Probability : A bag contains 12 pair socks . Four socks are picked up at random. Find the probability that there is at least one pair. Probability : A bag contains 12 pair socks . Four socks are picked up at random. Find the probability that there is at least one pair. 
My approach : 
Number of ways of selecting 4 socks from 24 socks is n(s) = $^{24}C_4$ 
The number of ways of selecting 4 socks from different pairs is n(E) = $^{12}C_4$
$$\Rightarrow P(E) = \frac{^{12}C_4 }{^{24}C_4}$$
Therefore, the probability of getting at least one pair is 
$$1- \frac{^{12}C_4 }{^{24}C_4}$$
But the answer is $$1- \frac{^{12}C_4 \times 2^4}{^{24}C_4}$$... Please guide the error .. thanks..
 A: Suppose the 12 pairs are different.
The number of ways of selecting 4 socks is $24*23*22*21$.
The number of ways of selecting 4 socks of no pair is $24*22*20*18$.
The chances of not getting a pair, then is $18*20 / 23*21$, or $120/161$. (by removing the common factors $22*24$.
The chances of getting at least one pair is then $41/161$
The error in your calculation, is that you are only counting one sock in each pair, similar to the second line of mine being $12*11*10*9$.  There are two socks in each pair, so you need to realise that either of the two can be fetched when a pair is selected.
A: Let the Event Probability of Selecting 4 Socks such that there is at least one Pair is $A$ 
Let the Event Probability of selecting 4 Socks such that there is exactly one pair is $B$ 
Let the  Event Probability of selecting 4 Socks such that there is exactly Two pairs is $C$.
Then
$$P(A)=P(B)+P(C)$$
$$P(B)= \frac{\binom{12}{1}\binom{11}{1}\binom{2}{1}\binom{10}{1}}{\binom{24}{4}}=\frac{40}{161}$$
$$P(C)=\frac{\binom{12}{2}}{\binom{24}{4}}=\frac{1}{161}$$
A: The error is in the second line.  The number of ways to select four different pairs must be multiplied by the number of ways to select just one sock from each of those four selected pairs. 
This is:  (12C4 × 24)
