从包含10个数字的集合 {0, 1, 2, . . . , 9} 中独立且随机地选择两个数字 $X$ 和 $Y$。事件 $A$ 和 $B$ 定义如下： A：$X = Y + 1$。 B：$X$ 和 $Y$ 都小于 6。 求： P(A)。 P(B)。 $P(A \cup B)$。 在6月的某个下午，5名女生和4名男生在邵氏图书馆（Shaw Library），6名女生和9名男生在霍莱特图书馆（Hullett Library）。一位老师从每个图书馆中随机挑选2名学生，分发4张免费音乐会门票。 计算收到门票的学生中，有2名女生和2名男生的概率。 已知收到门票的学生中有2名女生和2名男生，求从霍莱特图书馆选出的2名学生性别相同的概率。

1. Find the total number of outcomes and the number of favorable outcomes for event A

Since $X$ and $Y$ are chosen independently from 10 digits, the total number of possible pairs $(X, Y)$ is:
$10 \times 10 = 100$

For event A ($X = Y + 1$), the possible pairs are:
$(1, 0), (2, 1), (3, 2), (4, 3), (5, 4), (6, 5), (7, 6), (8, 7), (9, 8)$
There are 9 favorable outcomes.

2. Calculate the probability of event A

$\text{P}(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{9}{100}$

1. Find the probability that a single chosen digit is less than 6.

The digits less than 6 from the set of 10 digits are $\{0, 1, 2, 3, 4, 5\}$. There are 6 such digits.
$P(X < 6) = \frac{6}{10} = \frac{3}{5}$
Similarly, $P(Y < 6) = \frac{3}{5}$.

2. Calculate $P(B)$ using the independence of $X$ and $Y$.

Since $X$ and $Y$ are chosen independently:
$P(B) = P(X < 6) \times P(Y < 6)$
$P(B) = \frac{3}{5} \times \frac{3}{5} = \frac{9}{25}$

1. Concise calculation of $P(A)$ and $P(B)$ from parts (i) and (ii).

The total number of outcomes is $10 \times 10 = 100$.
For Event A ($X = Y + 1$), the pairs are $(1,0), (2,1), \dots, (9,8)$. There are $9$ pairs, so $P(A) = \frac{9}{100}$.
For Event B ($X < 6$ and $Y < 6$), there are $6 \times 6 = 36$ outcomes. Thus, $P(B) = \frac{36}{100}$.

2. Find the probability of the intersection, $P(A \cap B)$.

The intersection $A \cap B$ contains outcomes that satisfy both events: $X = Y + 1$ and $X, Y < 6$.
Listing these outcomes: $(1,0), (2,1), (3,2), (4,3), (5,4)$.
There are 5 such outcomes, so $P(A \cap B) = \frac{5}{100}$.

3. Calculate $P(A \cup B)$ using the addition rule.

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$P(A \cup B) = \frac{9}{100} + \frac{36}{100} - \frac{5}{100} = \frac{40}{100} = \frac{2}{5}$

1. Identify all mutually exclusive cases that result in 2 girls and 2 boys being selected in total.

Let $G$ represent a girl and $B$ represent a boy. Since 2 students are selected from each library, the possible distributions from the Shaw Library and the Hullett Library respectively are:

• Case 1: 2 $G$ from Shaw, 2 $B$ from Hullett
• Case 2: 1 $G$ and 1 $B$ from Shaw, 1 $G$ and 1 $B$ from Hullett
• Case 3: 2 $B$ from Shaw, 2 $G$ from Hullett

2. Calculate the probability of each independent case occurring.

The selections from the two libraries are independent. Using combinations $\binom{n}{r}$:

$P(\text{Case 1}) = \frac{\binom{5}{2}}{\binom{9}{2}} \times \frac{\binom{9}{2}}{\binom{15}{2}} = \frac{10}{36} \times \frac{36}{105} = \frac{2}{21}$

$P(\text{Case 2}) = \frac{\binom{5}{1}\binom{4}{1}}{\binom{9}{2}} \times \frac{\binom{6}{1}\binom{9}{1}}{\binom{15}{2}} = \frac{20}{36} \times \frac{54}{105} = \frac{2}{7}$

$P(\text{Case 3}) = \frac{\binom{4}{2}}{\binom{9}{2}} \times \frac{\binom{6}{2}}{\binom{15}{2}} = \frac{6}{36} \times \frac{15}{105} = \frac{1}{42}$

3. Sum the probabilities of the mutually exclusive cases.

$\text{Total Probability} = P(\text{Case 1}) + P(\text{Case 2}) + P(\text{Case 3})$
$= \frac{2}{21} + \frac{2}{7} + \frac{1}{42}$
$= \frac{4}{42} + \frac{12}{42} + \frac{1}{42} = \frac{17}{42}$

1. Calculate the total probability of selecting exactly 2 girls and 2 boys (required result from part b(i)).

Let $S$ be Shaw Library and $H$ be Hullett Library. The event of selecting 2 girls and 2 boys overall consists of three mutually exclusive cases: (2G from $S$, 2B from $H$), (2B from $S$, 2G from $H$), and (1G1B from $S$, 1G1B from $H$).

$\text{P(2G, 2B)} = \left(\frac{^5\text{C}_2}{^9\text{C}_2}\right)\left(\frac{^9\text{C}_2}{^{15}\text{C}_2}\right) + \left(\frac{^4\text{C}_2}{^9\text{C}_2}\right)\left(\frac{^6\text{C}_2}{^{15}\text{C}_2}\right) + \left(\frac{^5\text{C}_1 \times ^4\text{C}_1}{^9\text{C}_2}\right)\left(\frac{^6\text{C}_1 \times ^9\text{C}_1}{^{15}\text{C}_2}\right)$
$= \left(\frac{10}{36} \times \frac{36}{105}\right) + \left(\frac{6}{36} \times \frac{15}{105}\right) + \left(\frac{20}{36} \times \frac{54}{105}\right)$
$= \frac{360}{3780} + \frac{90}{3780} + \frac{1080}{3780}$
$= \frac{1530}{3780} = \frac{17}{42}$

2. Find the probability of the intersection: the 2 students from Hullett are of the same gender AND overall there are exactly 2 girls and 2 boys.

For the students from Hullett to be of the same gender while satisfying the 2G and 2B overall condition, the outcome must be either (2B from $S$, 2G from $H$) or (2G from $S$, 2B from $H$).

$\text{P(Hullett same gender} \cap \text{2G, 2B)} = \left(\frac{10}{36} \times \frac{36}{105}\right) + \left(\frac{6}{36} \times \frac{15}{105}\right)$
$= \frac{360}{3780} + \frac{90}{3780}$
$= \frac{450}{3780} = \frac{5}{42}$

3. Calculate the conditional probability.

Using the conditional probability formula $\text{P}(A|B) = \frac{\text{P}(A \cap B)}{\text{P}(B)}$:

$\text{P(Hullett same gender | 2G, 2B)} = \frac{\frac{5}{42}}{\frac{17}{42}}$
$= \frac{5}{17}$