How do we identify a probability problem as a conditional probability problem? NOT a conditional probability problem
Sixty percent of the students at a certain school
wear neither a ring nor a necklace. Twenty percent
wear a ring and 30 percent wear a necklace.
If one of the students is chosen randomly, what is
the probability that this student is wearing
(a) a ring or a necklace?
(b) a ring and a necklace?
A conditional probability problem 
All bags entering a research facility are screened. Ninety-seven percent of the bags
that contain forbidden material trigger an alarm. Fifteen percent of the bags that
do not contain forbidden material also trigger the alarm. If 1 out of every 1,000
bags entering the building contains forbidden material, what is the probability
that a bag that triggers the alarm will actually contain forbidden material?
How do we distinguish the two?
For the second problem(before I knew it was a conditional probability problem) I interpreted it as
$A$= triggers the alarm
$F$= contains forbidden material
$P(A \bigcap F)$ =.97
$P(A \bigcap F^C)$ =.15
$P(F)$ = $\frac{1}{1000}$
 A: 
NOT a conditional probability problem:
Sixty percent of the students at a certain school wear neither a ring nor a necklace. Twenty percent wear a ring and 30 percent wear a necklace. If one of the students is chosen randomly, what is the probability that this student is wearing (a) a ring or a necklace? (b) a ring and a necklace?

In this question, you are not talking about conditional probability because all the specific sub-groups referred to are "symmetric."  In other words you the subgroups of wearing a ring or necklace have equal status.  This can be demonstrated in a venn box diagram:  
\begin{array}
{|c|c|c|c} \hline & R & R' &  \\ \hline N& & &30\\ \hline N'& & 60&\\ \hline & 20 & & 100\\ 
\hline . \end{array}
You know the venn box diagram above is a magic table, so you can complete it as such:
\begin{array}
{|c|c|c|c} \hline & R & R' &  \\ \hline N& 10& 20&30\\ \hline N'&10 & 60 & 70\\ \hline & 20 & 80 & 100\\ 
\hline . \end{array}
So part (a) is asking:  $Pr(R \cup N)$.  By the Inclusion Exclusion Principle you know:
$Pr(R \cup N)=Pr(R) + Pr(N) - Pr(R\cap N)=20+30-10=40.$
part (b):  $Pr(R \cap N) = 10$ from the table above.
You see in both questions, they are just asking probabilities about specific groups.

A conditional probability problem
All bags entering a research facility are screened. Ninety-seven percent of the bags that contain forbidden material trigger an alarm. Fifteen percent of the bags that do not contain forbidden material also trigger the alarm. If 1 out of every 1,000 bags entering the building contains forbidden material, what is the probability that a bag that triggers the alarm will actually contain forbidden material?

In this case,  you can view the subgroups as a tree:

Observe that whether the Alarm gets setoff is a "subset" of F or F'.  In other words, there is a narrowing of specific subgroups within a group.  This is what makes this problem a conditional probability.  
A: the second example contains independent events both are separate and like an either/or situation the first example the evens are not independent, You can wear a ring and a necklace. 
