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. 

The hint should help. You can always think about probability from the opposite event, though it's not always simpler. One primary rule of probability is to remember:
$P(A) + P(\lnot A) = 1$
So to find $P(A)$ you can always calculate $1-P(\lnot A)$.
Here, the probability of at least one matching pair of socks is complicated, because we could have 1 matching pair of any of the colors and 4 non-matching socks. We could have 2 matching pairs of any two of the colors and 2 non-matching socks. OR we could have 3 matching pairs of 3 of the colors. Whenever we see many options like this (often typified in a problem by the phrases "at least" or "at most"), that's a clue that calculating $1-P(\lnot A)$ is something to consider.
So, in this case $P(\lnot A)$ is the probability that we have NO matching socks of the 6 that we drew. Let's look at each sock.
First sock: it doesn't matter what sock we draw, since we have nothing to match it to yet. Probability of getting a sock that's acceptable = 1. 15 socks left.
Second sock: Well, now we have a sock that we don't want to match, so we want one of the other 7 colors. That eliminates 1 sock from the desirable outcomes, so $P = \frac {14}{15}$ and we have 14 socks left in the drawer.
Third sock: Now two colors are out of contention, so there are two socks in the remaining pile that we don't want. $P = \frac {12}{14}$
Continue this same pattern until the sixth sock. Then multiply all the probabilities together, since you are drawing these socks out in series. Your end result will be:
P(no matching pairs) = $1 \cdot \frac {14}{15} \cdot \frac{12}{14} \cdot \frac{10}{13} \cdot \frac8{12} \cdot \frac6{11} = \frac{10 \cdot 8 \cdot 6}{15 \cdot 13 \cdot 11} = \frac{2 \cdot 8 \cdot 2}{13 \cdot 11} = \frac{32}{143}$
Take that result and subtract it from 1 to get your final answer! P(at least one matching pair) = $\frac {111}{143}$
