Respuesta :
Answer:
[tex] P(D|P)= \frac{0.11}{0.59}=0.186[/tex]
Step-by-step explanation:
For this case we have defined the following events:
D= "An international flight leaving the United States is delayed in departing"
P="An international flight leaving the United States is a transpacific flight "
And we have defined the probabilities:
[tex] P(D)= 0.29 , P(P) = 0.59[/tex]
And for the event: "an international flight leaving the U.S. is a transpacific flight and is delayed in departing" [tex] D \cap P [/tex] we know the probability:
[tex] P(D \cap P) =0.11[/tex]
We want to find this probability:
What is the probability that an international flight leaving the United States is delayed in departing given that the flight is a transpacific flight
So we want this probability:
[tex] P(D|P)[/tex]
And we can use the conditional formula from the Bayes theorem given two events A and B:
[tex] P(A|B) = \frac{P(A \cap B)}{P(B)}[/tex]
And if we use this formula for our case we have:
[tex] P(D|P)= \frac{P(D \cap P)}{P(P)}[/tex]
And if we replace the values we got:
[tex] P(D|P)= \frac{0.11}{0.59}=0.186[/tex]
