Answer:
a. The probability is 0.695
b. The probability is 0.0246
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