The probability that a train leaves on time is 0.9. The probability that the train arrives on time and leaves on time is 0.36. What is the probability that the train arrives on time, given that it leaves on time?

0.4
0.9
0.27
0.36



Answer :

P(arrives on time given leaves on time)=
P(B | A)= P(B n A) / P(A)
= P(A n B) / P(A)
= 0.36 / 0.9
= 0.4

The probability that the train arrives on time, given that it leaves on time is 0.4

How to determine the probability?

The given parameters are:

  • P(Leave on time) = 0.9
  • P(Arrive and Leave on time) = 0.36

The required probability is calculated using:

P(Arrive on time given that it leaves on time) = P(Arrive and Leave on time)/P(Leave on time)

So, we have:

P(Arrive on time given that it leaves on time) = 0.36/0.9

Evaluate

P(Arrive on time given that it leaves on time) = 0.4

Hence, the probability that the train arrives on time, given that it leaves on time is 0.4

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