inkyvoyd
  • inkyvoyd
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
dumbcow
  • dumbcow
use combinations \[\rightarrow \frac{(3C2)(3C2)}{(6C4)}\]
inkyvoyd
  • inkyvoyd
Yes, but why those answers?
anonymous
  • anonymous
The denominator is consistent to http://en.wikipedia.org/wiki/Vandermonde's_identity

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anonymous
  • anonymous
yes
anonymous
  • anonymous
and the numerator is is just common sense.
inkyvoyd
  • 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
  • inkyvoyd
TIL that I have no common sense D:
anonymous
  • 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
  • anonymous
Oh no she didnt!
anonymous
  • anonymous
i cant guess any other method..
dumbcow
  • 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
  • inkyvoyd
Actually, I think I understand it after Arnab explained it. Thanks everyone!
anonymous
  • 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
  • anonymous
Arnab is good teacher
anonymous
  • anonymous
teacher? me!! OMG..
inkyvoyd
  • 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
  • anonymous
Circular permutation.
anonymous
  • 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
  • inkyvoyd
Ah, thanks again :)
anonymous
  • anonymous
welcome. what does the username inkyvoyd signify?
inkyvoyd
  • 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
  • anonymous
lol, fair enough.
inkyvoyd
  • 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
  • anonymous
are you in high school?
inkyvoyd
  • 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
  • anonymous
That's great, this is actually a nice problem. Which I need to cogitate a bit.
inkyvoyd
  • 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
  • anonymous
Can you pass me the link ? Reddit
inkyvoyd
  • inkyvoyd
Also, I forgot to add in the problem, I randomly guess with no bias to any choice.
anonymous
  • anonymous
Okay so you are just attempting blindly? (not solving)
inkyvoyd
  • inkyvoyd
http://www.reddit.com/r/cheatatmathhomework/comments/re6fv/high_school_probability/
inkyvoyd
  • 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
  • anonymous
When you have this kind of problem, this is best place to get your answers : http://math.stackexchange.com/
inkyvoyd
  • inkyvoyd
Is it that hard?
anonymous
  • 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
  • anonymous
your*
perl
  • perl
fool, what was your sqrt x + sqrt y = sqrt a , problem, dealing with tangents , yesterday
anonymous
  • anonymous
The answer is a. perl
inkyvoyd
  • inkyvoyd
So, I could post it there without getting criticism for giving an elementary math problem?
inkyvoyd
  • inkyvoyd
nvm about that, I got the wrong site in mind
anonymous
  • anonymous
http://math.stackexchange.com/questions/81683/
anonymous
  • anonymous
http://math.stackexchange.com/questions/60726
anonymous
  • anonymous
http://math.stackexchange.com/questions/12587/
anonymous
  • anonymous
No lol users at math.stackexchange are very critical, be careful.
anonymous
  • anonymous
I don't think your problem is more elementary than them. Also probablity distribution ain't elementary.
anonymous
  • anonymous
Ishaan I am an user and I ain't critical :P
perl
  • perl
fool, im talking about the problem you posted yesterday, about the tangent or something
perl
  • perl
the question above is a hypergeometric distribution , fool
inkyvoyd
  • inkyvoyd
I'll just hope it'll be too hard for them, and they won't make fun of me. :D
anonymous
  • anonymous
yes perl I know I have answered it.
perl
  • perl
the question you posted yesterday, i think its one of your special questions
perl
  • perl
you posted like 10 open questions
anonymous
  • anonymous
I mean most of the users or most of the active users.
anonymous
  • anonymous
They never make fun of anyone, if the problem is bad we would close it down
anonymous
  • 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
  • anonymous
perl I know the answer is a.
anonymous
  • anonymous
Yes, you should post it.
anonymous
  • anonymous
You may like to add what have you tried so far. so that others may help you appropriately.
inkyvoyd
  • inkyvoyd
Done, http://math.stackexchange.com/questions/127549/probability-distribution-im-pretty-sure-its-higher-than-high-school-math-but
anonymous
  • anonymous
+1 from me :)
inkyvoyd
  • inkyvoyd
thanks :D
inkyvoyd
  • 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
  • anonymous
well pretty much however I am not sure if any with analysis.
perl
  • 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
  • anonymous
perl, for the n'th time the answer is a
inkyvoyd
  • inkyvoyd
wait, how do I have a profile without registering? That's an awesome system.
perl
  • perl
i didnt ask the answer,
perl
  • perl
i just post the question, see ?
anonymous
  • anonymous
That's an Unregistered profile with limited privileges.
perl
  • perl
but i will work on it, give me a moment
perl
  • perl
as far as your conics question, you took the partial deriv because the slope is zero there
perl
  • perl
i have a 3d grapher, im trying to get a feel for this
inkyvoyd
  • inkyvoyd
Wow, answer already!
anonymous
  • 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
  • anonymous
That's how I did it too.
inkyvoyd
  • inkyvoyd
It's "definitely basic", but I don't understand a word of the explanation.
anonymous
  • anonymous
Lol
perl
  • perl
what is xo,yo? in your problem
anonymous
  • 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
  • perl
you are not given any (xo,yo)
anonymous
  • anonymous
a year ruling MSE? !! Dude those people are professors
inkyvoyd
  • inkyvoyd
I think I'll go back there after I finish college.
anonymous
  • anonymous
I assumed them to be a point on the curve for the point of contact.
perl
  • perl
it says two distinct points
inkyvoyd
  • 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
  • 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
  • perl
so (xo,yo) is the point on the curve that the tangent goes through , ok
perl
  • perl
the intercept form is x/a + y/b = 1 , i believe
anonymous
  • 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
  • anonymous
Yes, that is the intercept form.
anonymous
  • anonymous
I mean't people with 10k+ rep.
anonymous
  • anonymous
and there are few exception here and there.
anonymous
  • anonymous
But 1k+ is not that hared.
perl
  • perl
so whats the next step
anonymous
  • anonymous
Oh hmm yeah probably 1k isn't that hard. I mean yeah, 10k+ and 1k+ huge difference. 10k+ is definitely hard.
inkyvoyd
  • inkyvoyd
Time to close this monster post xD
anonymous
  • anonymous
Get the equation of tangent.
anonymous
  • anonymous
Okay!
perl
  • perl
ok
perl
  • perl
how do you get a, i dont see

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