Hypothetical probability question I found this question on an online University of Washington course assignment related to Bayesian Probabilities:
You've lost contact with your safari leader and now you find yourself confronted by a legion of 100 marmosets. Your training tells you that you must give a present to the one king in order avoid being clawed. Fortunately, you have some raisinets and are a master of conditional probabilities. A lilliputian wearing a tophat approaches you directly. The king always wears a hat, and the custom has caught-on with 5% of the others. What is the probability that this is the king?
Before you get a chance to decide, a moustachioed marmoset pushes the first one out of the way. He too is wearing a hat. Good thing you memorized the 'M' section of the encyclopedia: the king is seen with a moustache half the time, whereas only 1% of ordinary monkeys have one. What is the probability that he is the king? Assume that moustaches and hats are independent of each other given the identity of the marmoset.

My probability skills are lacking. Would the first question be 1/6? Given that 5 lilliputians wear tophats and include one more for the king.
For the second question, I was having difficulty with the fact that there are 100 monkeys and only 1% have a moustache. Does that mean this has to be the king?
 A: Note: It is not the case that $6\%$ of marmosets wear hats. From the sample of $100$ marmosets, all we know is that $1$ of them (the king) definitely wears a hat and that $5\%$ of the remaining $99$ ordinary marmosets wear a hat (not $5\%$ of all $100$ marmosets).

Let $K \equiv$ the marmoset is a king, $H \equiv$ the marmoset wears a hat, and $M \equiv$ the marmoset has a mustache. We know that:


*

*$Pr(K)=1/100 \implies Pr(\overline{K})=99/100 $

*$Pr(H \mid K)=1$

*$Pr(H \mid \overline{K})=5/100=1/20$

*$Pr(M \mid K) = 1/2$

*$Pr(M \mid \overline{K}) = 1/100$


Hence, for the first question, we have:
$$ \begin{align*}
Pr(K \mid H)&=\dfrac{Pr(K,H)}{Pr(H)}\\
&=\dfrac{Pr(K)Pr(H \mid K)}{Pr(K)Pr(H \mid K) + Pr(\overline{K})Pr(H \mid \overline{K})}\\
&=\dfrac{\dfrac{1}{100} \cdot 1}{\dfrac{1}{100} \cdot 1 + \dfrac{99}{100} \cdot \dfrac{1}{20}} \\
&=\dfrac{20}{20 + 99} \\
&=\dfrac{20}{119} \\
\end{align*} $$

For the second question, we exploit the fact that mustaches and hats are independent of each other given the identity of the marmoset:
$$ \begin{align*}
Pr(K \mid H,M)&=\dfrac{Pr(K,H,M)}{Pr(H,M)}\\
&=\dfrac{Pr(K)Pr(H \mid K)Pr(M \mid K,H)}{Pr(K)Pr(H \mid K)Pr(M \mid K,H) + Pr(\overline{K})Pr(H \mid \overline{K})Pr(M \mid \overline{K},H)}\\
&=\dfrac{Pr(K)Pr(H \mid K)Pr(M \mid K)}{Pr(K)Pr(H \mid K)Pr(M \mid K) + Pr(\overline{K})Pr(H \mid \overline{K})Pr(M \mid \overline{K})}\\
&=\dfrac{\dfrac{1}{100} \cdot 1 \cdot \dfrac{1}{2}}{\dfrac{1}{100} \cdot 1 \cdot \dfrac{1}{2} + \dfrac{99}{100} \cdot \dfrac{1}{20} \cdot \dfrac{1}{100}}\\
&=\dfrac{1000}{1000 + 99}\\
&=\dfrac{1000}{1099}\\
\end{align*} $$
A: For the second question, 1% of ordinary monkeys have a moustache, so 1 ordinary monkey has one, as well as the king.  
The problem as written is rather silly; if you know both facts then you can rule out the first monkey immediately as not having a moustache.  If the two properties are independent, then each non-king has a $0.01\times 0.06=0.0006$ chance of having both; i.e. 6 monkeys out of ten thousand have both properties.
