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

$P(F)$ = $\frac{1}{1000}$

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Your interpretation that $P(A\cap F) = 0.97$, $P(A\mid F^c) = 0.15$ is incorrect. What you are told is that $P(A\mid F) = 0.97$, $P(A\mid F^c) = 0.15$. Given that a bag containd forbidden material, the chances that the alarm is triggered is $0.97$ etc. –  Dilip Sarwate Sep 30 '12 at 23:09
@DilipSarwate if they had used the keyword "Given" it would've been much easier, but here it doesn't use the keyword "Given". What other clues should i look for? –  user133466 Sep 30 '12 at 23:14
Yes, given would have been a better choice of wording. Conditional probabilities can be identified by the restriction on the sample space. It is not true that $97\%$ of all bags trigger an alarm; it is that $\967\%$ of all bags that contain forbidden material trigger an alarm; the universe has been restricted from the set of all bags to the set of bags containing forbidden material, and it is in this restricted sample space that $97\%$ bags trigger an alarm. Think also that your $P(A\cap F)=0.97$ means that $97\%$ of all bags contain forbidden material and trigger an alarm! –  Dilip Sarwate Sep 30 '12 at 23:27
@DilipSarwate minor math latex error –  user133466 Sep 30 '12 at 23:37

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.

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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.

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No, the second problem does not have any independent events in it. It does have complementary events (or mutually exclusive events), and your remark suggests that you are confusing between independent and mutually exclusive events. Here is a mantra that all beginning probabilists should be made to memorize: \begin{align*} &\text{Independent events are not mutually exclusive events}\\ &\text{Mutually exclusive events are not independent events}\end{align*} except for the trivial case when at least one event has probability $0$. –  Dilip Sarwate Sep 30 '12 at 23:17
opps! that is what I meant! –  user43126 Sep 30 '12 at 23:22