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20 May, 03:50

A business man is flying from San Francisco to New York with a connection in Chicago. He is concerned about having his luggage arriving safely in New York, but this will depend on whether or not his flight from San Francisco leaves on time. Using past flight information, he estimates the probability that the flight leaves on time is 0.15. If the flight leaves on time, the probability his luggage arrives safely is 0.95, but if the flight does not leave on time the probability his luggage arrived safely is only 0.65.

a. What is the probability that his luggage arrives safely in New York? b. When the businessman arrived in New York, his luggage was not there. What is the probability his flight from San Frcisco left on time?

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  1. 20 May, 04:23
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    a. The probability is 0.695

    b. The probability is 0.0246

    Step-by-step explanation:

    Let's call T the event that the flight leave on time, NT the event that the flight doesn't leave on time, S the luggage arrived safely and NS the luggage doesn't arrived safely.

    So, the probability P (S) that his luggage arrives safely in New York is:

    P (S) = P (T∩S) + P (NT∩S)

    P (S) = (0.15*0.95) + (0.85*0.65)

    P (S) = 0.695

    Where P (T∩S) is the multiplication of the probability that the flight leave on time and the probability that his luggage arrive safely given that the flight leave on time. At the same way, P (NT∩S) is the multiplication of the probability that the flight doesn't leave on time and the probability that his luggage arrive safely given that the flight doesn't leave on time.

    The probability P (T/NS) that his flight from San Francisco left on time given that his luggage was not there is:

    P (T/NS) = P (T∩NS) / P (NS)

    Where P (NS) = P (T∩NS) + P (NT∩NS)

    So, Calculating P (T∩NS), P (NT∩NS) and P (NS) at the same way that we calculate P (S), we get:

    P (T∩NS) = 0.15 * (1-0.95) = 0.0075

    P (NT∩NS) = 0.85 * (1-0.65) = 0.2975

    P (NS) = 0.0075 + 0.2975 = 0.305

    Then, if the businessman arrived in New York and his luggage was not there, the probability P (T/NS) that his flight from San Francisco left on time is:

    P (T/NS) = 0.0075/0.305

    P (T/NS) = 0.0246
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