A group of 100 students are asked if they are student leaders, in a sports CCA, or studying in a science faculty. The number of students who are student leaders is 25, the number of students who are in a sports CCA is 30 and the number of students studying in a science faculty is 40. There are 15 student leaders who are in a sports CCA. The number of students who are student leaders, in a sports CCA and studying in a science faculty is $x$. The number of students who are in a sports CCA and studying in a science faculty but not a student leader is $y$. One of the students is chosen at random. $A$ is the event that the student is a student leader.
$B$ is the event that the student is in a sports CCA.
$C$ is the event that the student is studying in a science faculty. It is given that $A$ and $C$ are independent.

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 ($x = 10$):
$y = 12 - 10 = 2$