zazabr3ath
Answered

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. [tex]$40 \%$[/tex]
C. About [tex]$82 \%$[/tex]
D. [tex]$60 \%$[/tex]



Answer :

GinnyAnswer
Certainly! Let's dissect the given problem using the relative frequency table provided:

[tex]\[ \begin{array}{|c|c|c|c|} \hline & \text{Siblings} & \text{No siblings} & \text{Total} \\ \hline \text{Pets} & 0.3 & 0.15 & 0.45 \\ \hline \text{No pets} & 0.45 & 0.1 & 0.55 \\ \hline \text{Total} & 0.75 & 0.25 & 1.0 \\ \hline \end{array} \][/tex]

The question asks for the probability that a student does not have a pet given that they have a sibling.

### Step-by-Step Solution:

1. Identify Relevant Probabilities from the Table:
- The probability that a student has siblings and no pets is given as \(0.45\).
- The probability that a student has siblings (regardless of pet status) is given as \(0.75\).

2. Apply Conditional Probability Formula:
- The conditional probability formula for our context is:
[tex]\[ P(\text{No pets} \mid \text{Siblings}) = \frac{P(\text{No pets and Siblings})}{P(\text{Siblings})} \][/tex]
- Substituting the values from the table:
[tex]\[ P(\text{No pets} \mid \text{Siblings}) = \frac{P(\text{No pets and Siblings})}{P(\text{Siblings})} = \frac{0.45}{0.75} \][/tex]

3. Calculate the Probability:
- Performing the division:
[tex]\[ P(\text{No pets} \mid \text{Siblings}) = \frac{0.45}{0.75} = 0.6 \][/tex]

4. Convert to a Percentage:
- To express this probability as a percentage:
[tex]\[ 0.6 \times 100\% = 60\% \][/tex]

Thus, the likelihood that a student does not have a pet given that he or she has a sibling is \(60\%\).

### Answer:
The correct answer is [tex]\( \boxed{60\%} \)[/tex]. This corresponds to option D.