Question

Suppose that a recent poll of American households about pet ownership found that for households with pets, 45% owned a dog, 34% owned a cat, and 10% owned a bird. Suppose that three households are selected randomly and with replacement and the ownership is mutually exclusive.

What is the probability that at most two of the three randomly selected households own either a cat or a dog? (Round to the nearest hundredth)

a. 0.39

b. 0.51

c. 0.49

d. 0.61

Answer 1

Here' the answer to the question. Please let me know in case you've doubts.

p(own dog) = .45

p(own cat) = .34

p(own bird)= .10

Lets tread this as a binomial distribution, where p = probability of success of the event

Event is defined as pet ownsership is of cat of dog.

Hence, parameter of binomial dist are given as:

p(either a cat or a dog) = .34+.45= .79

n = 3

P(atmost 2 of the 3 randomly selected households won either a cat or dog) = ?
= P(X<=2)
= 3C0*.79^0*.21^3 + 3C1.79^1*.21^2 + 3C2.79^2*.21^1
= 0.506961 or .51

Answer is 0.51

B is correct

answered by: anonymous