Answer to Question #90709 in Statistics and Probability for Lyra Dawson-McLundie

Question

Whilst shopping, the probability that Caroline buys fruit is 0.4.
The probability that Caroline independently buys a CD is 0.6.
Work out the probability that she buys fruit, a CD or both.

Answer 1

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)) = 0.6*0.4 = 0.24.

Required probability P(AB + AB^c + A^cB ) = P(AB) + P(AB^c) + P(A^cB) = 1- P(A^cB^c) = 1 - 0.24 = 0.76.

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