Probability to pick at least one pair of socks There are 10 pairs of socks. What is the probability that in 4 socks chosen at random there is at least one pair.
My try: Let $A$ be an event of choosing exactly one pair of socks among 4 socks and $B$ be an event of choosing exactly two pairs,
$$P(A)=\frac{\binom{10}{1}\left(1-\frac{\binom{9}{1}}{\binom{18}{2}}\right)}{\binom{20}{4}}$$ and 
$$P(B)=\frac{\binom{10}{2}}{\binom{20}{4}}$$
So the total probability is $P(A)+P(B)$.
But i know that some mistake is there in my solution... can any one help?
 A: There is no pair of socks among the 4 socks picked if and only if the 4 socks belong to different pairs. The number of possibilities for this to happen is $2^4\cdot\binom{10}{4}$ (first, choose which 4 pairs out of 10 are going to be hit; then for each pair, which of the two socks is picked).
In total, how many choices exist for choosing 4 socks out of 20? $\binom{20}{4}$. So the probability to "miss" every single pair of socks is $$\frac{2^4\cdot\binom{10}{4}}{\binom{20}{4}}=\frac{224}{323}$$, and the quantitity you are looking for is $1-\frac{2^4\cdot\binom{10}{4}}{\binom{20}{4}} = \frac{99}{323}.$
A: Calculate $1$ minus the probability of the complementary event:
The number of ways to choose $4$ out of $20$ socks is:


*

*Choose the $1$st sock out of $20$ socks

*Choose the $2$nd sock out of $19$ socks

*Choose the $3$rd sock out of $18$ socks

*Choose the $4$th sock out of $17$ socks


The number of ways to choose $4$ out of $20$ socks with no pairs is:


*

*Choose the $1$st sock out of $20$ socks

*Choose the $2$nd sock out of $18$ socks

*Choose the $3$rd sock out of $16$ socks

*Choose the $4$th sock out of $14$ socks


So the probability of choosing $4$ out of $20$ socks with at least one pair is:
$$1-\frac{20\cdot18\cdot16\cdot14}{20\cdot19\cdot18\cdot17}$$

Please note that I've essentially taken into account the order of the socks.
If I chose not to take it into account, then I would need to divide each result by $4!$.
But since this factor appears in both the numerator and the denominator, I can ignore it.
A: Your numerator in $P(A)$ is wrong—it isn't even an integer!  The numerator had to represent the number of ways that exactly one pair could be chosen:
$$^{10}\mathrm C_r\times\,^9\mathrm C_2\times2^2$$
Then you will get the same answer as CC: $\displaystyle\frac{99}{323}$
A: This is late, but here is a general answer (studying for an exam, so I figured this was good practice).
There are 
$2n$ shoes ($n$ pairs).  So the denominator is ${2n\choose 2r}$ ie, choosing $2r$ shoes from $2n$ total.  We find our numerator  by calculating how many ways we have no matching pairs.  This means we would choose our shoes from the total pairs, and thus choose $2r$ from $n$, ${n\choose 2r}$.  We then would choose ${2\choose 1}$ shoes, and multiply this result $2r$ times.  This is because if we want to have no pairs, we would only select one shoe from every two to ensure there are no pairs chosen.  ${n\choose 2r}$ different shoe types and 
$\underbrace{{2\choose 1}\cdot{2\choose 1}\cdot\ldots \cdot{2\choose 1}}_{2r \text{ of these}}$ ways of choosing one shoe from each pair ensuring no correct pairs.  Distinguishing the chosen specified shoes leads to 
$$\boxed{P(\text{no pairs})=\frac{{n\choose 2r}2^{2r}}{{2n\choose 2r}}}$$
If we want to find the probability of finding exactly one pair, then we note there are ${n\choose 1}$ ways to select one pair.  Then we want to select $2r-2$ pairs from $n-1$ shoes that remain, mathematically we can express this as ${n-1\choose 2r-2}$.  If we take one shoe from each pair, we would do this $2r-2$ times, so from the same reasoning as earlier, 
    $$\boxed{P(\text{one pair})=\frac{n{n-1\choose 2r-2}2^{2r-2}}{{2n\choose 2r}}}$$
Generalizing to getting exactly $i$ pairs
$$\boxed{P(\text{exactly i pairs})=\frac{{n\choose i}{n-i\choose 2r-2i}2^{2r-2i}}{{2n\choose 2r}}}$$
