Bobby is flying from city A to city C with a connection in city B. The probability his first flight arrives on time is 0.15. If the flight is on time, the probability that his luggage will make the connecting flight is 0.8, but if the flight is delayed, the probability that the luggage will make it is only 0.55. In either case, Bobby makes the flight. Complete parts (a) and (b).
a. What is the probability that Bobby's luggage is there to meet him in city C?

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- anonymous

- triciaal

|dw:1437365135548:dw|

- triciaal

one approach
probability of flight on time and connect = 0.15*0.8
probability of delayed and connect = 0.85* 0.55

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## More answers

- triciaal

add the probabilities

- anonymous

i mess up

- triciaal

did not see part (b)

- anonymous

(a) i mess up on

- anonymous

b. If Bobby'ss luggage is not there to meet him, what is the probability that Bobby was late in arriving in city B?

- anonymous

Helppp

- triciaal

|dw:1437366014392:dw|

- triciaal

@freckles can you help with this?

- anonymous

0.5175

- anonymous

what did

- freckles

someone correct me if I'm wrong I think it is
\[P(lug)=P(lug|A)P(A)+P(lug| \text{ not } A)P(\text{ not } A)\]
where we are given:
\[P(lug|A)=.8 \\ P(A)=.15 \\ P(lug| \text{not } A)=.55\]

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