Answer: The required probability is 0.8.
Step-by-step explanation: Let, 'E' denotes the event that a train leaves on time and 'F' denotes the event that the train arrives on time.
Then, according to the given information,
[tex]P(E)=0.7.[/tex]
The event that the train arrives on time and leaves on time will be [tex]F\cap E.[/tex]
So,
[tex]P(F\cap E)=0.56.[/tex]
We are to find the probability that the train arrives on time, given that it leaves on time, ie., P(F/E).
We have
[tex]P(F/E)=\dfrac{P(F\cap E)}{P(E)}=\dfrac{0.56}{0.7}=\dfrac{56}{70}=0.8.[/tex]
Thus, the required probability is 0.8.
