Conditional Probability matching socks A drawer contains eight different pairs of socks. If six socks are taken at random without replacement, compute the probability that there is at least one matching pair among these six socks.
Hint: compute the probability that there is not a matching pair first. 
Sorry guys, I'm really awful at this class, I'm trying to train my brain to think the right way to solve these on my own but it's hard. 

Hint
Number of ways you can choose 6 socks when there are no pairs in them  $A = 16 \cdot 14 \cdot 12 \cdot 10 \cdot 8 \cdot 6$
Total Number of ways you can choose 6 socks out of 8 pairs of socks B= $16 \cdot 15 \cdot 14 \cdot 13 \cdot 12 \cdot 11$
Probability that you will not find a match in the 6 drawn $= \frac{A}{B}$
Probability that you will find atleast one pair of socks in the 6 drawn $= 1-\frac{A}{B}$
Edit:
The first sock could be chosen in 16 ways, the second sock could only be chosen in $14$ ways, the third sockk could be chosen in 12 ways ( each time, you are deducting one that has already been taken and the pair sock that goes along with that). Thus the numerator is $16 \cdot 14 \cdot 12 \cdot 10 \cdot 8 \cdot 6$. For the denominator, you do not have any restriction, so, you pick the first sock in $16$ ways and $2^{\text{nd}}$ sock in $15$ ways, $3^{\text{rd}}$ sock in $14$ ways and until you pick the $6^{\text{th}}$th sock in $11$ ways.  Thus it is $16 \cdot 15 \cdot 14 \cdot 13 \cdot 12 \cdot 11$.
Thanks
Satish
