一个盒子包含 1 个红色圆盘、1 个蓝色圆盘和 $n$ 个黄色圆盘，其中 $n \ge 2$。从盒子中随机无放回地抽取两个圆盘。随机变量 $Y$ 表示从盒子中抽取的黄色圆盘的数量。

1. Show the probability of drawing exactly 1 yellow disc by considering the two possible sequences of draws.

There are $n$ yellow discs and $2$ non-yellow discs (1 red, 1 blue), making a total of $n+2$ discs.
$\text{P}(Y = 1) = \text{P}(\text{Yellow, Non-Yellow}) + \text{P}(\text{Non-Yellow, Yellow})$
$\text{P}(Y = 1) = \left(\frac{n}{n+2} \times \frac{2}{n+1}\right) + \left(\frac{2}{n+2} \times \frac{n}{n+1}\right)$
$\text{P}(Y = 1) = \frac{2n}{(n+1)(n+2)} + \frac{2n}{(n+1)(n+2)} = \frac{4n}{(n+1)(n+2)}$

2. Find the probability of drawing 0 yellow discs.

$\text{P}(Y = 0) = \text{P}(\text{Non-Yellow, Non-Yellow})$
$\text{P}(Y = 0) = \frac{2}{n+2} \times \frac{1}{n+1} = \frac{2}{(n+1)(n+2)}$

3. Find the probability of drawing 2 yellow discs.

$\text{P}(Y = 2) = \text{P}(\text{Yellow, Yellow})$
$\text{P}(Y = 2) = \frac{n}{n+2} \times \frac{n-1}{n+1} = \frac{n(n-1)}{(n+1)(n+2)}$

4. Present the probability distribution of $Y$ in a table.

The possible values for $Y$ are $0, 1,$ and $2$.

1. Apply the expectation formula using the probability distribution of $Y$.

From part (i), the required probabilities for the distribution of $Y$ are $\text{P}(Y=1) = \frac{4n}{(n+1)(n+2)}$ and $\text{P}(Y=2) = \frac{n(n-1)}{(n+1)(n+2)}$.
Apply the expectation formula $\text{E}(Y) = \sum y\,\text{P}(Y=y)$:
$\text{E}(Y) = 0 \cdot \text{P}(Y=0) + 1 \cdot \text{P}(Y=1) + 2 \cdot \text{P}(Y=2)$
$\text{E}(Y) = 1 \left( \frac{4n}{(n+1)(n+2)} \right) + 2 \left( \frac{n(n-1)}{(n+1)(n+2)} \right)$

2. Simplify the expression to show the given form and determine $A$.

Combine the fractions and factorise the numerator to simplify:
$\text{E}(Y) = \frac{4n + 2n(n-1)}{(n+1)(n+2)}$
$\text{E}(Y) = \frac{4n + 2n^2 - 2n}{(n+1)(n+2)}$
$\text{E}(Y) = \frac{2n^2 + 2n}{(n+1)(n+2)}$
$\text{E}(Y) = \frac{2n(n+1)}{(n+1)(n+2)}$
Cancel the common factor $(n+1)$:
$\text{E}(Y) = \frac{2n}{n+2}$
Comparing this result with the given form $\frac{An}{n+2}$, we obtain $A = 2$.

1. Use the probability distribution from part (i) to calculate E(Y²).

From the probability distribution found in part (i), $\text{P}(Y=2) = \frac{n(n-1)}{(n+1)(n+2)}$.

$\text{E}(Y^2) = \sum y^2 \text{P}(Y=y)$
$= (0^2)\text{P}(Y=0) + (1^2)\left(\frac{4n}{(n+1)(n+2)}\right) + (2^2)\left(\frac{n(n-1)}{(n+1)(n+2)}\right)$
$= \frac{4n + 4n^2 - 4n}{(n+1)(n+2)}$
$= \frac{4n^2}{(n+1)(n+2)}$

2. Apply the formula for variance using E(Y) from part (ii).

From part (ii), the constant is $A=2$, so $\text{E}(Y) = \frac{2n}{n+2}$.

Using the formula for variance:
$\text{Var}(Y) = \text{E}(Y^2) - [\text{E}(Y)]^2$
$= \frac{4n^2}{(n+1)(n+2)} - \left(\frac{2n}{n+2}\right)^2$
$= \frac{4n^2}{(n+1)(n+2)} - \frac{4n^2}{(n+2)^2}$

3. Simplify the expression to the required algebraic form.

Find a common denominator of $(n+1)(n+2)^2$ to combine the fractions:
$\text{Var}(Y) = \frac{4n^2(n+2) - 4n^2(n+1)}{(n+1)(n+2)^2}$
$= \frac{4n^2(n+2 - n - 1)}{(n+1)(n+2)^2}$
$= \frac{4n^2}{(n+2)^2(n+1)}$

Thus, $\text{f}(n) = 4n^2$, which is a quadratic polynomial as required.