袋子中装有 6 个红盘、6 个绿盘、6 个黄盘和 6 个蓝盘。除了颜色之外，这些圆盘无法区分。从袋子中不放回地依次取出 4 个圆盘。随机变量 $X$ 表示取出的 4 个圆盘中所获得的不同颜色的数量。

1. Calculate the total number of ways to select 4 discs from 24.

$\text{Total ways to draw 4 discs} = {}^{24}C_4 = 10626$

2. Calculate the number of ways to obtain 3 discs of one colour and 1 of another.

To get exactly 2 colours, the distribution of discs must be either a 3-1 split or a 2-2 split.

$\text{Ways for 3-1 split} = {}^{4}C_1 \times {}^{6}C_3 \times {}^{3}C_1 \times {}^{6}C_1$
$= 4 \times 20 \times 3 \times 6 = 1440$

3. Calculate the number of ways to obtain 2 discs of one colour and 2 of another.

$\text{Ways for 2-2 split} = {}^{4}C_2 \times {}^{6}C_2 \times {}^{6}C_2$
$= 6 \times 15 \times 15 = 1350$

4. Calculate the required probability by summing the valid outcomes and dividing by the total.

$\text{Total favourable ways} = 1440 + 1350 = 2790$

$\mathrm{P}(X = 2) = \frac{2790}{10626}$
$= \frac{465}{1771}$ (shown)

1. Identify the total number of possible outcomes for drawing 4 discs.

Total number of discs $= 6 \times 4 = 24$.
Number of ways to draw 4 discs $= \binom{24}{4} = 10626$.

2. Calculate the probability of drawing 4 discs of the same colour ($X=1$).

To get $X=1$, all 4 discs must be of the same colour.
We first choose $1$ of the $4$ colours, then select $4$ discs out of the $6$ available in that colour.

3. Calculate the probability of drawing 4 discs of completely different colours ($X=4$).

To get $X=4$, we must pick exactly $1$ disc from each of the $4$ colours.

4. Calculate the remaining probability ($X=3$) and compile the probability distribution table.

Since the sum of all probabilities must equal $1$, we can use the complement method along with the value of $\mathrm{P}(X=2)$ given in part (a).

(Alternatively, $\mathrm{P}(X=3) = \frac{\binom{4}{1}\binom{6}{2} \times \binom{3}{2}\binom{6}{1}^2}{10626} = \frac{1080}{1771}$)

The probability distribution table is:

1. Find the probability of winning a single round.

From part (b), the probability of getting exactly 1 colour ($X=1$) is:
$\mathrm{P}(X=1) = \frac{4 \times \binom{6}{4}}{\binom{24}{4}} = \frac{60}{10626} = \frac{10}{1771}$

For Joe to win a round, he must obtain at most 2 different colours ($X \le 2$). Using the given $\mathrm{P}(X=2)$ from part (a):
$\mathrm{P}(\text{win}) = \mathrm{P}(X \le 2) = \mathrm{P}(X=1) + \mathrm{P}(X=2)$
$\mathrm{P}(\text{win}) = \frac{10}{1771} + \frac{465}{1771} = \frac{475}{1771}$

2. Determine the minimum number of winning rounds required to win money overall.

Let $W$ be the number of winning rounds out of 8. We model this as $W \sim \mathrm{B}\left(8, \frac{475}{1771}\right)$.

For Joe to win some money, his total net winnings must be strictly positive:
$10W - 5(8 - W) > 0$
$15W - 40 > 0$
$W > \frac{40}{15} \approx 2.67$

Thus, Joe must win at least 3 rounds ($W \ge 3$).

3. Calculate the probability of Joe winning money at the end of the game.

We need to find $\mathrm{P}(W \ge 3)$:
$\mathrm{P}(W \ge 3) = 1 - \mathrm{P}(W \le 2)$

Using a graphing calculator's binomial CDF function:
$\mathrm{P}(W \ge 3) = 1 - 0.6337005$
$\mathrm{P}(W \ge 3) \approx 0.366 \text{ (to 3 s.f.)}$

1. Find the probability of winning a single round ($P(X \le 2)$).

From part (b), the probability of getting 1 colour is $P(X = 1) = \frac{4 \times \binom{6}{4}}{\binom{24}{4}} = \frac{10}{1771}$.

The probability of winning a round (getting at most 2 colours) is $P(X \le 2) = P(X=1) + P(X=2) = \frac{10}{1771} + \frac{465}{1771} = \frac{475}{1771}$.

2. Calculate the expected winnings in a single round.

Let $W$ be the winnings in a single round.
$E(W) = \char36{}10 \times P(X \le 2) - \char36{}5 \times P(X > 2)$

$E(W) = 10 \left(\frac{475}{1771}\right) - 5 \left(1 - \frac{475}{1771}\right)$
$E(W) = \frac{4750}{1771} - 5 \left(\frac{1296}{1771}\right) = -\frac{1730}{1771} \approx -\char36{}0.9768$

3. Determine the expected winnings for the whole game and conclude.

For the game of 8 rounds, the expected total winnings is $8 \times E(W) = 8 \left(-\frac{1730}{1771}\right) = -\frac{13840}{1771} \approx -\char36{}7.81$.

Since the expected total winnings is negative (an expected loss), it is not worthwhile for Joe to play this game.