Westonci.ca connects you with experts who provide insightful answers to your questions. Join us today and start learning! Explore our Q&A platform to find reliable answers from a wide range of experts in different fields. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
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
Read more about probability at:
https://brainly.com/question/25870256
#SPJ1
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.