Question

Suppose that a recent poll of American households about pet ownership found that for households with a pet, 39% owned a dog, 33% owned a cat, and 7% owned a bird. Suppose that three households are selected randomly and with replacement.

1. What is the probability that all three randomly selected households own a bird?

2. What is the probability that none of the three randomly selected households own a dog?

3. What is the probability that at least one of the three randomly selected households own a cat?

Answer 1

1) n = 3

    p = 0.07

It is a binomial distribution

P(X = x) = nCx * p^x * (1 - p)^{n - x}

P(X = 3) = 3C3 * (0.07)^3 * (0.93)^0 = 0.00034

2) n = 3

    p = 0.39

P(X = 0) = 3C0 * (0.39)^0 * (0.61)^3 = 0.22698

3) n = 3

    p = 0.33

P(X > 1) = 1 - P(X < 1)

             = 1 - P(X = 0)

            = 1 - (3C0 * (0.33)^0 * (0.67)^3)

            = 1 - 0.30076

           = 0.69924