A survey of pet ownership and car ownership among 120 people found: 45 own both a car and a pet, 30 own a car but no pet, 20 own a pet but…

This is a conditional probability, so restrict attention to the people who own a car. From the table, $45 + 30 = 75$ people own a car. Of those 75, 45 also own a pet. So $P(\text{pet} \mid \text{car}) = \frac{45}{75} = \frac{3}{5}$, choice B. Choice A ($\frac{3}{8}$) equals $\frac{45}{120}$, the chance of owning both out of all 120 people, which uses the wrong denominator. Choice C ($\frac{9}{24}$) simplifies to $\frac{3}{8}$ and has the same flaw. Choice D ($\frac{4}{9}$) comes from dividing by 65, the number of pet owners, or from using $\frac{20}{45}$, which is a different group entirely.