一个由25名学生组成的班级有15名女生和10名男生。学年初，将随机挑选4名学生组成临时班委会，包括“主席”、“副主席”、“财务”和“秘书”。求：

1. Set up the expression for the probability using combinations.

$P(\text{exactly 2 girls}) = \frac{{}^{15}C_2 \times {}^{10}C_2}{{}^{25}C_4}$

2. Evaluate the combinations to find the final probability.

1. Calculate the probability of choosing a girl as Chairperson and a boy as Vice-Chairperson.

$P(\text{Girl then Boy}) = \frac{15}{25} \times \frac{10}{24} = \frac{1}{4}$

2. Calculate the probability of choosing a boy as Chairperson and a girl as Vice-Chairperson, and sum the probabilities to find the final answer.

$P(\text{Boy then Girl}) = \frac{10}{25} \times \frac{15}{24} = \frac{1}{4}$

$P(\text{Opposite sex}) = P(\text{Girl then Boy}) + P(\text{Boy then Girl})$
$P(\text{Opposite sex}) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$

1. Identify the reduced pool of available students given the condition.

The condition states that the Treasurer and Secretary are of opposite sex. This means exactly $1$ boy and $1$ girl have been chosen from the class for these two specific roles.

Remaining total pupils $= 25 - 2 = 23$
Remaining boys $= 10 - 1 = 9$
Remaining girls $= 15 - 1 = 14$

2. Set up the probability expression for choosing two boys from the remaining pool.

We need to find the probability that the remaining two roles (Chairperson and Vice-Chairperson) are filled by boys. Since the slots are filled uniformly at random, this is equivalent to drawing $2$ boys from the remaining pool of $23$ students without replacement.

$P(\text{both are boys} \mid \text{condition}) = \frac{9}{23} \times \frac{8}{22}$

3. Simplify the probability to obtain the final fraction.

$P(\text{both are boys} \mid \text{condition}) = \frac{9}{23} \times \frac{4}{11} = \frac{36}{253}$

1. Calculate the conditional probability $P(A|B)$ as required from part (iii).

Let $A$ be the event 'Chairperson and Vice-Chairperson are both boys'.
Let $B$ be the event 'Treasurer and Secretary are of opposite sex'.
By symmetry, the unconditional probability of $B$ depends only on the two positions:
$P(B) = P(\text{T is boy, S is girl}) + P(\text{T is girl, S is boy})$
$P(B) = \left(\frac{10}{25} \times \frac{15}{24}\right) + \left(\frac{15}{25} \times \frac{10}{24}\right) = \frac{1}{2}$

To find $P(A|B)$, we first find $P(A \cap B)$, which requires C and VC to be boys, and T and S to be of opposite sex:
$P(A \cap B) = P(\text{B, B, B, G}) + P(\text{B, B, G, B})$
$P(A \cap B) = \left(\frac{10}{25} \times \frac{9}{24} \times \frac{8}{23} \times \frac{15}{22}\right) + \left(\frac{10}{25} \times \frac{9}{24} \times \frac{15}{23} \times \frac{8}{22}\right) = \frac{18}{253}$

From (iii), the conditional probability is:
$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{18/253}{1/2} = \frac{36}{253}$

2. Calculate the unconditional probability $P(A)$ and compare it with $P(A|B)$ to check for independence.

The unconditional probability of event $A$ is:
$P(A) = P(\text{C is boy}) \times P(\text{VC is boy}) = \frac{10}{25} \times \frac{9}{24} = \frac{3}{20}$

For events $A$ and $B$ to be independent, the condition $P(A|B) = P(A)$ must hold.
Since $\frac{36}{253} \neq \frac{3}{20}$ (or roughly $0.142 \neq 0.150$), $P(A|B) \neq P(A)$.
Thus, the events are not independent.