Respuesta :
Answer:
Hence, the probability is:
0.82
Step-by-step explanation:
Let A denote the event that the flight departs on time.
Let B denote the event that the flight arrives on schedule.
Also, A∩B denotes the event that the flight that departs on time also arrives on schedule.
Let P denotes the probability of an event.
We are asked to find:
P(B|A)
We know that:
[tex]P(B|A)=\dfrac{P(A\bigcap B)}{P(A)}[/tex]
Hence, we have:
P(A)=0.83 x.
P(A∩B)=0.68 x.
where x are the total number of flights.
Hence,
[tex]P(B|A)=\dfrac{0.68x}{0.83x}\\\\P(B|A)=0.82[/tex]
Hence, the probability that a flight that departs on schedule also arrives on schedule is:
0.82
