A bag contains exactly 15 marbles of which 3 are red, 5 are blue, and 7 are green. The marbles are chosen at random and removed one at a time from the bag until all of the marbles are removed. One colour of marble is the first to have 0 remaining in the bag. What is the probability that this colour is red?

## PART B

For each question in Part B, your solution must be well-organized and contain words of explanation or justification. Marks are awarded for completeness, clarity, and style of presentation. A correct solution, poorly presented, will not earn full marks.

## Useful Fact for Part B:

The sum of the first $k$ perfect squares is equal to $\frac{k(k+1)(2 k+1)}{6}$.

That is, $1^{2}+2^{2}+3^{2}+\cdots+k^{2}=\frac{k(k+1)(2 k+1)}{6}$.
We use $\mathrm{R}$ to represent a red marble, $\mathrm{B}$ to represent blue, and $\mathrm{G}$ to represent green.

Since there are 15 marbles of which 3 are red, 5 are blue, and 7 are green, there are $\frac{15 !}{3 ! 5 ! 7 !}$ orders in which the 15 marbles can be removed.

Since red is the first colour to have 0 remaining, the last of the 15 marbles removed must be B or $\mathrm{G}$. We look at these two cases.

Suppose that the final marble is B.

The remaining 4 B's are thus placed among the first 14 marbles. There are $\left(\begin{array}{c}14 \\ 4\end{array}\right)$ ways in which this can be done.

This leaves 10 open spaces. The last of these 10 marbles cannot be $R$, otherwise the last $G$ would be chosen before the last $\mathrm{R}$. Thus, the last of these 10 marbles is $\mathrm{G}$.

The remaining 6 G's are thus placed among the remaining 9 spaces. There are $\left(\begin{array}{l}9 \\ 6\end{array}\right)$ ways in which this can be done.

The $3 \mathrm{R}$ 's are then placed in the remaining 3 open spaces.

Therefore, there are $\left(\begin{array}{c}14 \\ 4\end{array}\right)\left(\begin{array}{l}9 \\ 6\end{array}\right)$ orders when the final marble is B.

Suppose that the final marble is G.

The remaining 6 G's are thus placed among the first 14 marbles. There are $\left(\begin{array}{c}14 \\ 6\end{array}\right)$ ways in which this can be done.

This leaves 8 open spaces. The last of these 8 marbles cannot be $\mathrm{R}$, otherwise the last $\mathrm{B}$ would be chosen before the last $\mathrm{R}$. Thus, the last of these 8 marbles is $\mathrm{B}$.

The remaining 4 B's are thus placed among the remaining 7 spaces. There are $\left(\begin{array}{l}7 \\ 4\end{array}\right)$ ways in which this can be done.

The 3 R's are then placed in the remaining 3 open spaces.

Therefore, there are $\left(\begin{array}{c}14 \\ 6\end{array}\right)\left(\begin{array}{l}7 \\ 4\end{array}\right)$ orders when the final marble is G.

Thus, the total number of ways in which the first colour with 0 remaining is red equals $\left(\begin{array}{c}14 \\ 4\end{array}\right)\left(\begin{array}{l}9 \\ 6\end{array}\right)+\left(\begin{array}{c}14 \\ 6\end{array}\right)\left(\begin{array}{l}7 \\ 4\end{array}\right)$.

This means that the desired probability, $p$, is

$$
\begin{aligned}
p & =\frac{\left(\begin{array}{c}
14 \\
4
\end{array}\right)\left(\begin{array}{l}
9 \\
6
\end{array}\right)+\left(\begin{array}{c}
14 \\
6
\end{array}\right)\left(\begin{array}{l}
7 \\
4
\end{array}\right)}{\frac{15 !}{3 ! 5 ! 7 !}}=\frac{\frac{14 !}{4 ! 10 !} \frac{9 !}{6 ! 3 !}+\frac{14 !}{6 ! 8 !} \frac{7 !}{4 ! 3 !}}{\frac{15 !}{3 ! 5 ! 7 !}} \\
& =\frac{14 !}{15 !} \frac{9 !}{10 !} \frac{5 ! 7 !}{4 !} \frac{3 !}{6 !}+\frac{14 !}{15 !} \frac{7 !}{8 !} \frac{7 !}{6 !} \frac{5 !}{4 !} \frac{3 !}{3 !} \\
& =\frac{1}{15} \cdot \frac{1}{10} \cdot 5 \cdot 7 \cdot 1+\frac{1}{15} \cdot \frac{1}{8} \cdot 7 \cdot 5 \cdot 1 \\
& =\frac{7}{30}+\frac{7}{24}=\frac{28}{120}+\frac{35}{120}=\frac{63}{120}=\frac{21}{40}
\end{aligned}
$$

## Part B