A faculty leader was meeting two students in Paris, one arriving by train from Amsterdam and the other arriving from Brussels at approximately the same time. Let A and B be the events that the trains are on time, respectively. If P (A) = 0.93, P (B) = 0.89 and P (A / B) = 0.87, then find the probability that at least one train is on time.

Answer: P (AUB) = 0.93 + 0.89 - 0.87 = 0.95

Therefore, the probability that at least one train is on time is 0.95.

Step-by-step explanation:

The probability that at least one train is on time is the probability that either train A, B or both are on time.

P (A) = P (A only) + P (A∩B)

P (B) = P (B only) + P (A∩B)

P (AUB) = P (A only) + P (B only) + P (A∩B)

P (AUB) = P (A) + P (B) - P (A∩B) ... 1

P (A) = 0.93

P (B) = 0.89

P (A∩B) = 0.87

Then we can substitute the given values into equation 1;

P (AUB) = 0.93 + 0.89 - 0.87 = 0.95

Therefore, the probability that at least one train is on time is 0.95.