To determine the correct answer, let's analyze the given probabilities.
First, we look at the probability that a randomly selected household will have 1 pet and no children. From the table, we can see that there are 53 such households out of the total 335 surveyed households.
The probability of having 1 pet and no children is calculated as follows:
[tex]\[
P(\text{1 pet and no children}) = \frac{53}{335} \approx 0.1582
\][/tex]
Next, let's consider the probability that a randomly selected household will have 2 pets and no children. From the table, there are 41 households with 2 pets and no children out of the total 335 surveyed households.
The probability of having 2 pets and no children is calculated as follows:
[tex]\[
P(\text{2 pets and no children}) = \frac{41}{335} \approx 0.1224
\][/tex]
Comparing these two probabilities, we can see that:
[tex]\[
P(\text{1 pet and no children}) \approx 0.1582 > P(\text{2 pets and no children}) \approx 0.1224
\][/tex]
Therefore, the customer is more likely to have 1 pet and no children than they are to have 2 pets and no children.
Complete the statement:
A customer is more likely to have 1 pet and no children than they are to have 2 pets and no children.
