Marbles in a Bag Probability Question

  • Context: High School 
  • Thread starter Thread starter ZZ Specs
  • Start date Start date
  • Tags Tags
    Probability
Click For Summary

Discussion Overview

The discussion revolves around a probability question involving drawing marbles from a bag. Participants explore the probability of drawing exactly 3 red marbles out of 5 total draws from a bag containing 19 red and 21 blue marbles. The conversation includes various approaches to solving the problem, including combinatorial methods and conditional probabilities.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant asks for the probability of drawing exactly 3 red marbles out of 5, prompting others to clarify the distinction between "exactly 3" and "at least 3" red marbles.
  • Another participant suggests using a tree diagram to visualize the problem but expresses confusion about calculating probabilities for multiple draws from a finite set.
  • A different participant advises against using conditional probabilities and instead recommends counting the combinations directly, emphasizing the need to choose 3 red from 19 and 2 blue from 21.
  • One participant calculates the total ways to choose 5 marbles from 40 but questions whether this method accounts for the changing probabilities after each draw.
  • Another participant explains the probability of drawing marbles in a specific order and notes that while the order affects the fractions, the overall probability remains consistent across different arrangements.
  • It is clarified that the number of different orders of drawing 3 red and 2 blue marbles is given by the combination formula, not by simply calculating 5!.

Areas of Agreement / Disagreement

Participants express differing views on the best approach to solving the problem, with some advocating for combinatorial counting and others suggesting conditional probabilities. The discussion remains unresolved as participants explore various methods without reaching a consensus on a single solution.

Contextual Notes

Some participants note the importance of distinguishing between "exactly" and "at least" in probability calculations, which may affect the approach taken. There are also mentions of the complexities involved in calculating probabilities with non-repetition, indicating that assumptions about independence may not hold in this context.

ZZ Specs
Messages
16
Reaction score
0
Let's pretend I have a bag of 40 marbles. 19 marbles are red, 21 marbles are blue. If I randomly pick 5 marbles out of the bag, what is the probability that 3 of those 5 marbles are red?
 
Physics news on Phys.org
What do YOU think the answer is? What is your reasoning?

On this forum, we try to help folks learn how to solve problems. We don't spoon feed answers.

And by the way, this is a homework type problem so you are supposed to use the homework template. Please read the forum rules.
 
Keep in mind that it makes a difference if you are looking for exactly 3 of 5, or at least 3 of 5.
 
Chronos said:
Keep in mind that it makes a difference if you are looking for exactly 3 of 5, or at least 3 of 5.

Keep this in mind, it does in deed make a difference!
 
Hello all! Sorry for the late response, hope you're all still around. Let's pretend I want exactly 3 out of 5, since it's not too hard (I think) to then add the probabilities of 3/5 +(or) 4/5 +(or) 5/5.

I can work it out on a tree, for instance:

19/40*(..conditional probabilities..) + 21/40*(conditional probabilities)

with 5 separate columns (if that makes sense) but I know there has to be a better way.

I understand how basic combinations could give us the chances of pulling 3 red marbles in 3 draws (i.e. all red, or all blue) but I get really messed up when I try to think about 3 red marbles out of 5 draws out of 40 marbles.

It's been a while since I did statistics haha, some hints would be nice! :D And sorry about the homework template / improper forum section, it's also been a bit since I've come to PF. No excuse, of course, totally my fault. I can re-post in the homework section if it's an issue.
 
Don't think about conditional probabilities, just count. You know how many ways there are to draw 5 from 40. Now if 3 are red then you must have chosen 3 from 19 and the remaining 2 from 21. See if that helps and post your answer. Simple counting and multiplication, no need to think harder than that.
 
Yes, 5C40 gives 658,008 ways of picking 5 marbles out of 40.

Does that not ignore certain conditions of non-repetition though? Once you draw the first marble, the probabilities of the next draw have already changed.

I believe it is correct that there are 5! = 120 ways of picking 3 red and 2 blue marbles, if that is anywhere near where you are getting at. Thanks for your help so far!
 
One way of picking 3 red and 2 blue marble is "RRRBB".

The probability the first marble you pick is red is, of course, 19/40. Now there are 39 marbles left and 18 are red. The probability that the second marble is red is 18/39. Now there are 38 marbles left and 17 are red. The probability that the third marble is red is 17/38. Now there are 37 marbles left and 21 of them are blue. The probability the fourth marble is blue is 21/37. There are now 36 marbles left and 20 of them are blue. The probability the fifth marble is blue is 20/36.

That will give you the probability of three red marbles followed by two blue marbles.

But if you calculate a different order, say RRBRB, you will see that while you get different fractions, the numerators and denominators are the same, just in different orders. That is, the probability of 'three red and two marbles' in any specific order is the same! You only need to multiply by the number of different orders.

But that is NOT 5!. Since all red are the same and all blue marbles are the same, the number of different orders is
_5C_3= \begin{pmatrix}5 \\ 3\end{pmatrix}= \frac{5!}{3!2!}= \frac{5(4)}{2}= 10.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 10 ·
Replies
10
Views
1K
Replies
2
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
8
Views
3K