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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]

Sagot :

GinnyAnsweridn
To determine the most appropriate choice to complete the given statement, we need to consider the probabilities of having different numbers of pets under different conditions. Let's analyze the probabilities step by step.

Given probabilities:

1. Probability of a household having 1 pet and no children: [tex]\( P(\text{1 pet, no children}) = 0.1582 \)[/tex]
2. Probability of a household having 2 pets and children: [tex]\( P(\text{2 pets, has children}) = 0.2537 \)[/tex]
3. Probability of a household having 2 pets and no children: [tex]\( P(\text{2 pets, no children}) = 0.1224 \)[/tex]
4. Probability of a household having 3 or more pets and children: [tex]\( P(\text{3 or more pets, has children}) = 0.1373 \)[/tex]
5. Probability of a household having 3 or more pets and no children: [tex]\( P(\text{3 or more pets, no children}) = 0.2149 \)[/tex]

We need to compare these probabilities to the probability of a household having 1 pet and no children to determine the appropriate statement.

From the probabilities listed above:

- [tex]\( P(\text{1 pet, no children}) = 0.1582 \)[/tex]
- [tex]\( P(\text{2 pets, has children}) = 0.2537 \)[/tex]
- [tex]\( P(\text{2 pets, no children}) = 0.1224 \)[/tex]
- [tex]\( P(\text{3 or more pets, has children}) = 0.1373 \)[/tex]
- [tex]\( P(\text{3 or more pets, no children}) = 0.2149 \)[/tex]

From these, we notice that:

- [tex]\( P(\text{2 pets, has children}) = 0.2537 \)[/tex] is greater than [tex]\( P(\text{1 pet, no children}) = 0.1582 \)[/tex]
- [tex]\( P(\text{2 pets, no children}) = 0.1224 \)[/tex] is less than [tex]\( P(\text{1 pet, no children}) = 0.1582 \)[/tex]
- [tex]\( P(\text{3 or more pets, has children}) = 0.1373 \)[/tex] is less than [tex]\( P(\text{1 pet, no children}) = 0.1582 \)[/tex]
- [tex]\( P(\text{3 or more pets, no children}) = 0.2149 \)[/tex] is greater than [tex]\( P(\text{1 pet, no children}) = 0.1582 \)[/tex]

Comparing these probabilities, it is evident that the probability of having 1 pet and no children is less than the probability of having 2 pets and children and less than the probability of having 3 or more pets and no children.

Thus, the correct answer to complete the statement is:

A customer is more likely to have 1 pet and no children than they are to have 2 pets and no children.
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