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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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    @DilipSarwate minor math latex error2012-09-30

2 Answers 2

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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: enter image description here

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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    opps! that is what I meant!2012-09-30