Given that the first child draws $10\$$ from his envelope, what is the probability that the second child has an envelope that contains a 20$ note? Three children each receive an envelope from their grandparents. It is known that each envelope contains three banknotes and that in the three envelopes together there are two $5\$$ notes, four $10\$$ notes and three $20\$$ notes. The first child opens his envelope and the first note that he draws from it is a $10\$$ bill. What is now the probability that the second child has a $20$$ note in his envelope?

I have difficulty thinking about this question as it seems very complex. My intuition tells me that probably conditional probabilities are involved and maybe even Bayes' rule. However, besides that, I have no clue where to start. Could anyone please help?
 A: Think of the bills as distinct (they have serial numbers). Imagine that the grandparents took the $9$ bills, arranged them in order at random, and slipped them into envelopes, preserving order. That gives $9!$ equally likely permutations. But $9!$ will not appear in our solution, since we want to avoid explicit use of conditional probability. 
The top bill in the first child's envelope is a $10$. Without loss of generality it is the $10$ with lowest serial number. That leaves $8!$ equally likely permutations.
How many of these leave the second child sad because she got no $20$'s? There are $(5)(4)(3)$ ways that non-$20$'s can be in her envelope (recall that we are counting order). And the remaining $5$ bills can be in any of $5!$ orders. Thus the probability that the second child receives no $20$'s is $(5)(4)(3)(5!)/8!$. This is $\frac{5}{28}$. The probability she gets at least one $20$ is $\frac{23}{28}$. 
Remark: There are many other ways to solve the problem. Here is one that could be thought simpler.   We use a slightly different but equivalent model. After placing the "top" bill in the first  envelope, the grandparents choose $3$ of the remaining bills to place in the second envelope. 
Given that the top bill was the $10$ dollar bill with lowest serial number, there are $\binom{8}{3}$ ways to choose $3$ bills from the rest. And there are $\binom{5}{3}$ ways of choosing $3$ non-$20$'s. So the (conditional) probability the second child gets no $20$'s is 
$$\frac{\binom{5}{3}}{\binom{8}{3}}.$$
As before, this is $\frac{5}{28}$. So the required probability is $\frac{3}{28}$. 
There are also decidedly more complicated ways to proceed. For instance, let $A$ be the event first bill drawn is a $10$, and let $B$ be the event the second envelope got no $20$'s. We find $\Pr(B|A)$, which is $\frac{\Pr(A\cap B)}{\Pr(A)}$. 
To find $\Pr(A)$, we could then divide into cases (i) first envelope got $1$ $10$ and the $10$ was on top; (ii) $2$ $10$'s, and a $10$ was on top; (iii) $3$ $10$'s. Not much fun. 
