一个 6 面骰子有三个面标有 '1'，两个面标有 '2'，一个面标有 '3'。另一个 6 面骰子有一个面标有 '1'，两个面标有 '2'，三个面标有 '3'。在一场游戏中，玩家同时掷这两个骰子。

• 如果朝上的数字相同，则从装有 3 张蓝色卡片和 2 张白色卡片的袋中无放回地随机抽取两张卡片。
• 如果朝上的数字不同，则从袋中随机抽取卡片，每次无放回地抽取一张，直到抽到 2 张白色卡片为止。

从袋中抽取的蓝色卡片数量记为 $X$。

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.

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