A bag contains 3 red balls and $n$ yellow balls, where $n \ge 2$. In a game, Joe removes 2 balls at random from the bag one at a time without replacement. The number of red balls Joe removes from the bag is denoted by $T$.

1. Identify the total number of balls and the possible values of $t$.

The total number of balls in the bag initially is $n + 3$.
Since Joe draws 2 balls and there are at least 2 red balls and 2 yellow balls (as $n \ge 2$), the number of red balls drawn, $t$, can be $0, 1,$ or $2$.

2. Calculate the probability of drawing 0 red balls (i.e. 2 yellow balls).

$\mathrm{P}(T=0) = \mathrm{P}(\text{Yellow, Yellow})$
$= \frac{n}{n+3} \times \frac{n-1}{n+2}$
$= \frac{n(n-1)}{(n+3)(n+2)}$

3. Calculate the probability of drawing exactly 1 red ball.

$\mathrm{P}(T=1) = \mathrm{P}(\text{Red, Yellow}) + \mathrm{P}(\text{Yellow, Red})$
$= \left(\frac{3}{n+3} \times \frac{n}{n+2}\right) + \left(\frac{n}{n+3} \times \frac{3}{n+2}\right)$
$= \frac{3n}{(n+3)(n+2)} + \frac{3n}{(n+3)(n+2)}$
$= \frac{6n}{(n+3)(n+2)}$

4. Calculate the probability of drawing 2 red balls.

$\mathrm{P}(T=2) = \mathrm{P}(\text{Red, Red})$
$= \frac{3}{n+3} \times \frac{2}{n+2}$
$= \frac{6}{(n+3)(n+2)}$

1. Determine the required probabilities from part (i)

From part (i), the probabilities of removing 1 or 2 red balls are:
$\mathrm{P}(T=1) = 2 \times \frac{3}{n+3} \times \frac{n}{n+2} = \frac{6n}{(n+3)(n+2)}$
$\mathrm{P}(T=2) = \frac{3}{n+3} \times \frac{2}{n+2} = \frac{6}{(n+3)(n+2)}$

2. Calculate the expected value $\mathrm{E}(T)$

$\mathrm{E}(T) = \sum t \cdot \mathrm{P}(T=t)$
$= 0 + 1 \cdot \frac{6n}{(n+3)(n+2)} + 2 \cdot \frac{6}{(n+3)(n+2)}$
$= \frac{6n + 12}{(n+3)(n+2)}$
$= \frac{6(n+2)}{(n+3)(n+2)} = \frac{6}{n+3}$

3. Calculate $\mathrm{E}(T^2)$ to prepare for the variance calculation

$\mathrm{E}(T^2) = \sum t^2 \cdot \mathrm{P}(T=t)$
$= 0 + 1^2 \cdot \frac{6n}{(n+3)(n+2)} + 2^2 \cdot \frac{6}{(n+3)(n+2)}$
$= \frac{6n + 24}{(n+3)(n+2)}$
$= \frac{6(n+4)}{(n+3)(n+2)}$

4. Calculate the variance $\mathrm{Var}(T)$ and simplify the algebraic fraction

$\mathrm{Var}(T) = \mathrm{E}(T^2) - [\mathrm{E}(T)]^2$
$= \frac{6(n+4)}{(n+3)(n+2)} - \left( \frac{6}{n+3} \right)^2$
$= \frac{6(n+4)(n+3) - 36(n+2)}{(n+3)^2(n+2)}$
$= \frac{6(n^2 + 7n + 12) - 36n - 72}{(n+3)^2(n+2)}$
$= \frac{6n^2 + 42n + 72 - 36n - 72}{(n+3)^2(n+2)}$
$= \frac{6n^2 + 6n}{(n+3)^2(n+2)}$
$= \frac{6n(n+1)}{(n+3)^2(n+2)}$