a survey asked students whether they have any siblings and pets. the survey data are shown in the relative…

a survey asked students whether they have any siblings and pets. the survey data are shown in the relative - frequency table.\n\n| | siblings | no siblings | total |\n|--|--|--|--|\n| pets | 0.3 | 0.15 | 0.45 |\n| no pets | 0.45 | 0.1 | 0.55 |\n| total | 0.75 | 0.25 | 1.0 |\n\ngiven that a student does not have a sibling, what is the likelihood that he or she has a pet?\n\na. about 33%\nb. 60%\nc. 75%\nd. 15%

a survey asked students whether they have any siblings and pets. the survey data are shown in the relative - frequency table.\n\n| | siblings | no siblings | total |\n|--|--|--|--|\n| pets | 0.3 | 0.15 | 0.45 |\n| no pets | 0.45 | 0.1 | 0.55 |\n| total | 0.75 | 0.25 | 1.0 |\n\ngiven that a student does not have a sibling, what is the likelihood that he or she has a pet?\n\na. about 33%\nb. 60%\nc. 75%\nd. 15%

Answer

Explanation:

Step1: Identify relevant frequencies

We want the probability of having a pet given no - siblings. The frequency of students with no siblings and pets is 0.15, and the total frequency of students with no siblings is 0.25.

Step2: Use conditional - probability formula

The formula for conditional probability $P(A|B)=\frac{P(A\cap B)}{P(B)}$. Here, $A$ is having a pet and $B$ is having no siblings. So the probability $P=\frac{0.15}{0.25}$.

Step3: Calculate the probability

$\frac{0.15}{0.25}=0.6$.

Step4: Convert to percentage

$0.6\times100% = 60%$.

Answer:

B. 60%