What's the probability to select socks of the same color? I'm sorry if this has been answered before already.
There's a drawer with 20 black socks and 20 white socks. We take one sock. What's the probability that the randomly chosen second sock has the same color as the second one?
I have two ways of thinking about it:


*

*The color of the first sock doesn't really matter, in any case, there're 19 socks of the same color left in the drawer and 39 in total. So that answer is $\frac{19}{39}$.

*We have two cases: the first sock was black or it was white. The probability of taking white sock after white sock is equal to the probability of taking black sock after black sock and is equal to $\frac{19}{39}$. The probability that the first sock is black is the same that it's white and is 0.5.
In general case we can apply The Law of Total Probability: 
\begin{equation}
\frac{19}{39} * 0.5 + \frac{19}{39} * 0.5 = \frac{19}{39}
\end{equation}
Does this reasoning make sense?
 A: Both the reasonings are correct. Consider the case when all balls socks $s_1,\dots, s_{40}$ are of different colors. First, the probability of selecting a sock of a particular color is $\frac{1}{40}$. Second, the probability of selecting a sock of a particular color from the rest of the 39 socks is $\frac{1}{39}$.  
Now apply counting principle. Since in $s_1,\dots, s_{40}$ there are 20 of the same color (black), the probability of the first event is $20 \times \frac{1}{40}=\frac{1}{2}$. Thus the probability of getting a black sock in the first event is $\frac{1}{2}$, similarly, the probability of getting a white sock in the first event is $\frac{1}{2}$. 
For the second event, you have two cases, either the sock in the first event is black or white. Thus the probability is $2\times\frac{1}{2}\times \frac{19}{40}=\frac{19}{40}$. The only difference is that your first answer is not using the counting principal whereas the second answer does.  
A: Let us denote by $B_{k}$ the event "a black sock has been chosen in the $k$-th selection". The same applies to $W_{k}$. Thus we are interested in the event $(B_{1}\cap B_{2})\cup(W_{1}\cap W_{2})$, where $B_{1}\cap B_{2}$ and $W_{1}\cap W_{2}$ are disjoint. Therefore we have
\begin{align*}
\textbf{P}((B_{1}\cap B_{2})\cup(W_{1}\cap W_{2})) & = \textbf{P}(B_{1}\cap B_{2}) + \textbf{P}(W_{1}\cap W_{2})\\
& = \textbf{P}(B_{2}|B_{1})\textbf{P}(B_{1}) + \textbf{P}(W_{2}|W_{1})\textbf{P}(W_{1})\\
& = \frac{C(19,1)}{C(39,1)}\times\frac{C(20,1)}{C(40,1)} + \frac{C(19,1)}{C(39,1)}\times\frac{C(20,1)}{C(40,1)}\\
& = \frac{19}{39}\times\frac{1}{2} + \frac{19}{39}\times\frac{1}{2} = \frac{19}{39}
\end{align*}
This is a more "systematic" way to describe the problem. Hope this helps.
