[Combinatorics]
Problem: A bag contains 3 red marbles and 6 blue marbles. Akshan removes one marble at a time until the bag is empty. Each marble that they remove is chosen randomly from the remaining marbles. Given that the first marble that Akshan removes is red and the third marble that they remove is blue, what is the probability that the last two marbles that Akshan removes are both blue?

Solution: Each possible order in which Akshan removes the marbles corresponds to a sequence of 9 colours, 3 of which are red and 6 of which are blue.

We write these as sequences of 3 R's and 6 B's.

Since are told that the first marble is red and the third is blue, we would like to consider all sequences of the form

$$
R \_B\_\_\_\_\_\_
$$

The 7 blanks must be filled with the remaining 2 R's and 5 B's.

There are $\left(\begin{array}{l}7 \\ 2\end{array}\right)=\frac{7 \cdot 6}{2}=21$ ways of doing this, because 2 of the 7 blanks must be chosen in which to place the R's. (We could count these 21 ways directly by working systematically through the possible pairs of blanks.)

Of these 21 ways, some have the last two marbles being blue.

These correspond to the sequences of the form

$$
R \_B \_\_\_\_ B B
$$

In these sequences, the 5 blanks must be filled with the remaining $2 \mathrm{R}$ 's and 3 B's.

There are $\left(\begin{array}{l}5 \\ 2\end{array}\right)=\frac{5 \cdot 4}{2}=10$ ways of doing this, because 2 of the 5 blanks must be chosen in which to place the R's.

Therefore, 10 of the 21 possible sequences end in two B's, and so the probability that the last two marbles removed are blue is $\frac{10}{21}$.

Final answer: $\frac{10}{21}$