A 6-sided die has three faces marked with a '1', two faces marked with a '2' and one face marked with a '3'. Another 6-sided die has one face marked with a '1', two faces marked with a '2' and three faces marked with a '3'. In a game, a player throws both dice simultaneously.

• If the numbers on the top faces are the same, two cards are drawn at random without replacement from a bag containing 3 blue cards and 2 white cards.
• If the numbers on the top faces are different, cards will be drawn at random from the bag, one at a time without replacement, until 2 white cards are taken out.

The number of blue cards drawn from the bag is denoted by X.

1. Formulate the sum of probabilities for the independent events of getting matching numbers (both '1', both '2', or both '3').

$P(\text{same}) = P(1,1) + P(2,2) + P(3,3)$
$= \left(\frac{3}{6} \times \frac{1}{6}\right) + \left(\frac{2}{6} \times \frac{2}{6}\right) + \left(\frac{1}{6} \times \frac{3}{6}\right)$

2. Evaluate the probability sum to show the target result.

$= \frac{3}{36} + \frac{4}{36} + \frac{3}{36}$
$= \frac{10}{36}$
$= \frac{5}{18} \text{ (Shown)}$

1. Identify the probabilities of the two dice outcome scenarios based on part (i).

Let $S$ be the event that the numbers on the top faces of the two dice are the same, and $S'$ be the event that they are different.
From part (i), $\mathrm{P}(S) = \frac{5}{18}$.
Therefore, $\mathrm{P}(S') = 1 - \frac{5}{18} = \frac{13}{18}$.

2. Calculate the conditional probabilities of drawing $X$ blue cards given that the top faces are the same.

If $S$ occurs, exactly 2 cards are drawn at random from 3 Blue (B) and 2 White (W) cards. The possible values for $X$ are 0, 1, and 2.
$\mathrm{P}(X=0 \mid S) = \mathrm{P}(WW) = \frac{2}{5} \times \frac{1}{4} = \frac{1}{10}$
$\mathrm{P}(X=1 \mid S) = \mathrm{P}(BW) + \mathrm{P}(WB) = 2 \times \left( \frac{3}{5} \times \frac{2}{4} \right) = \frac{3}{5}$
$\mathrm{P}(X=2 \mid S) = \mathrm{P}(BB) = \frac{3}{5} \times \frac{2}{4} = \frac{3}{10}$
$\mathrm{P}(X=3 \mid S) = 0$

3. Calculate the conditional probabilities of drawing $X$ blue cards given that the top faces are different.

If $S'$ occurs, cards are drawn one at a time until the 2nd white card is drawn. Since there are only 2 white cards in total, the sequence must always end with the 2nd W card. The possible values for $X$ are 0, 1, 2, and 3.
$\mathrm{P}(X=0 \mid S') = \mathrm{P}(WW) = \frac{2}{5} \times \frac{1}{4} = \frac{1}{10}$

For $X=1$, the sequence must have exactly 1 B and 1 W in the first two draws ($2$ arrangements: BW, WB), followed by W:
$\mathrm{P}(X=1 \mid S') = 2 \times \left( \frac{3}{5} \times \frac{2}{4} \times \frac{1}{3} \right) = \frac{1}{5}$

For $X=2$, the sequence must have 2 B and 1 W in the first three draws ($\binom{3}{2}=3$ arrangements: BBW, BWB, WBB), followed by W:
$\mathrm{P}(X=2 \mid S') = 3 \times \left( \frac{3}{5} \times \frac{2}{4} \times \frac{2}{3} \times \frac{1}{2} \right) = \frac{3}{10}$

For $X=3$, the sequence must have 3 B and 1 W in the first four draws ($\binom{4}{3}=4$ arrangements), followed by W:
$\mathrm{P}(X=3 \mid S') = 4 \times \left( \frac{3}{5} \times \frac{2}{4} \times \frac{1}{3} \times \frac{2}{2} \times \frac{1}{1} \right) = \frac{2}{5}$

4. Calculate the total (unconditional) probability for each value of $X$ to form the distribution.

Using the law of total probability, $\mathrm{P}(X=x) = \mathrm{P}(X=x \mid S)\mathrm{P}(S) + \mathrm{P}(X=x \mid S')\mathrm{P}(S')$:

$\mathrm{P}(X=0) = \left(\frac{1}{10}\right)\left(\frac{5}{18}\right) + \left(\frac{1}{10}\right)\left(\frac{13}{18}\right) = \frac{5 + 13}{180} = \frac{1}{10}$

$\mathrm{P}(X=1) = \left(\frac{3}{5}\right)\left(\frac{5}{18}\right) + \left(\frac{1}{5}\right)\left(\frac{13}{18}\right) = \frac{15}{90} + \frac{13}{90} = \frac{28}{90} = \frac{14}{45}$

$\mathrm{P}(X=2) = \left(\frac{3}{10}\right)\left(\frac{5}{18}\right) + \left(\frac{3}{10}\right)\left(\frac{13}{18}\right) = \frac{15 + 39}{180} = \frac{54}{180} = \frac{3}{10}$

$\mathrm{P}(X=3) = (0)\left(\frac{5}{18}\right) + \left(\frac{2}{5}\right)\left(\frac{13}{18}\right) = \frac{26}{90} = \frac{13}{45}$

