Question

researchers recorded data on pet ownership for randomly selected households in a large city. the venn diagram summarizes the data using events c: has a cat and d: has a dog.
if a household that owns a cat is selected at random, what is the probability that the household owns a dog?
(round to 4 decimal places. leave your answer in decimal form.)

Explanation:

Step1: Recall the formula for probability of an event

The probability of an event \(E\) is calculated by summing up the relevant probabilities from the Venn - diagram. For event \(D\) (owning a dog), we need to consider the parts of the Venn - diagram related to \(D\).

Step2: Sum the probabilities

The probability that a household owns a dog is the sum of the probability that a household only owns a dog (\(P(\text{only }D)\)) and the probability that a household owns both a cat and a dog (\(P(C\cap D)\)).
From the Venn - diagram, \(P(\text{only }D)=0.22\) and \(P(C\cap D) = 0.18\).

$$P(D)=0.22 + 0.18$$

$$P(D)=0.40$$

Answer:

\(0.40\)