Question

Use the following information to answer this question and the next. 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, 10% owned a bird, and 11% had no pets. Suppose that three households are selected randomly and with replacement and the ownership is mutually exclusive. What is the probability that none of the three randomly selected households own a cat?

Answer 1

Answer:

28.75% probability that none of the three randomly selected households own a cat

Step-by-step explanation:

For each household, there are only two possible outcomes. Either it owns a cat, or it does not. The probability of a household owning a cat is independent of other households. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

34% owned a cat

This means that $p = 0.34$

Three households are selected

This means that $n = 3$

What is the probability that none of the three randomly selected households own a cat?

This is P(X = 0).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{3,0}.(0.34)^{0}.(0.66)^{3} = 0.2875$

28.75% probability that none of the three randomly selected households own a cat