Let's say I have a friend named Dave. There was a murder committed next door. Dave is most likely the killer.
P(dave committed murder)=0.99
However, the probability that Dave would leave a blond hair at the scene is 0.1
Almost nobody is as careful as Dave and the overall probability that a blond hair would've been left at the crime scene is 0.99.
I found a hair at the scene. What's the probability Dave committed the murder?
The answer would be (.99*.1)/.99 which equals 0.1.
So in this case, P(B|A)=P(A|B). However I don't see how this is completely true. Dave almost definitely committed the murder. I get that it's a slim chance that he left the blond hair at the scene but does it really reduce the probability he did it from .99 to .1?
Thank you in advance.
Your intuition is correct. That does not sound, and it isn't, right.
Let $x$ be the conditional probability that a hair is left at the scene given that somebody else did it. By the law of total probability, the probability that a hair is left at the scene is $$ 0.99 = 0.99 \times 0.1 + 0.01 \times x $$
So $x=89.1>1$, which cannot be a probability. The problem statement implicitly contains a probability larger than one, so the problem is wrong (but not Bayes). Since $x\le1$, the maximum probability that a hair is left at the scene is $$0.99 \times 0.1 + 0.01 \times 1=0.109$$ which is far less than $0.99$.
More specifically, the problem with the construction is that it does not satisfy the additivity axiom. The sum of probability that dave did it and a hair was found and probability that someone else did it and a hair was found should be equal to probability that a hair was found, since the first two events are mutually exclusive.