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
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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