Problem:
A desk has eight drawers. There is a probability of 1/2 that someone placed a letter in one of the desk's eight drawers and a probability of 1/2 that this person didn't place a letter in any of the desk's eight drawers. You open the first 7 drawers and find that they are all empty. What is the probability that the 8th drawer has a letter in it?
Let "A" be the probability the letter is in the 8th drawer, and let B be the event that the first 7 are empty.
After getting this problem wrong and checking the answer, the solution states that the numerator is given by $P(A \cap B) = P(A) * P(B) = (1/2)(1/8)$. What I don't understand is why $P(A) = 1/2$. $P(A)$ is the probability that the letter is in the 8th drawer, which should be itself conditioned on whether the letter is in the desk, making it $P(A) = (1/2)(1/8) = 1/16$. Yet multiple solutions to this problem maintain that the joint probability is $(1/2)(1/8)$, where $A = 1/2$ and $B = 1/8$, instead of $P(A \cap B) = (1/16)(1/8)$.
Why is this so? What logic am I missing here?
First, let us call event $A$ something else than you have it. Let event $A$ be the event that there was a letter placed at all somewhere within the eight desks (as opposed to what you write with $A$ being the event that the eighth desk has a letter in it). This should be easier to conceptualize.
$\Pr(A\cap B)$ here being the probability that there was a letter placed in one of the desks (event $A$) and the first seven desks being empty (event $B$).
It is wrong to call this $\Pr(A\cap B)=\Pr(A)\times \Pr(B)$. This manipulation is only allowed when $A$ and $B$ are independent events. Here it is clear that they are not independent since if $A$ were false... i.e. if there was no letter placed anywhere... then of course the first seven desks would be empty. That is to say, $\Pr(B\mid A^c)=1$ which is not equal to $\Pr(B)<1$ since there exists nonzero probability that one of the first seven desks actually has something in it.
Thankfully, we can still manipulate things here using the "multiplication principle of probability", a rephrasing of the definition of conditional probability. We have that $\Pr(A\cap B) = \Pr(A)\times \Pr(B\mid A)$. Here, $\Pr(A)=\dfrac{1}{2}$ as stated in the problem statement with the phrase "There is a probability of 1/2 that someone placed a letter in one of the desk's eight drawers." Then, we can assume from the wording of the problem that had there been a letter placed in one of the drawers that each of the drawers should have equally likely been the one to have the letter placed in it. That makes $\Pr(B\mid A) = \dfrac{1}{8}$ since for the first seven to be empty given that there was a letter placed, this occurs exactly when the letter was placed in the eighth drawer.
To answer the question the problem was asking us to, "What is the probability that the eighth desk has a letter in it given the first seven did not?" that is asking us to find $\Pr(A\mid B)$. We use definition of conditional probability and/or Bayes' Theorem.
$$\Pr(A\mid B) = \dfrac{\Pr(A\cap B)}{\Pr(B)}=\dfrac{\Pr(A)\Pr(B\mid A)}{\Pr(A)\Pr(B\mid A)+\Pr(A^c)\Pr(B\mid A^c)}$$
$$=\dfrac{\frac{1}{2}\cdot\frac{1}{8}}{\frac{1}{2}\cdot\frac{1}{8}+\frac{1}{2}\cdot 1}$$