The question says: "For a recent year, 0.99 of the incarcerated population is adults and 0.07 is female. If an incarcerated person is picked at random, find the probability that the person is female given they are an adult."
I've been thinking about this for more than 4 hours and it just doesn't seem solvable to me.
We need the intersection(percentage of those that are females and adults) to use the following formula, but there is no way to find that.
${\mathbb{P(}F|A)= \mathbb{P}(F\cap A)}/{\mathbb{P}(A)}$
I tried to solve a similar but simpler problem that I made up such as:
In a population of 10 people, 8 of them(80%) are adults, and 4 of the total population are females(40%); what is the percentage of female adults?
I tried to visualize it as a set: $\{1, 3, 6, 8, 10, 12, 14, 16, 18, 20\}$ and ask if 80% of the set elements are positive and 40% of the set elements are divisible by 3, what is the percentage of positive numbers that are divisible by 3 in the set?
Is the original question really flawed? Or did my brain stop working because I can't think clearly anymore since I saw that question.
Best Answer
On the assumption of independence the answer is obviously $0.07$. That assumption is not necessarily reasonable. So indeed the question is poorly worded.
Imagine as an extreme case that no child females are put in jail. Then the probability a jailed person is female given the person is an adult is $\frac{0.07}{0.99}$.
As an opposite extreme case, suppose that no male children are put in jail. Then the probability is $\frac{0.06}{0.99}$. This is because the females account for all the child prisoners, leaving the proportion $0.06$ of the total that are both adult and female.