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anonymous

  • one year ago

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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  1. anonymous
    • one year ago
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    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?

  2. triciaal
    • one year ago
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    |dw:1437365135548:dw|

  3. triciaal
    • one year ago
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    one approach probability of flight on time and connect = 0.15*0.8 probability of delayed and connect = 0.85* 0.55

  4. triciaal
    • one year ago
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    add the probabilities

  5. anonymous
    • one year ago
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    i mess up

  6. triciaal
    • one year ago
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    did not see part (b)

  7. anonymous
    • one year ago
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    (a) i mess up on

  8. anonymous
    • one year ago
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    b. If Bobby'ss luggage is not there to meet him​, what is the probability that Bobby was late in arriving in city​ B?

  9. anonymous
    • one year ago
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    Helppp

  10. triciaal
    • one year ago
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    |dw:1437366014392:dw|

  11. triciaal
    • one year ago
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    @freckles can you help with this?

  12. anonymous
    • one year ago
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    0.5175

  13. anonymous
    • one year ago
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    what did

  14. freckles
    • one year ago
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    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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