A survey asked students whether they have any siblings and pets. The survey data are shown in the relative frequency table.

\begin{tabular}{|c|c|c|c|}
\hline & Siblings & No siblings & Total \\
\hline Pets & 0.3 & 0.15 & 0.45 \\
\hline No pets & 0.45 & 0.1 & 0.55 \\
\hline Total & 0.75 & 0.25 & 1.0 \\
\hline
\end{tabular}

Given that a student has a sibling, what is the likelihood that he or she does not have a pet?

A. [tex]$45\%$[/tex]

B. About [tex]$82\%$[/tex]

C. [tex]$60\%$[/tex]

D. [tex]$40\%$[/tex]



Answer :

To find the likelihood that a student who has a sibling does not have a pet, we need to use the concept of conditional probability. Specifically, we are interested in [tex]\( P(\text{No pets} \mid \text{Siblings}) \)[/tex], which represents the probability of a student not having a pet given that they have a sibling.

Using the data from the table, the steps are as follows:

1. Identify the total probability of having a sibling: From the table, we can see that the total probability of a student having a sibling is 0.75.

2. Identify the joint probability of having no pets and having a sibling: The table shows that the relative frequency of students who have no pets and have siblings is 0.45.

3. Use the formula for conditional probability:
[tex]\[ P(\text{No pets} \mid \text{Siblings}) = \frac{P(\text{No pets and Siblings})}{P(\text{Siblings})} \][/tex]
Substituting in the values:
[tex]\[ P(\text{No pets} \mid \text{Siblings}) = \frac{0.45}{0.75} \][/tex]

4. Calculate the values:

[tex]\[ P(\text{No pets} \mid \text{Siblings}) = 0.6 \][/tex]

5. Express the probability as a percentage:

[tex]\[ P(\text{No pets} \mid \text{Siblings}) \times 100 = 60\% \][/tex]

Therefore, the correct answer is:
C. [tex]\(60\%\)[/tex]