对一组 100 名学生进行了调查，询问他们是否是学生领袖、是否参加了体育类课外活动，或者是否在理学院学习。其中，学生领袖有 25 人，参加体育类课外活动的学生有 30 人，在理学院学习的学生有 40 人。有 15 名学生领袖参加了体育类课外活动。既是学生领袖、又参加了体育类课外活动且在理学院学习的学生人数为 $x$。参加了体育类课外活动且在理学院学习但不是学生领袖的学生人数为 $y$。 随机抽取一名学生。 $A$ 表示该学生是学生领袖的事件。
$B$ 表示该学生参加了体育类课外活动的事件。
$C$ 表示该学生在理学院学习的事件。 已知 $A$ 和 $C$ 相互独立。

1. Determine the total number of students in $A \cap C$ using the independence of events $A$ and $C$.

$\mathrm{P}(A) = \frac{25}{100} = 0.25$
$\mathrm{P}(C) = \frac{40}{100} = 0.40$
Since $A$ and $C$ are independent:
$\mathrm{P}(A \cap C) = \mathrm{P}(A) \times \mathrm{P}(C) = 0.25 \times 0.40 = 0.10$
Number of students in $A \cap C = 0.10 \times 100 = 10$.

2. Express the regions inside circle $A$ in terms of $x$.

Given $|A \cap B| = 15$ and $|A \cap B \cap C| = x$:
Number of students in $A \cap B$ only $= 15 - x$.

Using $|A \cap C| = 10$:
Number of students in $A \cap C$ only $= 10 - x$.

Number of students in $A$ only:
$|A| - |A \cap B \text{ only}| - |A \cap C \text{ only}| - |A \cap B \cap C|$
$= 25 - (15 - x) - (10 - x) - x = x$.

3. Express the regions for $B$ only and $C$ only in terms of $y$.

Number of students in $B$ only is:
$|B| - |A \cap B \text{ only}| - |A \cap B \cap C| - |B \cap C \text{ only}|$
$= 30 - (15 - x) - x - y = 15 - y$

Number of students in $C$ only is:
$|C| - |A \cap C \text{ only}| - |A \cap B \cap C| - |B \cap C \text{ only}|$
$= 40 - (10 - x) - x - y = 30 - y$

4. Express the region outside the three circles in terms of $y$.

The total number of students in $A \cup B \cup C$ is the sum of all $7$ non-overlapping inner regions:
$x + (15 - x) + (10 - x) + x + (15 - y) + y + (30 - y) = 70 - y$

Number of students outside all circles $(A \cup B \cup C)'$ is:
$100 - (70 - y) = 30 + y$

1. Find the number of students in $B \cap C$ and express $y$ in terms of $x$.

Since $B$ and $C$ are independent, $\mathrm{P}(B \cap C) = \mathrm{P}(B) \times \mathrm{P}(C)$.
$\mathrm{P}(B \cap C) = \frac{30}{100} \times \frac{40}{100} = 0.12$
Number of students in $B \cap C = 0.12 \times 100 = 12$.
From the Venn diagram, the region $B \cap C$ comprises $x$ and $y$, so:
$x + y = 12 \implies y = 12 - x$

2. Determine the maximum possible value of $x$ using the independence of $A$ and $C$.

Given $A$ and $C$ are independent (from the main stem):
$\mathrm{P}(A \cap C) = \mathrm{P}(A) \times \mathrm{P}(C) = \frac{25}{100} \times \frac{40}{100} = 0.10$
Number of students in $A \cap C = 10$.
The number of students who are in $A \cap C$ but not $B$ is $10 - x$.
Since the number of students in any region cannot be negative:
$10 - x \ge 0 \implies x \le 10$

3. Determine the minimum possible value of $x$ by evaluating all regions.

Since $x$ represents the number of students in $A \cap B \cap C$, we inherently have $x \ge 0$.
To ensure consistency, check the region of students strictly in $A$ only:
$n(A \text{ only}) = n(A) - n(A \cap B) - n(A \cap C) + n(A \cap B \cap C)$
$n(A \text{ only}) = 25 - 15 - 10 + x = x$
This region also requires $x \ge 0$. Thus, combining both bounds gives $0 \le x \le 10$.

4. Calculate the greatest and least possible values of $y$.

Using $y = 12 - x$ and the restricted domain $0 \le x \le 10$:
The greatest value of $y$ occurs when $x$ is at its minimum ($x = 0$):
$y = 12 - 0 = 12$
The least value of $y$ occurs when $x$ is at its maximum ($x = 10$):
$y = 12 - 10 = 2$