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I have come across this problem in a book I am studying out of interest.

A drawer contains 4 different pairs of socks. Find the probability that (a) if 2 socks are selected at random they will form a pair, (b) if 4 socks are selected at random they will form 2 pairs.

I solved (a) by saying the number of different selections of socks is $\binom{8}{2} = 28$

Number of different matching combinations = 4

$\implies probability = \frac{4}{28} = \frac{1}{7}$

To solve (b) I have looked at similar questions on this forum but can find none which I have been successfully able to apply.

I know that when choosing the second sock there is a $\frac {1}{7}$ probability of it matching the first. But then I get confused.

The answer in the book is $\frac{3}{35}$

Steblo
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    For (a) you got it right. For (b) if you were looking elsewhere on this site you probably won't find it phrased with socks but you will find many similar problems phrased with poker cards. Imagine all pairs being different colors (so you have two blue, two black, two white, and two striped socks). Pick the colors used for your two pairs simultaneously. In doing so, we have all the information needed for the numerator. Divide by number of ways of picking four socks. $\dfrac{\binom{4}{2}}{\binom{8}{4}}$ – JMoravitz Feb 16 '21 at 14:14

2 Answers2

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So for (a) you want the second sock to match the first with probability $\frac17$.

Do something similar for (b):

  • Second sock matches first and fourth sock matches third with probability $\frac17\times \frac15$

  • Second sock does not match first, third sock matches first and fourth sock matches second with probability $\frac67\times \frac16\times \frac15$

  • Second sock does not match first, third sock matches second and fourth sock matches first with probability $\frac67\times \frac16\times \frac15$

and add these up to give $\frac3{35}$

Henry
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  • I’m trying to follow the line of reasoning and even trying diagrams but it’s not clear to me. – Steblo Feb 16 '21 at 18:09
  • Why do we need to bother about the order in which the socks are drawn ? – true blue anil Feb 19 '21 at 05:09
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    @trueblueanil In this case you do not, but in others you do (e.g. what is the probability that two coin tosses match), and if you do not then you additionally need to show that it does not make a difference – Henry Feb 19 '21 at 07:22
  • Isn't probability of two coins matching analogous,, $1\cdot\frac 1 2 $, we aren't bothered whether the first toss (coin) is head or tail ? – true blue anil Feb 19 '21 at 12:19
  • @trueblueanil: I was think more of times when some people say that ignoring order the possibilities are two heads, one of each, or two tails, and then wrongly conclude that the probability of matching is $\frac23$ – Henry Feb 19 '21 at 12:27
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Another way is to imagine all the socks randomly laid out in a line, and you pull them out sequentially.

Suppose the first sock you draw out is type $A,\; A - - - | - - - -$

You need another type $A$ in the next $3$ places from remaining $7$,

say $\;A--A|---- \;$with$\; Pr = 3/7, $

Again, suppose one of the remaining two socks is type $C,\; A\,C - A | - - - -$, the remaining sock must also be $C$ with $Pr = 1/5$

Putting it together, $Pr = \dfrac 3 7\cdot\dfrac1 5 = \dfrac3 {35}$

Clarification to OP's query

We are looking at filling of four slots by two pairs

Assume that you draw the socks blind, they remain hidden, and then someone reveals one slot to have type $A$.

It is now essential that one of the other $3$ slots must have type $A$. Each remaining sock has equal probability of being in these $3$ slots, thus the probability that $A$ is in one of these slots is $3/7$. And so on.

The approach can be easily adapted for larger problems, eg for getting $3$ pairs from $6$ pairs,

$\frac 5{11}\cdot\frac 3 9\cdot \frac 1 7 = \frac 5{231}$

true blue anil
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  • Where does the $P_r = \frac {3}{7}$ come from? There is only one A left. – Steblo Feb 16 '21 at 22:20
  • I have explained in some detail in an addendum to the answer – true blue anil Feb 17 '21 at 05:33
  • You can, of course, use the approach of (ways to form pairs) / (unrestricted ways). For Pr of getting $3$ pairs from $6$ pairs, eg, you could compute $\frac {\binom 6 3}{\binom{12} 6}$, but as you will appreciate,it will need more clearing of fractions than $\frac 5{11}\cdot\frac 3 9\cdot \frac 1 7$ – true blue anil Feb 17 '21 at 06:46
  • Thank you for your clarification. Very helpful. – Steblo Feb 17 '21 at 08:46
  • I hope you have absorbed the minimalistic simplicity of the approach, and the ease with which it can be extended. – true blue anil Feb 19 '21 at 08:17