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Anonymous Tue 21 Jul 04:34:33 2020 No.11921481 View Reply Original Report

Quoted By: >>11921494 >>11921502 >>11921598 >>11921701 >>11921759 >>11922245 >>11922823

Anons, I can't figure out this problem.
>Given a bag of n different groups, each group contains m identical balls, what is the probability of drawing two balls from the bag with replacement such that the drawn two balls belong to the same group?
What is the most intuitive way to think of this problem?

Anonymous

Anonymous Tue 21 Jul 2020 04:39:44 No.11921494 Report

Quoted By: >>11921737

>>11921481
based thread incoming

Anonymous

Anonymous Tue 21 Jul 2020 04:41:50 No.11921499 Report

Quoted By: >>11921683 >>11921740

You draw one ball, you replace it and now you need to draw from the same group that you drew the first place, seems like the answer is just 1/n since with replacement the number of balls in each group doesn't matter.

Anonymous

Anonymous Tue 21 Jul 2020 04:43:18 No.11921502 Report

Quoted By:

>>11921481
50% either you do or you dont

Anonymous

Anonymous Tue 21 Jul 2020 05:28:35 No.11921598 Report

Quoted By:

>>11921481
>each group contains m identical balls

If each group has the same number of balls, then the odds are the same as if each group has just one ball. Are you sure that that is actually what the question is asking?

Anonymous

Anonymous Tue 21 Jul 2020 06:01:32 No.11921683 Report

Quoted By:

>>11921499
This.

Let's say your groups are the ROY G BIV rainbow colors, and each group has five balls. (I.e., n=7,m=5)

The first draw doesnt matter, so let's say you draw a green ball, which is replaced. So what are your odds of drawing a green ball next time? There are 5 green balls out of 35 total, so 5/35 or 1/7, which is 1/n.

Repeat it for any m and/or n and you get the same result. Basic, VERY basic statistics.

Anonymous

Anonymous Tue 21 Jul 2020 06:11:41 No.11921701 Report

Quoted By: >>11921740 >>11921841

>>11921481
Since this is for any m, think of the different groups as being in bags of their own, such that you draw ALL the balls of that group at the same time - they're being replaced anyway, so it doesnt change anything. Same for the second draw, you're stopping the experiment after the second draw so it doesn't matter that you draw the whole group.

So it is really just a case of drawing from a bag with n objects, with replacement. And the odds of that are always 1/n.

Anonymous

Anonymous Tue 21 Jul 2020 06:22:15 No.11921737 Report

Quoted By:

>>11921494
absolutely based thread

Anonymous

Anonymous Tue 21 Jul 2020 06:24:29 No.11921740 Report

Quoted By: >>11921768 >>11921789 >>11921922

>>11921499
>>11921701

Not OP, but why do I struggle with probability, permutations and combinatorics while calculus is easy as shit for me?

Anonymous

Anonymous Tue 21 Jul 2020 06:30:49 No.11921759 Report

Quoted By:

>>11921481

if the bag has 2 red balls and 2 green balls then it has 4 balls and 2 groups with 2 balls on each group.

Probability of selecting a red ball is 2/4 and a green is 2/4. Since you replace the ball, your second selection will also have a probability of 2/4.

2/4 * 2/4 is 25 percent. This makes sense because if you took a green ball, and then replaced it, you only have a 2/4 percent chance of getting the a green ball again. question is unclear but it sounds like the events are completely independent so you use the multiplication rule.

If you had 4 groups of 5 balls you might have 5 red, 5 green. and 5 black and 5 white.

p(red)=5/20
p(green)=5/20
p(black)=5/20
p(white)=5/20

5/20 * 5/20 is .0625 shows that you have a 5/20 chance of picking a ball from any group. Once you replace it and need to get a ball from the same group you will have a 5/20 or 1/4 chance of picking the same group. If you didn't have replacement you'd have a 4/19 chance of picking the same group, which is a bit less than 5/20

assuming all groups have the same number of balls then
if n = groups and m = balls in groups

so the formula would be (m/(nm))^2

Anonymous

Anonymous Tue 21 Jul 2020 06:33:52 No.11921768 Report

Quoted By: >>11921786

>>11921740
the questions in low level probability classes are very poorly worded and stats textbooks and professors tend to be bad at explaining the ideas.

