What is the Probability of a Delayed Flight Given Luggage Arrived in Vancouver?

[PirateFan308]

Homework Statement

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.

Homework Equations

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

$P(A\cap B)=P(A)P(B|A)=P(B)P(A|B)$

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 $P(A\cap B)=0$ so then $P(A)P(B|A)=P(B)P(A|B)=0$ but this doesn't seem as if it could be true...

 

[Mark44]

What events do A and B represent, and are these independent events?

 

[Ray Vickson]

Homework Statement

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.

Homework Equations

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

$P(A\cap B)=P(A)P(B|A)=P(B)P(A|B)$

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 $P(A\cap B)=0$ so then $P(A)P(B|A)=P(B)P(A|B)=0$ but this doesn't seem as if it could be true...

What do A and B represent in this case?

Sometimes (not always) people find it easier to think about such problems in the following manner: imagine that Sarah makes the trip 10,000 times. In how many trips is her first flight on time? How many times late? For all the on-time trips, in how many does her luggage arrive? For all the late trips, in how many does her luggage arrive? Now look at all the cases in which her luggage arrives. In how many of those was the first flight on time?

RGV

 

[PirateFan308]

Wow, I really over thought that. Thanks!

 

[Ellx]

sorry didn't mean to submit, only meant to use to view this other post with latex correctly. There's no delete button?