Answer to Question #195345 in Statistics and Probability for In need

Question

Each time Caroline goes shopping she decides whether or not to buy fruit.

The probability that she does buy fruit is 0.4.

Independently, she then decides whether or not to buy a CD, with a probability of 0.2 that she does buy a CD.

Work out the probability that she buys fruit or buys a CD or both.

Answer 1

Solution:

Let event A = Caroline buys fruit, event B = Caroline buys CD, A^c and B^c are complementary events.

Events AB, AB^c, A^cB and A^cB^c are jointly exhaustive and disjoint, hence P(AB) + P(AB^c) + P(A^cB) +P(A^cB^c) =1.

Events A and B independent, hence A^c and B^c independent too and probability P(A^cB^c) = P(A^c)*P(B^c) = (1 - P(A))(1-P(B)) = 0.2*0.4 = 0.08.

Required probability P(AB + AB^c + A^cB ) = P(AB) + P(AB^c) + P(A^cB) = 1- P(A^cB^c) = 1 - 0.08 = 0.92