A class of twenty-five pupils consists of 15 girls and 10 boys. At the beginning of the year four pupils are to be chosen at random to form an interim class committee comprising of "Chairperson", "Vice-Chairperson", "Treasurer" and "Secretary". Find

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.