1. Establish the probability distribution of $X$ from part (ii).

We first succinctly establish the probability distribution of $X$ required from part (ii).
Using the probabilities of the dice faces being the same ($\frac{5}{18}$) or different ($\frac{13}{18}$) and calculating the respective conditional probability of drawing $x$ blue cards:
\begin{align*}
P(X=0) &= \frac{5}{18}\left(\frac{1}{10}\right) + \frac{13}{18}\left(\frac{1}{10}\right) = \frac{1}{10} \
P(X=1) &= \frac{5}{18}\left(\frac{6}{10}\right) + \frac{13}{18}\left(\frac{2}{10}\right) = \frac{14}{45} \
P(X=2) &= \frac{5}{18}\left(\frac{3}{10}\right) + \frac{13}{18}\left(\frac{3}{10}\right) = \frac{3}{10} \
P(X=3) &= \frac{5}{18}(0) + \frac{13}{18}\left(\frac{4}{10}\right) = \frac{13}{45}
\end{align*}

2. Relate $P(|X_1 - X_2| \ge 1)$ to its complementary event.

The event $|X_1 - X_2| \ge 1$ occurs when the number of blue cards obtained in the two games are different.
Using the complementary event $X_1 = X_2$:
$P(|X_1 - X_2| \ge 1) = 1 - P(X_1 = X_2)$

3. Compute $P(X_1 = X_2)$ using the fact that the two games are independent.

Since the outcomes of the two games are independent, $P(X_1 = X_2)$ is the sum of the squares of each probability in the distribution:
\begin{align*}
P(X_1 = X_2) &= P(X=0)^2 + P(X=1)^2 + P(X=2)^2 + P(X=3)^2 \
&= \left(\frac{1}{10}\right)^2 + \left(\frac{14}{45}\right)^2 + \left(\frac{3}{10}\right)^2 + \left(\frac{13}{45}\right)^2 \
&= \frac{1}{100} + \frac{196}{2025} + \frac{9}{100} + \frac{169}{2025} \
&= \frac{10}{100} + \frac{365}{2025} \
&= \frac{1}{10} + \frac{73}{405} \
&= \frac{227}{810}
\end{align*}

4. Subtract from 1 to find the final probability.

Now, subtract from $1$ to find the desired probability:
\begin{align*}
P(|X_1 - X_2| \ge 1) &= 1 - \frac{227}{810} \
&= \frac{583}{810}
\end{align*}

1. Deduce the probability distribution of $X$ (as required from part ii).

$X$ is the number of blue cards drawn from a bag with 3 Blue, 2 White cards.
Case 1: Same dice numbers ($P = \frac{5}{18}$). 2 cards drawn without replacement.

• $P(X=0 \mid \text{same}) = \frac{2}{5} \times \frac{1}{4} = \frac{1}{10}$
• $P(X=1 \mid \text{same}) = 2 \times \left( \frac{3}{5} \times \frac{2}{4} \right) = \frac{6}{10}$
• $P(X=2 \mid \text{same}) = \frac{3}{5} \times \frac{2}{4} = \frac{3}{10}$

Case 2: Different dice numbers ($P = \frac{13}{18}$). Cards drawn until 2 white cards are out.

• $P(X=0 \mid \text{diff}) = \frac{2}{5} \times \frac{1}{4} = \frac{1}{10}$
• $P(X=1 \mid \text{diff}) = 2 \times \left(\frac{3}{5} \times \frac{2}{4} \times \frac{1}{3}\right) = \frac{2}{10}$
• $P(X=2 \mid \text{diff}) = 3 \times \left(\frac{3}{5} \times \frac{2}{4} \times \frac{2}{3} \times \frac{1}{2}\right) = \frac{3}{10}$
• $P(X=3 \mid \text{diff}) = 4 \times \left(\frac{3}{5} \times \frac{2}{4} \times \frac{1}{3} \times \frac{2}{2} \times 1\right) = \frac{4}{10}$

By the Law of Total Probability:
$P(X=0) = \frac{5}{18}(\frac{1}{10}) + \frac{13}{18}(\frac{1}{10}) = \frac{1}{10}$
$P(X=1) = \frac{5}{18}(\frac{6}{10}) + \frac{13}{18}(\frac{2}{10}) = \frac{56}{180} = \frac{14}{45}$
$P(X=2) = \frac{5}{18}(\frac{3}{10}) + \frac{13}{18}(\frac{3}{10}) = \frac{3}{10}$
$P(X=3) = \frac{5}{18}(0) + \frac{13}{18}(\frac{4}{10}) = \frac{13}{45}$

2. Calculate the expectation and variance of $X$.

$E(X) = 0\left(\frac{1}{10}\right) + 1\left(\frac{14}{45}\right) + 2\left(\frac{3}{10}\right) + 3\left(\frac{13}{45}\right) = \frac{16}{9}$

$E(X^2) = 0^2\left(\frac{1}{10}\right) + 1^2\left(\frac{14}{45}\right) + 2^2\left(\frac{3}{10}\right) + 3^2\left(\frac{13}{45}\right) = \frac{37}{9}$

$Var(X) = E(X^2) - [E(X)]^2$
$= \frac{37}{9} - \left(\frac{16}{9}\right)^2 = \frac{77}{81}$

3. Use the Central Limit Theorem to state the distribution of the sum of observations.

Let $S = \sum_{i=1}^{30} X_i$.
$E(S) = 30 \times E(X) = 30\left(\frac{16}{9}\right) = \frac{160}{3}$
$Var(S) = 30 \times Var(X) = 30\left(\frac{77}{81}\right) = \frac{770}{27}$

Since $n = 30 \ge 30$, by the Central Limit Theorem, the sum $S$ is approximately normally distributed:
$S \sim N\left(\frac{160}{3}, \frac{770}{27}\right)$

4. Calculate the required probability.

$P(S > 45) \approx P\left(Z > \frac{45 - \frac{160}{3}}{\sqrt{\frac{770}{27}}}\right)$
$= P(Z > -1.5605)$
$= 0.94068... \approx 0.941 \text{ (to 3 s.f.)}$