations of this happening, (ex. woman #1, w#4, man#2; w#2, m#1, m#3; etc.) In only one of those combinations do all the men vote no. So 1/56.

I hope that explains it. (and I hope that's right)
edit: well I'm about 5 minutes too late

That is absolutely correct. Thank you. The answer is 1/56.

 

[Lager]

One more stats question.

How do I figure out "A intersect B" and "B intersect C" if the given are P(A) = .4 and P(B) = .8 and P(C) = .4 ?? Thanks.