# Conditional propability question

So I am having a statistics test tomorrow and was looking through my done assignments and notes, and stumbled on this particular question that I can't seem to understand.

It goes like: A lost key is in one of the five drawers with a propability of 0.5 It appears that it is not in the first four drawers, what is the propability of it being in the fifth if it wasn't in the first four?

Correct answer is 1/6 if someone could give me detailed explanation on how to get there I would be grateful. I understand you have to use the formula of conditional propability but I can't seem to wrap my head around the upper "A and B" part of the formula.

Event $K$ is 'the key is in one of the five drawers' and event $A$ is 'the key is not in the first four drawers'. We know that $P(K)=\frac 12$ and that $P(A|K)=\frac 15$ (I assume the probability is $\frac 15$ for each drawer.) $P(X|Y)$ is the probability of event $X$ given that we know that $Y$ happened. The question asks for the probability $P(K|A)$ (because if it is in one of the drawers and is not in the first four, it must be in the fifth drawer).
We know that $$P(K|A)=\frac{P(K\cap A)}{P(A)}=P(A|K)\frac{P(K)}{P(A)}$$ $$P(A)=P(A|K)P(K)+P(A|K^C)P(K^C)=\frac 15\frac 12+ 1\frac 12=\frac 35$$ where $K^C$ is the complement (negation) of $K$. Now, we get $$P(K|A)=\frac{P(K\cap A)}{P(A)}=P(A|K)\frac{P(K)}{P(A)}=\frac 15\frac{\frac 12}{\frac 35}=\frac 16$$ Thus, the probability that the key is in the fifth drawer is $\frac 16$, as given.
There are $10$ drawers each having equal probability to contain the key. Let's say $5$ real drawers and $5$ virtual drawers. This gives a probability of $0.5$ that the key will be contained by one of the real drawers. After opening $4$ of the real drawers - that do not appear to contain the key - there are $6$ drawers left. Again they all have equal probability to contain the key so there is a chance of $\frac{1}{6}$ that it will be found in the fifth real drawer that is opened.