(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Suppose Sarah is flying from Regina to Vancouver with a connection in Edmonton. The probability that her first flight leaves on time is 0.77. If the first flight is on time, the probability that her luggage will make the connection flight in Edmonton is 0.92. But if the first flight is delayed, the probability that the luggage will make it is only 0.63.

Suppose that her luggage arrived in Vancouver with her, what is the probability that the fist flight was delayed?

The answer for this is 0.17, but I'm not sure how our teacher got this.

2. Relevant equations

[itex]P(A|B) = \frac{P(A\cap B)}{P(B)}[/itex]

[itex]P(A\cap B)=P(A)P(B|A)=P(B)P(A|B)[/itex]

3. The attempt at a solution

I found that the probability her luggage arrives in Vancouver with her is 0.8533.

I drew the following:

Probability first flight leaves on time: 0.77

- Probability baggage arrives: 0.92

- Probability baggage does not arrive: 0.04

Probability first flight leaves late: 0.23

- Probability baggage arrives: 0.63

- Probability baggage does not arrive: 0.370

I thought that [itex]P(A\cap B)=0[/itex] so then [itex]P(A)P(B|A)=P(B)P(A|B)=0[/itex] but this doesn't seem as if it could be true...

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# Probability if A then B

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