Discover the answers you need at Westonci.ca, a dynamic Q&A platform where knowledge is shared freely by a community of experts. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.

Each time Caroline goes shopping she decides whether or not to buy fruit.
The probability that she does buy fruit is 0.7.
Independently, she then decides whether or not to buy a CD, with a probability of 0.6 that she does buy a CD.
Work out the probability that she buys fruit or buys a CD or both.


Sagot :

Let event A = Caroline buys fruit, event B = Caroline buys CD, Ac and Bc are complementary events.

Events AB, ABc, AcB and AcBc are jointly exhaustive and disjoint, hence P(AB) + P(ABc) + P(AcB) +P(AcBc) =1.

Events A and B independent, hence Ac and Bc independent too and probability P(AcBc) = P(Ac)*P(Bc) = (1 - P(A))(1-P(B)) = 0.6*0.4 = 0.24.

Required probability P(AB + ABc + AcB ) = P(AB) + P(ABc) + P(AcB) = 1- P(AcBc) = 1 - 0.24 = 0.76.

Answer: Probability that Caroline buys fruit, a CD or both is 0.76.

Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.