The bag contains 6 red discs, 6 green discs, 6 yellow discs and 6 blue discs. The discs are indistinguishable aside from their colour. 4 discs are taken one after another from the bag without replacement. The random variable $X$ is the number of different colours obtained among the 4 discs.

1. Calculate the total possible combinations for drawing 4 discs.

$\text{Total number of ways to draw 4 discs from 24} = \binom{24}{4} = 10626$

2. Identify the possible distributions of colours that result in exactly 2 colours.

To obtain exactly 2 colours among 4 discs, the breakdown of the discs must be either:

• 3 discs of one colour and 1 disc of a second colour (a 3-1 split)
• 2 discs of one colour and 2 discs of a second colour (a 2-2 split)

3. Calculate the number of ways to obtain a 3-1 split.

Select 1 colour out of 4 for the 3 discs, and pick 3 discs from its 6. Then select 1 colour out of the remaining 3 for the 1 disc, and pick 1 disc from its 6.
$\text{Ways for 3-1 split} = \binom{4}{1}\binom{6}{3} \times \binom{3}{1}\binom{6}{1}$
$= 4(20) \times 3(6) = 1440$

4. Calculate the number of ways to obtain a 2-2 split.

Select 2 colours out of 4. For each selected colour, pick 2 discs from its 6.
$\text{Ways for 2-2 split} = \binom{4}{2}\binom{6}{2}\binom{6}{2}$
$= 6(15)(15) = 1350$

5. Sum the favourable outcomes and divide by the total to prove the given probability.

$P(X=2) = \frac{1440 + 1350}{10626}$
$P(X=2) = \frac{2790}{10626} = \frac{465}{1771} \text{ (shown)}$

1. Identify the possible values of $X$ and calculate $\mathrm{P}(X = 1)$.

The possible values for the number of different colours, $X$, are 1, 2, 3, and 4.

To obtain $X = 1$, all 4 discs must be of the same colour. We choose 1 colour out of the 4, and then pick 4 discs out of the 6 of that chosen colour.
$\mathrm{P}(X = 1) = \frac{\binom{4}{1} \times \binom{6}{4}}{\binom{24}{4}}$
$\mathrm{P}(X = 1) = \frac{4 \times 15}{10626} = \frac{60}{10626} = \frac{10}{1771}$

2. Calculate $\mathrm{P}(X = 4)$.

To obtain $X = 4$, the 4 discs must all be of different colours. We must pick exactly 1 disc of each of the 4 colours.
$\mathrm{P}(X = 4) = \frac{\binom{6}{1}^4}{\binom{24}{4}}$
$\mathrm{P}(X = 4) = \frac{6^4}{10626} = \frac{1296}{10626} = \frac{216}{1771}$

3. Calculate $\mathrm{P}(X = 3)$ using the sum of probabilities, and compile the probability distribution.

Since the sum of all probabilities in a discrete probability distribution must equal 1, and we are given $\mathrm{P}(X = 2) = \frac{465}{1771}$ from part (a):
$\mathrm{P}(X = 3) = 1 - \mathrm{P}(X = 1) - \mathrm{P}(X = 2) - \mathrm{P}(X = 4)$
$\mathrm{P}(X = 3) = 1 - \frac{10}{1771} - \frac{465}{1771} - \frac{216}{1771}$
$\mathrm{P}(X = 3) = \frac{1080}{1771}$

The probability distribution of $X$ is:

1. Calculate the probability of winning a single round.

From part (b), we first find $P(X=1)$, the probability of all 4 discs being the same colour. There are 4 possible colours to choose from, and we must choose 4 discs from the 6 available of that colour:
$P(X=1) = \frac{4 \times \binom{6}{4}}{\binom{24}{4}} = \frac{60}{10626} = \frac{10}{1771}$
The probability of winning a single round is the probability of obtaining at most 2 different colours ($X \le 2$):
$P(\text{Win}) = P(X \le 2) = P(X=1) + P(X=2)$
$P(\text{Win}) = \frac{10}{1771} + \frac{465}{1771} = \frac{475}{1771}$

2. Determine the minimum number of rounds Joe must win to have a positive net gain.

Let $Y$ be the number of rounds Joe wins out of 8. His total winnings in dollars can be expressed as:
$\text{Net Winnings} = 10Y - 5(8 - Y) = 15Y - 40$
For Joe to win some money at the end of the game, his net winnings must be strictly greater than 0:
$15Y - 40 > 0 \implies Y > \frac{40}{15} = 2.67$
Since $Y$ must be a whole number, Joe needs to win at least 3 rounds ($Y \ge 3$).

3. Calculate the probability that Joe wins at least 3 rounds using the Binomial distribution.

The number of winning rounds follows a binomial distribution, $Y \sim \text{B}\left(8, \frac{475}{1771}\right)$.
We want to find $P(Y \ge 3)$:
$P(Y \ge 3) = 1 - P(Y \le 2)$
Using a graphing calculator:
$P(Y \ge 3) \approx 1 - 0.631736 = 0.368264$
$P(Y \ge 3) \approx 0.368 \text{ (to 3 s.f.)}$

1. Find the probability of obtaining at most 2 different colours in a round.

Relying on the distribution of $X$ derived in part (b), the probability of winning a single round is:
$P(X \le 2) = P(X = 1) + P(X = 2)$
Since $P(X=1) = \frac{\binom{4}{1} \times \binom{6}{4}}{\binom{24}{4}} = \frac{10}{1771}$, we have:
$P(X \le 2) = \frac{10}{1771} + \frac{465}{1771} = \frac{475}{1771}$

2. Calculate the expected winnings for the 8 rounds.

Let $W$ be Joe's total winnings over 8 rounds. The expected winnings per round is calculated by multiplying each payout by its corresponding probability.
$E(W) = 8 \times \left( 10 \times P(X \le 2) - 5 \times P(X > 2) \right)$
$E(W) = 8 \times \left( 10 \left(\frac{475}{1771}\right) - 5 \left(1 - \frac{475}{1771}\right) \right)$
$E(W) = 8 \times \left( \frac{4750}{1771} - 5 \left(\frac{1296}{1771}\right) \right)$
$E(W) = 8 \times \left( \frac{4750 - 6480}{1771} \right) = -\frac{13840}{1771} \approx -\$7.81$

3. State the conclusion based on the expected winnings.

Since the expected total winnings is negative (he expects to make a loss of $\char36{}7.81$ on average per game), it is not worthwhile for Joe to play this game to win money.