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.
