Question

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 1

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.

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.