Answer:
Option B.
Step-by-step explanation:
The given table represents the relative frequency.
Let A and B represent the following events.
A : Student does not have a sibling
B : Students has a pet.
Then intersection of A and B (A∩B) denotes the event that the student does not have a sibling but has a pet.
Using the given table it is clear that
P(A∩B) = 0.15
We need to find the probability that the student does not have a pet given that student has a sibling. It means we need to find the value of P(B|A).
Formula for conditional probability:
[tex]P(B|A)=\dfrac{P(A\cap B)}{P(A)}[/tex]
Substitute the given values.
[tex]P(B|A)=\dfrac{0.15}{0.25}[/tex]
[tex]P(B|A)=\dfrac{3}{5}[/tex]
[tex]P(B|A)=0.6=60\%[/tex]
The probability that the student does not have a pet given that student has a sibling is 60%.
Therefore, the correct option is B.
