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 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

Let the event 'arrive on time' be AO.
Let the event 'leave on time' be LO.

`P(AO\cap LO)=0.8`
P(AO) = 0.84
P(LO) = 0.86

`P(LO|AO)=(P(AO\cap LO))/(P(AO))=(0.8)/(0.84)=0.95`

Answer 2

Answer: Our required probability is 0.952.

Explanation:

Since we have given that

Probability that train arrive on time and leaving on time = 0.8

Probability that train arrive on time = 0.84

Probability that train leave on time = 0.86

We need to find that probability that it will on time given that train arrived on time.

We will use "Conditional probability":

`P(\text{Leave on time}|\text{arrive on time})=\frac{P(\text{Arrive}\cap \text{Leave})}{P(Arrive)}=(0.8)/(0.84)=0.952`

Hence, our required probability is 0.952.