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Someone teach me how to count

Anonymous Mon 06 Apr 11:04:58 2020 No.11538184 View Reply Original Report

Quoted By: >>11538282 >>11538294 >>11538474 >>11540393 >>11540446 >>11540450

Please I am so fucking dumb

Every day, the 15 students in Mr. Evan's class are randomly divided into 5 groups of 3 students each. What is the probability that three of the students–Adam, John, and Lance–are in the same lab group today?

If someone could walk me through the number of ways to think about calculating the numbers of steps we have, I'd really appreciate it.

Anonymous

Anonymous Mon 06 Apr 2020 11:06:26 No.11538188 Report

Quoted By:

Also will accept book suggestions on what I should read. Please no super advanced books since I have no brain.

Anonymous

Anonymous Mon 06 Apr 2020 11:25:24 No.11538206 Report

Quoted By: >>11538214

Khan academy videos about combinatorics

Anonymous

Anonymous Mon 06 Apr 2020 11:28:13 No.11538211 Report

Quoted By: >>11538214 >>11538221

Isn't it just a 1 in 3 chance?

Anonymous

Anonymous Mon 06 Apr 2020 11:29:15 No.11538214 Report

Quoted By:

>>11538206

Have you seen them? They're literally like "This is how you can make an outfit from 3 shirts and 2 pants."

They don't help.

>>11538211

I don't know. But I am looking to learn how to count, the answer doesn't really matter.

Anonymous

Anonymous Mon 06 Apr 2020 11:34:15 No.11538221 Report

Quoted By: >>11538241

>>11538211
How would it be 1/3? lol. That makes no sense.

Anonymous

Anonymous Mon 06 Apr 2020 11:45:22 No.11538235 Report

Quoted By: >>11538244

why do we need to count? isn't it just 3!*12!/15!
?

Anonymous

Anonymous Mon 06 Apr 2020 11:47:53 No.11538241 Report

Quoted By: >>11538244 >>11540481

>>11538221
Oh I saw 3 groups of 5 students. Lol
It's 1 in 5.

Anonymous

Anonymous Mon 06 Apr 2020 11:48:42 No.11538244 Report

Quoted By: >>11538247

>>11538241
>>11538235

No, it's not. The answer is 1/91 if you want to know.

Anonymous

Anonymous Mon 06 Apr 2020 11:49:16 No.11538247 Report

Quoted By: >>11538250 >>11538251 >>11540481

>>11538244
why

Anonymous

Anonymous Mon 06 Apr 2020 11:51:44 No.11538250 Report

Quoted By:

>>11538247

idk I don't know how to count. That's why I made the thread?

Anonymous

Anonymous Mon 06 Apr 2020 11:51:55 No.11538251 Report

Quoted By:

>>11538247
oh right

Anonymous

Anonymous Mon 06 Apr 2020 12:10:18 No.11538282 Report

Quoted By: >>11538290

>>11538184
I don't think you're dumb OP. I think that's a tricky question. How much of the problem do you understand? Can you explain your thinking?

Anonymous

Anonymous Mon 06 Apr 2020 12:14:49 No.11538290 Report

Quoted By: >>11538292 >>11538314

>>11538282
In this question, I am not sure how to make the problem more manageable. Of course, there are 15! ways to arrange all of the people, but I am not and there are (3!)^5 ways to arrange all of the people in each group individually, but I am not sure where to go from here.

Anonymous

Anonymous Mon 06 Apr 2020 12:15:51 No.11538292 Report

Quoted By:

>>11538290
I typo'd, I didn't mean to say the "but I am not" in the middle of the sentence.

Anonymous

Anonymous Mon 06 Apr 2020 12:18:27 No.11538294 Report

Quoted By: >>11538297

>>11538184
1/5*1/5*1/5?

