Solution: Let P(A) denotes the probability that the plane arrives on time and let P(D) denotes the probability that the plane departs on time. Therefore, we have:
[tex]P(A) = 0.82, P(D)=0.83, P(A \cap D)=0.78[/tex]
Now the probability that a plane arrives on time given that it departs on time is:
[tex]P(A|D) = \frac{P(A \cap D)}{P(D)}[/tex]
[tex]=\frac{0.78}{0.83} =0.94[/tex]
Therefore, the probability that a plane arrives on time given that it departs on time = 0.94