Select the correct answer from the drop-down menu.

The table shows the results from a survey of 335 randomly selected households with pets. This survey was conducted by a new pet store that is opening nearby.

\begin{tabular}{|l|l|l|l|}
\hline & Have Children & Do Not Have Children & Total \\
\hline 1 pet & 38 & 53 & 91 \\
\hline 2 pets & 85 & 41 & 126 \\
\hline 3 or more pets & 46 & 72 & 118 \\
\hline Total & 169 & 166 & 335 \\
\hline
\end{tabular}

The pet store uses the data to make decisions about inventory. Complete the given statement:

A customer is more likely to have 1 pet and no children than they are to have [tex]$\square$[/tex].



Answer :

To solve this problem, we need to analyze the probabilities of different scenarios based on the given data. Here is the step-by-step solution:

1. Understand the Problem Statement:
- We need to compare the likelihood that a customer has 1 pet and no children to the likelihoods of having other numbers of pets and no children.

2. Extract the Given Data:
- Households with 1 pet and no children: 53
- Households with 2 pets and no children: 41
- Households with 3 or more pets and no children: 72
- Total number of households: 335

3. Calculate the Probabilities:
- Probability of having 1 pet and no children:
[tex]\[ \text{P(1 pet, no children)} = \frac{53}{335} \approx 0.1582 \][/tex]
- Probability of having 2 pets and no children:
[tex]\[ \text{P(2 pets, no children)} = \frac{41}{335} \approx 0.1224 \][/tex]
- Probability of having 3 or more pets and no children:
[tex]\[ \text{P(3 or more pets, no children)} = \frac{72}{335} \approx 0.2149 \][/tex]

4. Compare the Probabilities:
- [tex]\(\text{P(1 pet, no children)} \approx 0.1582\)[/tex]
- [tex]\(\text{P(2 pets, no children)} \approx 0.1224\)[/tex]
- [tex]\(\text{P(3 or more pets, no children)} \approx 0.2149\)[/tex]

5. Conclusion:
- The probability of a household having 1 pet and no children (approximately 0.1582) is higher than the probability of having 2 pets and no children (approximately 0.1224).
- However, the probability of having 3 or more pets and no children (approximately 0.2149) is higher than the probability of having 1 pet and no children.

So, completing the statement:
A customer is more likely to have 1 pet and no children than they are to have 2 pets and no children.