Ok my suggestion is the following.
You surely know how to use the hint: If you denote the probability of not having a matching term by $P$ then the probability of having at least one matching pair is ... (I'll leave it to you to think about this one)
Now you might want to use the following formula.
Let $\Omega$ be a (finite) probability space and $A\subset \Omega$ an event, then for the probability of $A$ holds$$
P(A)=\frac{|A|}{|\Omega|}
$$
That means the probability of $A$ is the number of all results such that $A$ has happened divided by the total number of possible results.
Now you have to compute those cardinalities: $\binom{n}{k}$ denotes the number of possibilities to draw $k$ items out of $n$ items when you do not care about the order in which the items were drawn (which is the case in your problem).
So $|\Omega|=\binom{16}{6}$ in your case, because you draw 6 socks out of 16 socks in total.
Furthermore there are $16 \cdot 14 \cdot 12 \cdot 10 \cdot 8 \cdot 6$ possibilities to draw $6$ socks without having a matching pair when taking into account the order (why?). Furthermore you have $6!$ possibilities to arrange $6$ objects on $6$ slots, so you have to divide by that number. I'll leave the rest to you. I hope this was helpful (and I didn't make a mistake somewhere).
EDIT: I just read your comment asking to explain the $16 \cdot 14 \cdot 12 \cdot 10 \cdot 8 \cdot 6$ part. Well this is just counting socks. I'll call the socks that result in the event "no matching pairs" as good socks and any sock preventing that result a bad sock.
- In the first round you cannot get a matching pair obviuosly (as you'd
need $2$ socks for a pair) so the number of good socks is 16.
- In the second round you have only 15 socks left, however one of those is of the same color as the one you already have and therefore a bad sock. Makes in total 15-1=14 good socks
- round three: you have 14 socks left. Among those are 2 bad socks (the corresponding sock to your first and second pick) so you have 14-2=12 socks left.
- and so on.
You also asked for a way to write this with binomials. Look at the following$$
\frac{\binom{2}{1} \cdot \binom{2}{1} \cdot \binom{2}{1} \cdot \binom{2}{1} \cdot \binom{2}{1} \cdot \binom{2}{1} \cdot \binom{2}{0} \cdot \binom{2}{0} \cdot \binom{8}{2}}{\binom{16}{6}}
$$
What do I want to tell you with that fraction? Well we now consider pairs. One pair consists of $2$ socks and we draw $6$ socks in total. That means we want exactly one sock out of $6$ pairs (therefore $6$ times $\binom{2}{1}$) and no sock out of the resulting $2$ pairs (therefore the two $\binom{2}{0}$). Of course you could just omit the $\binom{2}{0}$ term as it is $1$ anyway, I just put it there for illustration. Ok and now you have $\binom{8}{2}$ possibilities to choose "which pair you don't get any sock from" ( I hope that sounds remotely logical).
If you type it in a calculator you'll find that the two results are the same.
You are completely correct.
Simultaneously or not is completely irrelevant.
There are $\binom62=15$ possible choices and $3$ of them are matching choices.
That gives probability $\frac3{15}=\frac15$.
Also your logic is correct and in my view even more elegant.
addendum.
The correct interpretation of the probability question might have been: draw $3$ times a pair of socks without replacement. What is the probability that at least one of the pairs is matching?
It is handsome to calculate the probability of the complement. As shown above the probability that the first pair is not matching is $\frac45$. Assume that this happens and let's say that from socks $A_1,A_2,B_1,B_2,C_1,C_2$ we have drawn $A_1B_1$. Then $A_2,B_2,C_1,C_2$ are left and no matching pair will appear if the second draw will be one of $4$ equiprobable possibilities: $A_2C_1$, $A_2C_2$, $B_2C_1$ and $B_2C_2$. There are $6$ possibilities in total so there is chance of $\frac46=\frac23$ that this happens.
Then we come out on probability $\frac45\frac23=\frac8{15}$ that no matching pair is drawn. The probability that at least one matching pair has been drawn is $1-\frac8{15}=\frac7{15}$.
Best Answer
There are three cases to consider:
Since these cases are mutually exclusive, you can add the probabilities to determine the probability that Wednesday is the first day that Sandy selects a matching pair of socks.