<s>[INST] Probability and Conditional Probability Could someone please provide a worked solution to this question?
A box contains 5 amber, 7 blue and 9 green balls. Six of the balls are removed from the box at random and without replacement. 
Find the probability that
(a) three out of the six balls are blue;
(b) four of the balls are blue and two are green;
(c) the second ball to be selected is amber, given that the final ball to be selected is blue. 
I was wondering if there is a combinatorial argument for these types of problems,  because multiplying fractions here seems a little time consuming and messy. 
My answers:
(a) $\frac{7}{21}\times\frac{6}{20}\times\frac{5}{19}\times\frac{14}{18}\times\frac{13}{17}\times\frac{12}{16}\times\frac{6!}{(3!)^2}$
(b) $\frac{7}{21}\times\frac{6}{20}\times\frac{5}{19}\times\frac{4}{18}\times\frac{9}{17}\times\frac{8}{16}\times\frac{6!}{(4!)(2!)}$
Any help very much appreciated, as always. 
 [/INST] Well, for such combinatorical questions there are typically multiple different "counting schemes" to get to the same answer. Here is the thought process I would have to arrive at those. Hope it helps:
a) There are $\binom{5+7+9}{6}$ ways to choose the $6$ balls. There are $\binom{7}{3}$ ways to choose $3$ blue balls and $\binom{5+9}{3}$ ways to choose $3$ non-blue balls. So the Answer is
$\begin{align}\frac{\binom{5+9}{3}\binom{7}{3}}{\binom{5+7+9}{6}}\end{align}$
b) very similar: $\binom{7}{4}$ ways to choose $4$ blue balls. $\binom{9}{2}$ ways to choose $2$ green balls. So the answer is
$\begin{align}\frac{\binom{7}{4}\binom{9}{2}}{\binom{5+7+9}{6}}\end{align}$
Generally: It is always about the "binomial coefficient" $\binom{n}{k}=\frac{n!}{k!(n-k)!}$ which is the number of ways to select $k$ objects (without replacement) out of a total of $n$ objects.
c) Here, order is important, so you can't use binomial coefficients in the same way. The best way here is to multiply fractions for each step like this:
$P(\text{last is blue}) = \frac{7}{5+7+9}$.
$P(\text{second is amber and last is blue}) = \frac{5}{5+7+9} \cdot\frac{7}{4+7+9}$
$P(\text{second is amber}|\text{last is blue}) = \frac{P(\text{second is amber and last is blue})}{P(\text{last is blue})} = \frac{5\cdot7\cdot21}{7\cdot21\cdot20}$
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