Hypothetical probability question

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?

Answer 1

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: $$ \begin{align*} Pr(K \mid H,M)&=\dfrac{Pr(K,H,M)}{Pr(H,M)}\\ &=\dfrac{Pr(K)Pr(H \mid K)Pr(M \mid K,H)}{Pr(K)Pr(H \mid K)Pr(M \mid K,H) + Pr(\overline{K})Pr(H \mid \overline{K})Pr(M \mid \overline{K},H)}\\ &=\dfrac{Pr(K)Pr(H \mid K)Pr(M \mid K)}{Pr(K)Pr(H \mid K)Pr(M \mid K) + Pr(\overline{K})Pr(H \mid \overline{K})Pr(M \mid \overline{K})}\\ &=\dfrac{\dfrac{1}{100} \cdot 1 \cdot \dfrac{1}{2}}{\dfrac{1}{100} \cdot 1 \cdot \dfrac{1}{2} + \dfrac{99}{100} \cdot \dfrac{1}{20} \cdot \dfrac{1}{100}}\\ &=\dfrac{1000}{1000 + 99}\\ &=\dfrac{1000}{1099}\\ \end{align*} $$

Answer 2

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.