Earlier, someone asked a question of the form,
Given A girls and B boys
what is the probability of selecting X girls and Y boys?
For example, there are 3 girls and 3 boys. What is the probability of randomly selecting 2 boys and 2 girls (without replacement)?
I am curious as to what formulas for probability I am missing here.

- inkyvoyd

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- chestercat

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- dumbcow

use combinations
\[\rightarrow \frac{(3C2)(3C2)}{(6C4)}\]

- inkyvoyd

Yes, but why those answers?

- anonymous

The denominator is consistent to http://en.wikipedia.org/wiki/Vandermonde's_identity

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## More answers

- anonymous

yes

- anonymous

and the numerator is is just common sense.

- inkyvoyd

(by formulas I mean something like formula for event A and B, disjoint A and B, etc. Or even just something like binomial prob)

- inkyvoyd

TIL that I have no common sense D:

- anonymous

at first.. we have to select x+y from a+b
this can be done by a+bCx+y ways
and to fulfill the condition, we have aCx * bCy ways
so, probability: \[^{a} C _{x} * ^{b} C _{y }/^{a+b}C _{x+y}\]

- anonymous

Oh no she didnt!

- anonymous

i cant guess any other method..

- dumbcow

probability = desired outcomes/ total outcomes
6C4 is total ways you can pick 4 out of 6
picking 2 boys out of 3 AND picking 2 girls out of 3
2 independent events so multiply those outcomes

- inkyvoyd

Actually, I think I understand it after Arnab explained it. Thanks everyone!

- anonymous

I am sure there must be another one using one of the convoluted probability distribution but this is the easiest in my opinion.

- anonymous

Arnab is good teacher

- anonymous

teacher? me!! OMG..

- inkyvoyd

btw, how do you solve round table problems? Like, there are 5 people sitting at a round table... What do I divide 5P/Cr by?

- anonymous

Circular permutation.

- anonymous

n linear arrangement consists of 1 circular arrangement. So, when you have n people sitting round a table then you have n!/n =(n-1)! circular permutation. Note the clockwise and anticlockwise are considered distinct here. other wise you need to multiply by 1/2

- inkyvoyd

Ah, thanks again :)

- anonymous

welcome. what does the username inkyvoyd signify?

- inkyvoyd

It signifies the inability to come up with a creative username/ign, so I just picked two random works and misspelled one so it would be less likely to be taken.

- anonymous

lol, fair enough.

- inkyvoyd

Btw, one last question (I came up with this after I took thought about the probability of taking guesses and getting a score on an actual test)
Let's say I have 60 multiple choice questions each with 4 choices and that the total score is 100(%).
Now, 20 of those questions are 3 points each, and 40 are 1 point each. Let's say that answer distribution is equal among the four choices (A,B,C,D). How would I find the probability of getting a score n?

- anonymous

are you in high school?

- inkyvoyd

Freshman year. I self studied trig, precalc concepts, and calc one (not very well). I'm trying to learn calc 2, but it's going at a snail's pace. I've never taken a formal statistics course pre or post calculus, but I learned a little about normal distribution in an algebra 2 class.

- anonymous

That's great, this is actually a nice problem. Which I need to cogitate a bit.

- inkyvoyd

Cool. I got answers on reddit, but they weren't very detailed (I just got a distribution graph of a similar problem which I posted, and the problem was quickly forgotten)

- anonymous

Can you pass me the link ? Reddit

- inkyvoyd

Also, I forgot to add in the problem, I randomly guess with no bias to any choice.

- anonymous

Okay so you are just attempting blindly? (not solving)

- inkyvoyd

http://www.reddit.com/r/cheatatmathhomework/comments/re6fv/high_school_probability/

- inkyvoyd

For now, attempting blindly.
(I'll probably make up a bunch of variations on this problem, but I have no idea how to solve the first, so I'll just try to keep it simple)

- anonymous

When you have this kind of problem, this is best place to get your answers : http://math.stackexchange.com/

- inkyvoyd

Is it that hard?

- anonymous

It could be you are adding lots of variations of you own, I often do that and sometime end up scratching open problems.

- anonymous

your*

- perl

fool, what was your sqrt x + sqrt y = sqrt a , problem, dealing with tangents , yesterday

- anonymous

The answer is a. perl

- inkyvoyd

So, I could post it there without getting criticism for giving an elementary math problem?

