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 $\frac{3}{35}$
Best Answer
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 $\frac3{35}$