Conditional Probability Example A drawer contains 8 different pairs of socks.   If 6 socks are taken at random and without replacement, compute the probability that there is at least one matching pair among these 6 socks.  
 A: It is little known, but socks have individual identities. There are $\binom{16}{6}$ equally likely ways to choose $6$ socks from the $16$.
Now we find the number of ways to choose $6$ socks, so that there is no pair among them. There are $\dbinom{8}{6}$ ways to choose the "types" of sock we will have.  For each choice of $6$ types, there are $2^6$ ways to choose the actual socks. For at each chosen  "type" of sock, we have $2$ choices as to which of the two socks of that type to take.
Thus the probability there is no pair is $p=\dfrac{\binom{8}{6}2^6}{\binom{16}{6}}$.
The probability there is at least one pair is therefore $1-p$.
A: Select your first sock.  Now you have 15 choices left for your 2nd sock, and 14 of them will allow you to avoid getting a pair.  
For your 3rd sock, you have a total of 14 choices remaining, and you can choose any sock other than the first two chosen and their mates to avoid a pair, so you have 12 choices to do this.
Continuing in this manner, we get $\frac{14}{15}\cdot\frac{12}{14}\cdot\frac{10}{13}\cdot\frac{8}{12}\cdot\frac{6}{11}$ for the probability of not getting a matching pair,  so $$1-\frac{14}{15}\cdot\frac{12}{14}\cdot\frac{10}{13}\cdot\frac{8}{12}\cdot\frac{6}{11}$$ gives the probability of getting at least one pair.
A: Hints:


*

*In how many ways can you pull 6 socks out, one at a time, so that there are no matching pairs among them?  (How many choices do you have for the first sock?  The second?  The third?)

*In how many ways total can you pull 6 socks out, one at a time?

*What does this mean about the probability that there is a matching pair?