Anonymous

Anonymous Mon 06 Apr 2020 12:20:18 No.11538297 Report

Quoted By: >>11540481

>>11538294
or maybe 1/15 * 1/14 * 1/13

Anonymous

Anonymous Mon 06 Apr 2020 12:23:10 No.11538303 Report

Quoted By:

15x14x13 gives the amount of possible ordered combinations. Then you have to account for the fact that a group of 3 can be ordered in 6 different ways, so the amount of groups with unique combinations of students can be given by (15x14x13)/6. One of this number of groups has the three specific students. Because there are five trials in one day, multiply this by 5, then by 100 to find the percent chance, which is 1.0989%.

Anonymous

Anonymous Mon 06 Apr 2020 12:27:20 No.11538314 Report

Quoted By: >>11538393

>>11538290
I can see how they got the answer of 1/91. But I'm not entirely sure I can explain why they got that.

The way they did is to use 15C3 to determine the number of combinations of any one group. This gives 455.

Then, they reason that there are five groups made, and so there are five opportunities for this to occur. Thus they give the probability as 5/455 which is 1/91. However, I'm having a little trouble justifying to myself why this is good reasoning.

Anonymous

Anonymous Mon 06 Apr 2020 13:00:07 No.11538374 Report

Quoted By: >>11540481

Is it not 1/125?

Anonymous

Anonymous Mon 06 Apr 2020 13:10:05 No.11538393 Report

Quoted By:

>>11538314
Could you look at it like it's 5 separate tries? The probability that you suceed is the probability that you succeed at one try. On first try, you draw three students randomly in the group of 15. You have 1/(15C3) chance that you get the target triple. If not, you try again, but among the remaining students. Does that describe the situation better for you?

Anonymous

Anonymous Mon 06 Apr 2020 13:57:30 No.11538474 Report

Quoted By:

>>11538184
Make sure that your result reflects these facts:
1.) You form 5 groups each day.
2.) The three people may possibly be in any of these five groups.
3.) The composition of groups within a day are interlinked. i.e. people already in one group can not be in another group. (that day)

Anonymous

Anonymous Mon 06 Apr 2020 20:44:19 No.11539616 Report

Quoted By: >>11539638 >>11539643

Posts like this show how mathematically disabled /sci is.

You only care about Adam's group. The fact that you have four other groups is completely inconsequential.
Adam is with 2 other people. How many ways are there for this to happen? 14C2, which is 91. How many ways are good? The one where he is with the pair {John, Lance}. Thus giving 1/91.

Anonymous

Anonymous Mon 06 Apr 2020 20:50:25 No.11539638 Report

Quoted By:

>>11539616

What do you mean? I literally said I am dumb. Unless you're talking about the replies.

Anonymous

Anonymous Mon 06 Apr 2020 20:51:28 No.11539643 Report

Quoted By: >>11540408 >>11540481

>>11539616
Also, could you please recommend a book or something to learn this stuff? I really want to get better at it.

Anonymous

Anonymous Tue 07 Apr 2020 01:08:27 No.11540393 Report

Quoted By: >>11540481

>>11538184
15!/(3!(15?3)!)
= 455
You are dividing by five because there are five groups and they can only be in one of those groups at a time.

Anonymous

Anonymous Tue 07 Apr 2020 01:12:31 No.11540408 Report

Quoted By:

>>11539643
Bumping in case anyone has a book recommendation for problems like this.

Anonymous

Anonymous Tue 07 Apr 2020 01:29:22 No.11540446 Report

Quoted By:

>>11538184
This is fucking easy
Its \frac{\frac{12!}{3!3!3!3!4!}}{\frac{15!}{3!3!3!3!3!5!}}

Anonymous

Anonymous Tue 07 Apr 2020 01:31:36 No.11540450 Report

Quoted By: >>11540451

>>11538184
Its $\frac{\frac{12!}{3!3!3!3!4!}}{\frac{15!}{3!3!3!3!3!5!}}$

Anonymous

Anonymous Tue 07 Apr 2020 01:32:24 No.11540451 Report

Quoted By: >>11540458

>>11540450
No logic doesn't count.

Anonymous

Anonymous Tue 07 Apr 2020 01:35:07 No.11540458 Report

Quoted By: >>11540485

>>11540451
Well, then learn the logic.
Why does the multinomial theorem work?
Why do you divide a factorial term by multiple, smaller factorial terms?

