Conditional Probability The inhabitants of Liarland tell the truth $40\%$ of the time. One-third of the inhabitants have six toes on their right foot.
You ask four inhabitants of Liarland whether the King of Liarland has six toes on his right foot. They've all seen the King's feet, since the King always goes barefoot, and they all say yes.
What is the probability that the King's right foot has six toes?

My train of thought, I believe, is oversimplified. 
Since $4$ people are asked and there is a $40\%$ probability they are telling the truth, the probability is $(.4)^4$.
 A: Let $A$ be the event the king has six toes on his right foot, and let $B$ be the event they all say yes. We want $\Pr(A|B)$, which by definition is equal to $\frac{\Pr(A\cap B)}{\Pr(B)}$.
What is $\Pr(B)$? The event $B$ can happen in two ways (i) the king has $6$ toes on his right foot, and they all tell the truth or (ii) the king does not have $6$ toes on his right foot, and they all lie.
The probability of (i) is $\frac{1}{3}\cdot (0.4)^4$, and the probability of (ii) is $\frac{2}{3}\cdot (0.6)^4$. Add.
Now you should find it easy to find $\Pr(A\cap B)$ and complete the conditional probability calculation.
Remark: In our calculation, we have assumed that the people lie or tell the truth independently, and that their choice of whether to lie or tell the truth is independent of toe facts about the king. So in answering the question, they each toss an appropriately weighted coin in order to decide whether to tell the truth or to lie.  We have also assumed, possibly unreasonably, that the probability the king has $6$ toes on his right foot is $\frac{1}{3}$.
