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Sagot :
To determine the likelihood that a student who has a sibling does not have a pet, we need to analyze the given relative frequency table:
[tex]\[ \begin{tabular}{|c|c|c|c|} \hline & \text{Siblings} & \text{No siblings} & \text{Total} \\ \hline \text{Pets} & 0.3 & 0.15 & 0.45 \\ \hline \text{No pets} & 0.45 & 0.1 & 0.55 \\ \hline \text{Total} & 0.75 & 0.25 & 1.0 \\ \hline \end{tabular} \][/tex]
Let's break down the steps:
1. Identify the probability that a student does not have a pet given that they have a sibling.
From the table:
- The probability that a student has siblings and no pets is 0.45.
- The total probability that a student has siblings (regardless of pets) is 0.75.
2. Use the conditional probability formula:
[tex]\[ P(\text{No pets} \mid \text{Siblings}) = \frac{P(\text{Siblings and No pets})}{P(\text{Siblings})} \][/tex]
3. Substitute the values from the table:
[tex]\[ P(\text{No pets} \mid \text{Siblings}) = \frac{0.45}{0.75} \][/tex]
4. Simplify the fraction:
[tex]\[ P(\text{No pets} \mid \text{Siblings}) = \frac{0.45}{0.75} = 0.6 \][/tex]
5. Convert the result to a percentage:
[tex]\[ 0.6 \times 100 = 60\% \][/tex]
So, the likelihood that a student with siblings does not have a pet is [tex]\(\boxed{60\%}\)[/tex].
Therefore, the correct answer is:
B. [tex]\(60 \% \)[/tex]
[tex]\[ \begin{tabular}{|c|c|c|c|} \hline & \text{Siblings} & \text{No siblings} & \text{Total} \\ \hline \text{Pets} & 0.3 & 0.15 & 0.45 \\ \hline \text{No pets} & 0.45 & 0.1 & 0.55 \\ \hline \text{Total} & 0.75 & 0.25 & 1.0 \\ \hline \end{tabular} \][/tex]
Let's break down the steps:
1. Identify the probability that a student does not have a pet given that they have a sibling.
From the table:
- The probability that a student has siblings and no pets is 0.45.
- The total probability that a student has siblings (regardless of pets) is 0.75.
2. Use the conditional probability formula:
[tex]\[ P(\text{No pets} \mid \text{Siblings}) = \frac{P(\text{Siblings and No pets})}{P(\text{Siblings})} \][/tex]
3. Substitute the values from the table:
[tex]\[ P(\text{No pets} \mid \text{Siblings}) = \frac{0.45}{0.75} \][/tex]
4. Simplify the fraction:
[tex]\[ P(\text{No pets} \mid \text{Siblings}) = \frac{0.45}{0.75} = 0.6 \][/tex]
5. Convert the result to a percentage:
[tex]\[ 0.6 \times 100 = 60\% \][/tex]
So, the likelihood that a student with siblings does not have a pet is [tex]\(\boxed{60\%}\)[/tex].
Therefore, the correct answer is:
B. [tex]\(60 \% \)[/tex]
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