- inkyvoyd

nvm about that, I got the wrong site in mind

- anonymous

http://math.stackexchange.com/questions/81683/

- anonymous

http://math.stackexchange.com/questions/60726

- anonymous

http://math.stackexchange.com/questions/12587/

- anonymous

No lol users at math.stackexchange are very critical, be careful.

- anonymous

I don't think your problem is more elementary than them.
Also probablity distribution ain't elementary.

- anonymous

Ishaan I am an user and I ain't critical :P

- perl

fool, im talking about the problem you posted yesterday, about the tangent or something

- perl

the question above is a hypergeometric distribution , fool

- inkyvoyd

I'll just hope it'll be too hard for them, and they won't make fun of me. :D

- anonymous

yes perl I know I have answered it.

- perl

the question you posted yesterday, i think its one of your special questions

- perl

you posted like 10 open questions

- anonymous

I mean most of the users or most of the active users.

- anonymous

They never make fun of anyone, if the problem is bad we would close it down

- anonymous

If I were you I would have posted it. I dont' really care what others thinks about me as long as I am learning ...;)

- anonymous

perl I know the answer is a.

- anonymous

Yes, you should post it.

- anonymous

You may like to add what have you tried so far. so that others may help you appropriately.

- inkyvoyd

Done,
http://math.stackexchange.com/questions/127549/probability-distribution-im-pretty-sure-its-higher-than-high-school-math-but

- anonymous

+1 from me :)

- inkyvoyd

thanks :D

- inkyvoyd

By the way, how intertwined are statistics and other mathematics? (like graduate level I mean, I just want to have a general idea of what I will be learning if I major or minor in math)

- anonymous

well pretty much however I am not sure if any with analysis.

- perl

If the tangent the to the curve x√+y√=a√ at any point on cuts x-axis and y-axis at two distinct points. Can you find the sum of the intercepts ? Good luck!

- anonymous

perl, for the n'th time the answer is a

- inkyvoyd

wait, how do I have a profile without registering? That's an awesome system.

- perl

i didnt ask the answer,

- perl

i just post the question, see ?

- anonymous

That's an Unregistered profile with limited privileges.

- perl

but i will work on it, give me a moment

- perl

as far as your conics question, you took the partial deriv because the slope is zero there

- perl

i have a 3d grapher, im trying to get a feel for this

- inkyvoyd

Wow, answer already!

- anonymous

Assume the point to be \((x_0,y_0)\)
\[\frac{dy}{dx} = -\sqrt{\frac{y_o}{x_0}}\]Equation of tangent at point \((x_0,y_0)\) must be,
\[y - y_0 = -\sqrt{\frac{y_0}{x_0}}(x - x_0)\]
Get the intercept form of the equation and then add them up.
That's how I did it, but I would love to see a better way to do this.

- anonymous

That's how I did it too.

- inkyvoyd

It's "definitely basic", but I don't understand a word of the explanation.

- anonymous

Lol

- perl

what is xo,yo? in your problem

- anonymous

Happens a lot to me, I don't understand most of the question on math.stackexchange :/
But don't worry, you just need to work hard and then in a year, you will be ruling math.stackexchange.

- perl

you are not given any (xo,yo)

- anonymous

a year ruling MSE? !! Dude those people are professors

- inkyvoyd

I think I'll go back there after I finish college.

- anonymous

I assumed them to be a point on the curve for the point of contact.

- perl

it says two distinct points

- inkyvoyd

I think a year might get me finished with calculus and differential equations (hopefully discrete and linear), but I want to learn so much more than just that.

- perl

maybe i misunderstand the question.
If the tangent the to the curve x√+y√=a√ at any point on cuts x-axis and y-axis at two distinct points. Can you find the sum of the intercepts ? ,
this is not grammatically correct

- perl

so (xo,yo) is the point on the curve that the tangent goes through , ok

- perl

the intercept form is
x/a + y/b = 1 , i believe

- anonymous

Not all, I have seen several guys just aged 16 or 17 with more than 1k+ points.
I think if one prepares for the IMO, he can easily answer most of the questions on math.stackexchange. Serious preparation.

- anonymous

Yes, that is the intercept form.

- anonymous

I mean't people with 10k+ rep.

- anonymous

and there are few exception here and there.

- anonymous

But 1k+ is not that hared.

- perl

so whats the next step

- anonymous

Oh hmm yeah probably 1k isn't that hard. I mean yeah, 10k+ and 1k+ huge difference. 10k+ is definitely hard.

- inkyvoyd

Time to close this monster post xD

- anonymous

Get the equation of tangent.

- anonymous

Okay!

- perl

ok

- perl

how do you get a, i dont see

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