Probability- Guessing names of people 
In the newspaper "Pop World," an old photo of well-known pop singers was published from when they were kids.
The names of the singers are known, and one needs to identify the singers. If the match is done randomly, what is the probability of identifying at least one of the first 2 singers?

So it is like ordering 4 people in a row to create a 1-1 and onto function between the names and the people, so $|\Omega|=4!$.
Now let $A,B,C,D$ be the event of guessing the name right, so we are looking for $(A\cup B)\cap(C^{c}\cap D^{c})$, so using the complement would not help? as it $(A^{c}\cap B^{c})\cup(C\cup D)$?
That is why we need to to calculate $P(A\cup B)=P(A)+P(B)-P(A\cap B)=\frac{3!}{4!}+\frac{3!}{4!}-P(A\cap B)$.  What should $P(A\cap B)$ be? Intuitively, it is $2!\over 4!$, but why isn't it $4*3\over 4!$?
(Maybe I should look at it this way: knowing the names of 2 people is equivalent to ordering the 2 that are left, so I have $2!$ ways of doing it. So that's why it is $2!\over 4!$, but still why is it $1*1*2*1$?  Why does knowing the first and second remove higher numbers?)

You almost solved the problem yourself:
$\begin{align}
P(A\cup B) &= P(A)+P(B)-P(A\cap B)\\
P(A)=P(B)&= \frac{1}{4}\\
P(A\cap B)&=P(A)P(B\mid A)\\
&=\frac{1}{4}\frac{1}{3}=\frac{2!}{4!}\\
P(A\cup B)&=\frac{5}{12}
\end{align}$
The reason why $P(B\mid A)=\frac{1}{3}$ is simply that name used for the first guy is clearly not the one of the second one, since the first guess was right.
