Probability of socks selection

  • Context: High School 
  • Thread starter Thread starter weibing86
  • Start date Start date
  • Tags Tags
    Probability
Click For Summary

Discussion Overview

The discussion revolves around calculating the probability of selecting 4 socks from a total of 12 socks (6 pairs of different colors) such that 2 socks are of the same color and the other 2 are of different colors. The conversation includes attempts to derive the solution using both combinatorial reasoning and probability distributions.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant expresses uncertainty about how to start the probability question regarding sock selection.
  • Several participants inquire about the initial attempts made by the original poster, emphasizing the importance of identifying the sample space and relevant events.
  • One participant suggests narrowing the problem by using a smaller sample size of 3 pairs of socks to explore combinations.
  • Another participant introduces the hyper-geometric distribution as a potential model for solving the problem, recommending that the original poster learn about its derivation and assumptions.
  • A participant presents a calculation involving combinations to derive the probability, but expresses uncertainty about the correctness of their approach.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct method to solve the problem, and multiple approaches and models are discussed without resolution.

Contextual Notes

There are limitations regarding the assumptions made in the calculations and the applicability of different probability models, which remain unresolved.

Who May Find This Useful

This discussion may be useful for individuals interested in probability theory, combinatorial problems, and those seeking to understand different approaches to solving probability-related questions.

weibing86
Messages
3
Reaction score
0
I don't even know how to start this question:

I have 6 pairs of socks (12 socks) of different colors.

If I choose 4 socks randomly, what is the probability that I will pick 2 socks of same color, and the other 2 are different colors?
 
Physics news on Phys.org
Hey there webing86 and welcome to the forums.

Have you made an attempt at answering the question? If so could you show whatever you've tried so that we can help you see what you understand, and that you have attempted the problem on your own (its the forum policy to do this).

In saying that, what have you identified as your sample space and the appropriate subset for any or all of the events? (ie probability of one sock any color, two socks same color, two socks same with one with other color, or all four socks)?

Its best to break up your probability space into its simplest events and then use probability axioms to figure out the total space of the event.
 
chiro said:
Hey there webing86 and welcome to the forums.

Have you made an attempt at answering the question? If so could you show whatever you've tried so that we can help you see what you understand, and that you have attempted the problem on your own (its the forum policy to do this).

In saying that, what have you identified as your sample space and the appropriate subset for any or all of the events? (ie probability of one sock any color, two socks same color, two socks same with one with other color, or all four socks)?

Its best to break up your probability space into its simplest events and then use probability axioms to figure out the total space of the event.


Hi,

I have tried to solve this by narrowing my sample size.

I used 3 pairs of socks (red, red), (blue, blue), and (yellow, yellow).

The combination of picking 2 socks of same or different colors will be 6 (rr, rb, ry, bb, by, yy). I can list out all the combinations as the sample size is small, but I just do not how to derive it using formula.
* If all the socks are of different colors, of course the combination will be 6C2.
 
weibing86 said:
Hi,

I have tried to solve this by narrowing my sample size.

I used 3 pairs of socks (red, red), (blue, blue), and (yellow, yellow).

The combination of picking 2 socks of same or different colors will be 6 (rr, rb, ry, bb, by, yy). I can list out all the combinations as the sample size is small, but I just do not how to derive it using formula.
* If all the socks are of different colors, of course the combination will be 6C2.

Its not a good idea to try and think of things in terms of formulas. Formulas can be useful if you have a system with some pretty common assumptions (i.e. there are a lot of real world phenomena that use these assumptions, so its handy to use that particular model), but its a lot better to use the foundational concepts and axioms of probability.

Having said that, there is a model known as the hyper-geometric distribution:

http://en.wikipedia.org/wiki/Hypergeometric_distribution

Your kind of problem looks like it can be solved using this kind of distribution. This distribution has a probability density function that is used to calculate probability given a total multivariate sample size (N categories of things - in your case sock colors) and from this you use the formula to get your probability.

If I were you I would try to find a derivation of the distribution and look at any assumptions that have been made to get to that result. Its probably a good idea to learn the binomial first and work your way up to the multivariate hyper-geometric distribution.
 
say the 6 colors are a,b,c,d,e and f.

then, the given cases will be:

aa-bc
aa-bd
aa-be
aa-bf
aa-cd
aa-ce
... and so on.

(1) 2 same color and 2 different colors : 6 * 5C2 = 6 * 10 = 60
(2) all the cases : 12C4 = 55x9

P = (1)/(2) = 60 / (55*9) = 4 / 33


I am not 100% sure though... :p
 

Similar threads

  • · Replies 36 ·
2
Replies
36
Views
5K
  • · Replies 3 ·
Replies
3
Views
7K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K