A team of 15 students was selected for an outdoor education trip. One student volunteered to be the trip leader while another volunteered as the assistant trip leader. They decided to have some ice-breaker games, where all 15 students sat in a circle.

1. Find the total number of ways to arrange the 15 students in a circle

Total unrestricted ways $= (15 - 1)! = 14!$

2. Find the number of ways where the two leaders are seated together

Treat the two leaders as a single grouped entity.
Number of entities $= 15 - 2 + 1 = 14$.
Ways to arrange these 14 entities in a circle $= (14 - 1)! = 13!$
The two leaders can swap positions within their entity in $2!$ ways.
Number of ways they sit together $= 13! \times 2!$

3. Calculate the probability that they are not seated together using the complement rule

$\text{Probability (together)} = \frac{13! \times 2!}{14!} = \frac{2}{14} = \frac{1}{7}$

$P(\text{not together}) = 1 - P(\text{together})$
$= 1 - \frac{1}{7} = \frac{6}{7}$

1. State the conditional probability formula and identify the total number of arrangements where the leaders are separated.

Let $L$ be the event that the 2 leaders are separated, and $S$ be the event that the 3 specific students are separated.
We require $P(S \mid L) = \frac{n(S \cap L)}{n(L)}$.
From part (a), the number of ways the 2 leaders are separated is:
$n(L) = 14! - 2! \times 13! = 13!(14 - 2) = 12 \times 13!$
$\phantom{n(L)} = 12 \times (13 \times 12 \times 11 \times 10!) = 20592 \times 10!$

2. Calculate the total number of arrangements where the 3 specific students are separated (ignoring the leaders' condition).

Arrange the other $15 - 3 = 12$ students (including the leaders) in a circle:
Number of ways = $(12 - 1)! = 11!$
There are 12 spaces between these 12 students. We choose 3 spaces to insert the 3 specific students so they do not sit next to each other.
Number of ways to insert = $^{12}\mathrm{P}_3 = 12 \times 11 \times 10 = 1320$
$n(S) = 11! \times 1320 = (11 \times 10!) \times 1320 = 14520 \times 10!$

3. Calculate the number of arrangements where the 3 specific students are separated AND the 2 leaders are seated together.

Group the 2 leaders as 1 unit. There are 10 'other' students, giving $10 + 1 = 11$ entities to arrange in a circle.
Number of ways = $(11 - 1)! \times 2! = 10! \times 2$
There are 11 spaces between these entities. Choose 3 spaces to insert the 3 specific students.
Number of ways to insert = $^{11}\mathrm{P}_3 = 11 \times 10 \times 9 = 990$
$n(S \cap L') = (2 \times 10!) \times 990 = 1980 \times 10!$

4. Find the number of arrangements where BOTH conditions are satisfied (students separated and leaders separated).

Subtract the arrangements where leaders are together from the total arrangements where the students are separated:
$n(S \cap L) = n(S) - n(S \cap L')$
$\phantom{n(S \cap L)} = 14520 \times 10! - 1980 \times 10!$
$\phantom{n(S \cap L)} = 12540 \times 10!$

5. Calculate the final conditional probability.

$P(S \mid L) = \frac{n(S \cap L)}{n(L)}$
$\phantom{P(S \mid L)} = \frac{12540 \times 10!}{20592 \times 10!}$
$\phantom{P(S \mid L)} = \frac{12540}{20592} = \frac{95}{156}$

1. Find the total number of ways to divide the students into the two groups.

The team has $15$ students in total. We need to choose $7$ students to form the training program group (the remaining $8$ will naturally form the administrative matters group).

$\text{Total possible ways} = {}^{15}C_7 = 6435$

2. Find the number of ways where the trip leader and assistant leader are in their specific required groups.

For the trip leader to be in the training group and the assistant leader to be in the administrative group, their places are fixed.

We must select the remaining $6$ members for the training program group from the remaining $13$ unassigned students.

$\text{Number of favourable ways} = {}^{13}C_6 = 1716$

3. Calculate the required probability.

$\text{Probability} = \frac{\text{Number of favourable ways}}{\text{Total possible ways}} = \frac{1716}{6435} = \frac{4}{15}$