Tricky probability puzzle Here is the 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?
Solution: Let S = Snowfall occurred, and C = Claim. 
Probability of (Snowfall given Claim) $= P(S | C) = P(C|S)\times P(S)/P(C)$
Now, $P(C|S) = 1/6$, $P(S) = 1/8$. 
$P(C ) = P(\text{true claim}) + P(\text{False Claim}) = P(C|S)\times P(S) + P( \text{false claim}|\text{no snow})\times P(\text{no snow})$. 
This is same as $[1/6*1/8]/[ 1/6*1/8 + 7/8*5/6] = 1/36$
Doubts: 


*

*What is exactly this event C? C= Claim doesn't seems like an event to me. C = 
Claim is true/false is an event but what's C = Claim?

*Is there a better explanation to this problem? This one seems less understandable.

 A: I assume in this problem the daughters' claims are to be taken as true, and the father's truthfulness is independent of the weather. 
In that case, with probability $1/8$ there was snow last night (event S).
With probability $5/6$ the father lies about whether event S occurred (otherwise he tells the truth).
Event C is the event that the father claims it snowed. This can occur in two ways: either it snowed and he told the truth (probability $\frac18\times\frac16=\frac1{48}$), or it did not snow and he lied (probability $\frac78\times\frac56=\frac{35}{48}$). 
So $\Pr(C)=\frac{36}{48}$, but $\Pr(C\cap S)=\frac{1}{48}$. Thus the conditional probability $\Pr(S\mid C)=\frac1{36}$.
A: $P(C)$ is the probability that the father would make that claim (whether it is true or false). Either it was snowing and he is telling the truth ($\frac{1}{8} \cdot \frac{1}{6}$) or it was not snowing and he is lying ($\frac{7}{8} \cdot \frac{5}{6}$). The probability that the first instance happend is therefore
$$\frac{\frac{1}{8} \cdot \frac{1}{6}}{\frac{1}{8} \cdot \frac{1}{6} + \frac{7}{8} \cdot \frac{5}{6}}.$$
