# Self study of basic probability: Can't solve exercise

I am currently working through 'Concepts of probability theory by Pfeiffer' and can't solve the following problem:

A carnival man hides a pea under one of three nut shells. By a series of complicated movements, he attempts to confuse the bystander so that he no longer knows which shell covers the pea. If he follows the sleight of hand, he will guess correctly. If he cannot detect the proper shell throughout the series of movements, he will pick a shell at random. There is a probability of 0.10 that he will know the correct shell. What is the probability (conditional) that he has detected the correct shell and not merely guessed the answer randomly?

How I tried it:
Let $$C$$ be the event of a correct guess and $$F$$ the event that the bystander successfully followed the movement. We then have \begin{align}P(C|F) &= 1\\P(C|F^C) &= \frac 1 3\\P(C)&=0.10.\end{align} Bayes' rule states that $$P(A|B) = \frac{P(B|A)P(A)}{P(B)}$$ and as we are interested in $$P(F^c|C)$$ this results in \begin{align} P(F|C) = \frac{P(C|F)P(F)}{P(C)}. \end{align} In the text $$P(C|F)$$ and $$P(C)$$ are given.

I am unsure how to calculate $$P(F)$$. I tried to partition $$P(C)$$ and solve for P(F) but this results in a negative result: \begin{align} P(C) &= P(C|F)P(F) + P(C|F^c)P(F^c) \\&= P(C|F)P(F) + P(C|F^c)(1- P(F)) \end{align} Solving for $$P(F)$$ gives $$-.35$$ which can not be right.

Edit: The solution in the book is $$\frac 1 4$$.
I think $$P(F)=0.1$$ and $$P(C)$$ is unknown; it's actually knowing, not correctly guessing. So, $$P(C)=0.1+0.9\times\frac{1}{3}=0.4$$ And, $$P(F|C)=\frac{1}{4}$$, which is the question. The question is not $$P(F^c|C)$$; read the last sentence more carefully.