Question

The probability of a train arriving on time and leaving on time is 0.8. The probability of the same train arriving on time is 0.84. The probability of this train leaving on time is 0.86. Given the train arrived on time, what is the probability it will leave on time?

Answer 1

No answer is possible, because the given information is inconsistent.

If the arrival probability is 0.84 and the departure probability is 0.86,
then the probability of the same train arriving AND leaving on time
is

(0.84) * (0.86) = 0.7224

NOT 0.8 , like the first sentence says.

Answer 2

To help you with this topic, I would suggest you revise probability trees.

P(train arriving) = 0.8
P(train arriving on time) = 0.84
P(train arriving late) = 0.86

0.8 x 0.86 = 0.688

Answer: 0.688

Hope it helped :)