Father of lies problem

Question

Question: A father claims about snowfall last night. First daughter tells that the probability of snowfall on a particular night is 1/8. Second daughter tells that 5 out of 6 times the father is lying! What is the probability that there actually was a snowfall?

Let $S$ be an event that it snows and $C$ an event that a claim of a snowfall is made

From the answer here. The statement "Second daughter tells that 5 out of 6 times the father is lying" supposed to imply that $P(C|S) = \frac{1}{6}$ and $P(C\mid\lnot S) = \frac{5}{6}$ why is that the case?

My interpretation would be that $\frac{5}{6}$ times the claim is made, there was no snowfall i.e $P(S|C) = \frac{1}{6}$

Answer 1

Second daughter tells that 5 out of 6 times the father is lying" supposed to imply that $P(C|S) = \frac{1}{6}$ and $P(C|S^c) = > \frac{5}{6}$ why is that the case?

To be fair, the question has quite a bit of ambiguity. The statement

5 out of 6 times the father is lying

seems to be a claim about the father's general habits and not specific to the scenario. If we assume snowfall takes a similar probability (the father's lying habits with respect snowfall are the same as his general lying habit), then we can say $P(C|S^c) = \frac{5}{6}$. Why? Because there is a probability of $\frac{5}{6}$ he claims something when it is not true. In a similar manner, if we assume it is snowing, $S$, then there is a probability of $\frac{5}{6}$ he claims it won't snow, $C^c$. Thus $P(C^c | S) = \frac{5}{6}$, and it follows that $P(C | S) = 1 - \frac{5}{6} = \frac{1}{6}$

Answer 2

Let $S$ denotes it snows, and $C$ denotes the father claims it snows, and $S^c$ it does not snow, and $C^c$ the father claims it didn't snow.

Bayes rule states:

$$P(S|C) = \frac{P(C|S)P(S)}{P(C)} = \frac{P(C|S)P(S)}{P(C|S)P(S) + P(C|S^c)P(S^c)}$$

$$ = \frac{(1/6) (1/8)}{(1/6)(1/8) + (5/6)(7/8)} = \frac{1}{37}.$$