Anonymous

Anonymous Tue 21 Jul 2020 06:40:21 No.11921786 Report

Quoted By: >>11921792 >>11921802 >>11921831 >>11921925

>>11921768
>the "it's textbooks/profs' fault!" meme
or
he is just retarded
chances are, "calculus" is taking a derivative of cosx
so the answer is - whatever "calculus" he is taught, it's a set of rules, that's why it's easy, because it's designed for monkeys
combinatorics require reformulating the problem and that's too hard for a typical 4channeler

Anonymous

Anonymous Tue 21 Jul 2020 06:41:11 No.11921789 Report

Quoted By:

>>11921740
In combinatorics every problem can be completely different than the other and it's rarely obvious how to approach it (or even to be sure exactly what is being asked)
Calculus (at least at the undergrad level) is about technique, and most of the times it's obvious what technique you need to employ, that is to say, it is a more "routine" and straightforward approach. In combo often it is not clear at all how to approach a problem, it gets easier with practice though

Anonymous

Anonymous Tue 21 Jul 2020 06:42:49 No.11921792 Report

Quoted By: >>11921806

>>11921786
>chances are, "calculus" is taking a derivative of cosx
I've done partial derivative, integrations and multi order calculus.

Anonymous

Anonymous Tue 21 Jul 2020 06:47:24 No.11921802 Report

Quoted By: >>11921813

>>11921786
>he is just retarded
I think you're right. I'm just retarded.

Anonymous

Anonymous Tue 21 Jul 2020 06:50:54 No.11921806 Report

Quoted By: >>11921813 >>11921925

>>11921792
>I've done partial derivative, integrations and multi order calculus.
... and all of these are just "follow these rules and you are set"
the problems are always formulated in such a way that there is nothing left but to compute (like "Find the integral of the flow of this vector field over this surface, here are all the required equations"

In combinatorics or intro prob you almost never get a problem like "Find the number of permutations with n=.., k=..". You have to think first whether these are permutations or combinations or whatever

Regardless, both calculus and combinatorics/intro probability have been successfully taught to high schoolers
The latter is more important in principle as it illustrates how well you can interpret contexts and translate them into mathematical notation

Anonymous

Anonymous Tue 21 Jul 2020 06:53:20 No.11921813 Report

Quoted By:

>>11921806
>>11921802
And Oh... I flunked math, if it wasn't obvious enough already.

Anonymous

Anonymous Tue 21 Jul 2020 07:05:12 No.11921831 Report

Quoted By:

>>11921786
>it requires reformulating the question
no it really doesn't, its no more difficult than an optimization problem lol

Anonymous

Anonymous Tue 21 Jul 2020 07:09:41 No.11921841 Report

Quoted By:

>>11921701

Wrong.

You are suggesting that the chances of picking a ball from the same group 100 times in a row is the same as picking it 1 time in a row. That's obviously not true.

Anonymous

Anonymous Tue 21 Jul 2020 07:56:51 No.11921922 Report

Quoted By: >>11921924

>>11921740
Relatable. I am pretty jealous of guys good at such stuff

Anonymous

Anonymous Tue 21 Jul 2020 07:58:19 No.11921924 Report

Quoted By: >>11921925

>>11921922
there's literally nothing to be jealous of its not an impressive ability. probability is built off of analysis and calculus not combinatorics

Anonymous

Anonymous Tue 21 Jul 2020 08:01:53 No.11921925 Report

Quoted By:

>>11921924
>>11921806
>>11921786

Anonymous

Anonymous Tue 21 Jul 2020 10:24:33 No.11922245 Report

Quoted By: >>11922313

>>11921481
mc2 / (n*m)c2

Anonymous

Anonymous Tue 21 Jul 2020 11:01:31 No.11922313 Report

Quoted By:

>>11922245
based chad Einsteinian

Anonymous

Anonymous Tue 21 Jul 2020 15:17:08 No.11922823 Report

Quoted By:

>>11921481
oh good question i have no clue since i gave up that class a month ago and i dont want to think. you should ask in a better place, not in underage ifls reddit

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