<s>[INST] Probability of picking three marbles in order A bag contains 9 red marbles, 5 blue marbles and 7 green marbles. Need to select one marble at a time without replacement. What is the probability that the first marble is blue, the second is blue or green an the third is red.
My Solution:
$= P(\text{blue}) * P(\text{blue} \cup \text{green}) * P(\text{red})$
$\displaystyle = {5\over 21} * ({4\over 20} + {7\over 20}) * ({9\over 19})$
Does that sound that right?
 [/INST] That looks like the right answer.  These are conditional probabilities, so
\begin{align*}
& \mathrm{Pr}(\text{marble 1 is blue} \cap \text{marble 2 is blue or green} \cap \text{marble 3 is red}) \\
&= \mathrm{Pr}(\text{marble 1 is blue}) \times \mathrm{Pr}(\text{marble 2 is blue or green} \mid \text{marble 1 is blue}) \times \mathrm{Pr}(\text{marble 3 is red} \mid \text{marble 1 is blue} \cap \text{marble 2 is blue or green}) \\
&= \frac{5}{9+5+7} \times \frac{(5+7)-1}{(9+5+7)-1} \times \frac{9}{(9+5+7)-2}\\
&= \frac{33}{532}.
\end{align*}

Alternatively, we can use a counting argument.  There are $3! \binom{9+5+7}{3}=7980$ ordered $3$-tuples of $\{r_1,\ldots,r_9,b_1,\ldots,b_5,g_1,\ldots,g_7\}$.  Of these, precisely $5 \times (5+7-1) \times 9=495$ satisfies the blue-blue/green-red pattern.  We have $$\frac{5 \times (5+7-1) \times 9}{3! \binom{9+5+7}{3}}=\frac{33}{532}.$$
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