Anonymous

Anonymous Tue 07 Apr 2020 01:44:36 No.11540481 Report

Quoted By: >>11540485

There is a 2/14 chance that John is with Adam, because there are 14 slots besides Adam's slot which John can take, and 2 of those 14 are with Adam.
There are 13 slots besides Adam's and John's slot, one of which is Lance's slot. Of those 13 slots, only 1 is with Adam and John, if Adam and John are together. Therefore, if Adam and John are together (which itself is a 2/14 chance), there is a 1/13 chance that Lance is also with them.
Therefore, there is a (2/14)(1/13) chance that all 3 are together which is 1/91.
This is 12 year old math.
>>11538374
no you're just dumb
>>11538297
you're also dumb
>>11538241
you're even more dumb
>>11538247
cause people that aren't dumb said so
>>11540393
It doesn't matter which of the 5 groups they are in, just that they are all 3 in the same group, so it is actually 5x as likely as you deemed. 455/5=91 so it's 1/91.
>>11539643
I assume that it would be covered in a high school level statistics/probability textbook, but I never used such a book or took a statistics class so idk. Check the wiki in the sticky for specific textbooks.

Anonymous

Anonymous Tue 07 Apr 2020 01:46:45 No.11540485 Report

Quoted By: >>11540497 >>11540523

>>11540458

The multinomial has nothing to do with this? What?

>>11540481

Do you have a recommendation that you would give? Wiki books seem complicated.

Anonymous

Anonymous Tue 07 Apr 2020 01:50:29 No.11540497 Report

Quoted By: >>11540499

>>11540485
As I said I haven't used a statistics book. Just go to the /sci/ wiki (check sticky for URL), it lists textbook recommendations. Scroll through the list of math textbooks, find a intro level statistics/probability one, then buy it if you're a good goy or download for free on libgen if doing so is legal in your country, and see if the book covers this stuff

Anonymous

Anonymous Tue 07 Apr 2020 01:51:19 No.11540499 Report

Quoted By: >>11540513

>>11540497

Could you tell me where you learned this?

Anonymous

Anonymous Tue 07 Apr 2020 01:54:51 No.11540513 Report

Quoted By: >>11540527

>>11540499
You mean the method I used to solve it? I don't recall learning it anywhere. I didn't have some method or set of steps memorized. I just used logic when I came across this thread and figured out for myself how to solve it.
If you're looking for what kind of textbook or class would cover it, probably statistics as I said or maybe high school algebra 2 or college algebra? I haven't taken an algebra class since 9th grade and and I don't remember if this stuff was covered there.

Anonymous

Anonymous Tue 07 Apr 2020 01:59:28 No.11540523 Report

Quoted By: >>11540527

>>11540485
Yes, it does.
What does the multnomial theorem claim about coefficients? You should be able to explain this combinatorially (without induction). Apply this explanation to multi-tiered combinations (like lab group membership).

Anonymous

Anonymous Tue 07 Apr 2020 02:01:01 No.11540527 Report

Quoted By: >>11540535 >>11540540

>>11540513

So you have no formal education in math, and you never read a book on this and were able to solve it? Well, I have no hope for the future. :)

>>11540523

The multinomial theorem does not explain the logic behind what you did at all. You still need to be able to count to use it.

Anonymous

Anonymous Tue 07 Apr 2020 02:04:22 No.11540535 Report

Quoted By:

>>11540527
>you have no formal education in math
I studied algebra and calculus in school but that was a decade ago and I don't remember if or when I did problems like this. That said I am very naturally good at math, i.e. I took a calculus class in 10th grade (age 15)
My advice would be to focus more on logic, reasoning, figuring out how to solve the problem yourself and understand the math, instead of just trying to memorize steps, methods, and formulas without understanding them.

Anonymous

Anonymous Tue 07 Apr 2020 02:05:39 No.11540540 Report

Quoted By:

>>11540527
There you go, that is what you need to study. You don't even need to be able to count to understand the multinomial theorem, o=you just need a vague idea of what counting is. Once you can explain the multinomial theorem to a Boomer, then you can understand the logic relevant to this problem.

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