Exercise: The probability that a regularly scheduled flight departs on time is PD) = 0.83; the
probability that it arrives on time is P(A) =0.82; and the probability that it departs and arrives on
time is PDA) = 0.78. Find the probability that a plane (a) arrives on time, given that it departed
on time, and (b) departed on time, given that it has arrived on time.
To find the probability that a plane arrives on time, given that it departed on time, we can use the concept of conditional probability. In this case, we want to find P(A|D), which represents the probability of the plane arriving on time (A) given that it departed on time (D).
We know that:
P(D) = 0.83 (probability of departing on time)
P(A) = 0.82 (probability of arriving on time)
P(D&A) = 0.78 (probability of departing and arriving on time)
To find P(A|D), we can use the formula for conditional probability:
P(A|D) = P(D&A) / P(D)
Substituting the given values:
P(A|D) = 0.78 / 0.83
P(A|D) ≈ 0.941 (rounded to three decimal places)
Therefore, the probability that a plane arrives on time, given that it departed on time, is approximately 0.941.
Now, let's find the probability that a plane departed on time, given that it has arrived on time, denoted as P(D|A).
Using the formula for conditional probability:
P(D|A) = P(D&A) / P(A)
Substituting the given values:
P(D|A) = 0.78 / 0.82
P(D|A) ≈ 0.951 (rounded to three decimal places)
Therefore, the probability that a plane departed on time, given that it has arrived on time, is approximately 0.951.