Question

6. If a person adopts a pet, it's either a cat or a dog. 25% of the time, it's a cat and 75% of the time, it's a dog. If you meet someone at the shelter that adopted three pets, what's the probability that they got either: three cats or three dogs? Show your calculation and give your single answer as a decimal.

Answer 1

Answer:

0.0156 probability that they got three cats.

0.4219 probability that they got three dogs.

Step-by-step explanation:

For each person there are only two possible outcomes. Either they adopted a dog, or a cat. The probability of a person adopting a cat or a dog is independent of any other person, which means that the binomial probability distribution is used 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.

Three people:

This means that $n = 3$

What's the probability that they got three cats?

25% of the time it's a cat adopted, which means that $p = 0.25$

This is P(X = 3). So

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

$P(X = 3) = C_{3,3}.(0.25)^{3}.(0.75)^{0} = 0.0156$

0.0156 probability that they got three cats.

What's the probability that they got three dogs?

75% of the time it's a dog adopted, which means that $p = 0.75$

This is P(X = 3). So

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

$P(X = 3) = C_{3,3}.(0.75)^{3}.(0.25)^{0} = 0.4219$

0.4219 probability that they got three dogs.