Question

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.

Answer 1

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.