Find the probability
The probability of a bus arriving on time and leaving on time is 0.8. The same bus has a probability of 0.84 for arriving on time and 0.86 for leaving on time. If the bus arrived on time, what is the probability that it will leave on time?
OPTIONS
- 0.9524
- 0.8
- 0.84
- 0.688
ANSWER
0.9524
EXPLANATION
The solution to this problem involves conditional probability. The probability of Event B occurring, given Event A has occurred is P(B/A) = P (A∩B)/P(A) Let event B be the probability of the bus leaving on time. Let event A be the probability of the bus arriving on time. The interaction of Events A and B is the probability of the bus arriving and leaving on time. Thus, the probability of event B given Event A is 0.8/0.84 = 0.9524
