# Confusing Conditional Probability Question

• May 11th 2009, 06:48 PM
jstluise
Confusing Conditional Probability Question
Hello all!
I'm trying to work through a problem involving conditional probability and I am more than confused. I need some guidance. Here is a brief description of the problem statement:
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Someone makes frequent airplane trips from LA to Wash. DC; She flies 50% of the time on airline #1, 30% on airline #2, and 20% on airline #3.

-For airline #1, flights are late into DC 30% of the time and late into LA 10% of the time.
-For airline #2, flights are late into DC 25% of the time and late into LA 20% of the time.
-For airline #3, flights are late into DC 40% of the time and late into LA 25% of the time.

Question: On a particular trip she arrived LATE at exactly 1 of 2 destinations. What are the posterior probabilites of having flown on airlines #1, #2, and #3?
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My professor hinted to use Baye's Theorem: P(A|B)P(B)=P(B|A)P(A).

So I know the basic question we want to ask is: What is the probablility that she flew #1 given she was late at one location? etc for #2 and #3.

I'll say: Event A=she flew #1, Event B=she flew #2, Event C=she flew #3. And say Event X=late at exactly 1 destination.

I want to compute: P(A|X), P(B|X), and P(C|X).

We know: P(A)=.5, P(B)=.3, and P(C)=.2

Now it is this event X that is getting me confused. Is X based on all 3 different situations? Or is X different depending on what airline we are looking at? I think it this problem can be done by looking at each airline seperately but I'm not really sure.

Any input would be much appreciated. I'll be working through it so I'll post any progress I make to see if I'm heading in the right direction.

Thanks!
• May 11th 2009, 07:42 PM
pickslides
Have you considered...

$P(A/B) = \frac{P(A\cap B)}{P(B)}$
• May 11th 2009, 07:49 PM
jstluise
Quote:

Originally Posted by pickslides
Have you considered...

$P(A/B) = \frac{P(A\cap B)}{P(B)}$

Thanks for the response. I have considered that, but still I guess my question remains what event X (that I mentioned above) am I looking for?

Am I to create a single event X so that P(X) is the probability that she'll be late to one destination regardless of what airline she takes, or should there be an event for each airline that describes that she'll be late to one destination. That's probably confusing.

And then the problem states "exactly" one destination...so does that mean I can't use P(late at one or the other)?
• May 12th 2009, 07:51 AM
Plato
Quote:

Originally Posted by jstluise
Someone makes frequent airplane trips from LA to Wash. DC; She flies 50% of the time on airline #1, 30% on airline #2, and 20% on airline #3.
-For airline #1, flights are late into DC 30% of the time and late into LA 10% of the time.
-For airline #2, flights are late into DC 25% of the time and late into LA 20% of the time.
-For airline #3, flights are late into DC 40% of the time and late into LA 25% of the time.

Question: On a particular trip she arrived LATE at exactly 1 of 2 destinations. What are the posterior probabilites of having flown on airlines #1, #2, and #3?

This reply may be entirely too late, but this is an interesting question
First, from the wording of the question we must assume it is asking about a round trip on the same airline.
Second, we must assume that being late exactly once means being late going or being late on the return but not both.
Third, we must assume that being late going or being late on the return are independent events.

Say that L stands for “being late going or being late on the return but not both”.
Then $P(L) = P\left( {L \cap \# 1} \right) + P\left( {L \cap \# 2} \right) + P\left( {L \cap \# 3} \right)$

Let’s look at $P\left( {L \cap \# 1} \right) = P\left( {L|\# 1} \right)P(\# 1)$.
We need to consider two separate cases, because $L = \left( {G \cup \overline R } \right) \cup \left( {\overline G \cap R} \right)$.
That is, late going and not late returning; or not late going but late on the return.
So $P\left( {L \cap \# 1} \right) = \left[ {(0.3)(0.9) + (0.7)(0.1)} \right](0.5)$.

Now you find $P\left( {L \cap \# 2} \right)~\&~P\left( {L \cap \# 3} \right)$.

Then $P(L)$ is the denominator of each of the answers.
Or $P\left( {\# k|L} \right) = \frac{{P(L \cap \# k)}}{{P(L)}},\;k = 1,2,3$.