A class of twenty four pupils consists of 11 girls and 13 boys. To form the class committee, four of the pupils are chosen at random as "Chairperson", "Vice Chairperson", "Treasurer" and "Secretary".

1. Calculate the probability that all 4 selected committee members are girls.

$\mathrm{P}(\text{all girls}) = \frac{11}{24} \times \frac{10}{23} \times \frac{9}{22} \times \frac{8}{21} = \frac{5}{161}$

2. Calculate the probability that all 4 selected committee members are boys.

$\mathrm{P}(\text{all boys}) = \frac{13}{24} \times \frac{12}{23} \times \frac{11}{22} \times \frac{10}{21} = \frac{65}{966}$

3. Use the complement rule to find the probability of having at least one girl and at least one boy.

$\mathrm{P}(\text{at least 1 girl and 1 boy}) = 1 - \mathrm{P}(\text{all girls}) - \mathrm{P}(\text{all boys}) \\ = 1 - \frac{5}{161} - \frac{65}{966} \\ = 1 - \frac{30}{966} - \frac{65}{966} \\ = 1 - \frac{95}{966} = \frac{871}{966}$

1. Set up the probability expression for the two specific roles being filled by girls

Since the remaining two roles (Chairperson and Vice Chairperson) can be filled by anyone and do not restrict the Treasurer and Secretary positions, we can focus entirely on picking 2 girls from the class for these 2 specific roles.

$\text{P(Treasurer and Secretary are girls)} = \text{P(1st is a girl)} \times \text{P(2nd is a girl given 1st is a girl)}$
$= \frac{11}{24} \times \frac{10}{23}$

2. Evaluate and simplify the fraction