How to solve this six-fingered man problem on conditional probability? This example is taken from Blitzstein and Hwang, Introduction to Probability (2019, 2nd Edition), pg 58. The second part to this question has been solved here.
The question:

A crime has been committed in a certain country. The perpetrator is
one (and only one) of the n men who live in the country. Initially,
these n men are all deemed equally likely to be the perpetrator. An
eyewitness then reports that the crime was committed by a man with six
fingers on his right hand.
Let $p_0$ be the probability that an innocent man has six fingers on
his right hand, and $p_1$ be the probability that the perpetrator has
six fingers on his right hand, with $p_0 < p_1$. (We may have $p_1<1$,
since eyewitnesses are not 100% reliable.) Let $a =p_0/p_1$ and $b =
> (1−p_1)/(1−p_0)$. Rugen lives in the country. He is found to have six
fingers on his right hand.
Now, given this information, what is the probability that Rugen is the
perpetrator?

My approach:
Let $R$ is the event that Rugen is the perp and $F$ is the event that the crime was committed by a man with 6 fingers on his right hand.
So, $$P(R|F) = \frac{P(RF)}{P(F)} \\
=\frac{P(F|R).P(R)}{P(F|R).P(R) + P(F|R^c).P(R^c)} \\
=\frac{p_1 . 1/n}{p_1 . 1/n + p_1 . (n-1)/n}
$$
However the solution in the book goes like this:
Let $R$ is the event that Rugen is guilty and $M$ be the event that he has six fingers on his right hand.
$$P(R|M) = \frac{P(RM)}{P(M)} \\
=\frac{p_1 . 1/n}{p_1 . 1/n + p_0 . (n-1)/n}
$$
My Questions:

*

*Why did they take $M$ to be defined as such? Shouldn't the event on
which the probability is conditioned upon be the fact that the crime
has been committed by a man with 6 fingers on his right hand? Instead of the fact that Ruben has 6 fingers on his right hand?

*Why is $P(M|R^c) = p_0$ ?

 A: First of all, we don't know that the crime was committed by someone with six fingers on his right hand; we only know that a witness said that.
More to the point though, we want to know how the fact that Rugen has six finger on one hand changes our estimate of the likelihood that he did the crime.  The denominator is the probability that a random man selected from the population has six figures on his right hand.  Either he is the guilty party ($\frac1n$) and the guilty party has six fingers ($p_1$) or he is innocent ($\frac{n-1}{n}$) and he is an innocent man with six fingers ($p_0$).
Simplification of your answer leads to $\frac1n$ which can't be right -- the witness's statement and the fact that Rugen has six fingers make no different in our assessment of the likelihood of his guilt.  Furthermore, your calculation doesn't use $p_0$ at all, so it makes no allowance for the possibility that a man with six fingers might be innocent, even if we believe the witness to be $100\%$ reliable.
