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 $C$.

Since choosing a student from $A$ and choosing a student from $C$ are independent events:
$\mathrm{P}(A \cap C) = \mathrm{P}(A) \times \mathrm{P}(C)$
$\mathrm{P}(A \cap C) = \left(\frac{25}{100}\right) \times \left(\frac{40}{100}\right) = \frac{1}{10}$
$n(A \cap C) = 100 \times \frac{1}{10} = 10$

2. Find the expressions for the overlapping regions involving exactly two events.

We are given $n(A \cap B) = 15$, $n(A \cap B \cap C) = x$, and $n(B \cap C \text{ only}) = y$.

The region for $A \cap B$ only (excluding $C$):
$n(A \cap B \cap C') = n(A \cap B) - n(A \cap B \cap C) = 15 - x$

The region for $A \cap C$ only (excluding $B$):
$n(A \cap C \cap B') = n(A \cap C) - n(A \cap B \cap C) = 10 - x$

3. Deduce the expressions for the regions representing exactly one event, and the region outside all three.

For $A$ only:
$n(A \text{ only}) = 25 - (15 - x) - (10 - x) - x = x$

For $B$ only:
$n(B \text{ only}) = 30 - (15 - x) - x - y = 15 - y$

For $C$ only:
$n(C \text{ only}) = 40 - (10 - x) - x - y = 30 - y$

For the region outside $A, B, \text{ and } C$:
$\text{Total in } A \cup B \cup C = 25 + (15 - y) + y + (30 - y) = 70 - y$
$n((A \cup B \cup C)') = 100 - (70 - y) = 30 + y$

1. Find the number of students in the intersection of B and C using independence.

Since $B$ and $C$ are independent events:
$P(B \cap C) = P(B) \times P(C)$
$P(B \cap C) = \frac{30}{100} \times \frac{40}{100} = 0.12$
The number of students in $B \cap C$ is:
$0.12 \times 100 = 12$

2. Express $y$ in terms of $x$ using the Venn diagram.

From the Venn diagram, the region representing $B \cap C$ consists of the overlapping parts $x$ and $y$.
$x + y = 12$
$y = 12 - x$

3. Determine the constraints on $x$ to find the bounds for $y$.

The number of students in any region of the Venn diagram must be a non-negative integer ($\ge 0$).
For the central intersection ($A \cap B \cap C$):
$x \ge 0$
For the region in $A$ and $C$ but not $B$:
$10 - x \ge 0 \implies x \le 10$
So, the domain for $x$ is $0 \le x \le 10$.

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

Since $y = 12 - x$ and $0 \le x \le 10$:
The greatest value of $y$ occurs when $x$ is at its minimum ($x = 0$):
$y_{\text{max}} = 12 - 0 = 12$
The least value of $y$ occurs when $x$ is at its maximum ($x = 10$):
$y_{\text{min}} = 12 - 10 = 2$
(Checking the other regions ensures $15 - y \ge 0$ and $30 - y \ge 0$ are both safely satisfied for $y \in [2, 12]$.)