Two families plan to go on a hiking trip together. There are 10 people from the Lee family, consisting of 3 married couples, 3 single men and 1 single woman. There are 14 people from the Tan family, consisting of 4 married couples, 1 single man and 5 single women. Two people are randomly chosen to organise the hiking trip.

1. Determine the total number of people available to be chosen.

$\text{Total people} = 10 + 14 = 24$

2. Calculate the probability of selecting two people from the Lee family using the multiplication rule for dependent events.

$\text{P(both from Lee family)} = \frac{10}{24} \times \frac{9}{23} = \frac{15}{92} \text{ (shown)}$

1. Determine the number of men and women in each family.

A married couple consists of $1$ man and $1$ woman.
Lee family:
Men = $3$ married $+ 3$ single $= 6$
Women = $3$ married $+ 1$ single $= 4$
Tan family:
Men = $4$ married $+ 1$ single $= 5$
Women = $4$ married $+ 5$ single $= 9$
Total people $= 10 + 14 = 24$.

2. Calculate the probability of randomly selecting a man and a woman from the Lee family.

This selection can happen in two mutually exclusive orders: (Man, then Woman) or (Woman, then Man).
$P(\text{Lee man, Lee woman}) = \left(\frac{6}{24} \times \frac{4}{23}\right) + \left(\frac{4}{24} \times \frac{6}{23}\right) = \frac{48}{552}$

3. Calculate the probability for the Tan family and find the total combined probability.

$P(\text{Tan man, Tan woman}) = \left(\frac{5}{24} \times \frac{9}{23}\right) + \left(\frac{9}{24} \times \frac{5}{23}\right) = \frac{90}{552}$
The total probability is the sum of the probabilities of these mutually exclusive events:
$P(\text{man and woman from same family}) = \frac{48}{552} + \frac{90}{552} = \frac{138}{552} = \frac{1}{4}$

1. Calculate the probability of choosing a man and a woman from the same family (as required by part b).

Let $A$ be the event that a man and a woman from the same family are chosen.
From the family compositions, the Lee family has $3+3=6$ men and $3+1=4$ women, while the Tan family has $4+1=5$ men and $4+5=9$ women.
$P(A) = \frac{6 \times 4 + 5 \times 9}{^{24}C_2} = \frac{24 + 45}{276} = \frac{69}{276}$

2. Find the probability of choosing a married couple.

Let $B$ be the event that a married couple is chosen. Since all married couples consist of a man and a woman from the same family, $B$ is a subset of $A$, meaning $A \cap B = B$.
Total number of married couples = $3 \text{ (Lee)} + 4 \text{ (Tan)} = 7$.
$P(A \cap B) = P(B) = \frac{7}{276}$

3. Calculate the conditional probability.

$P(B \mid A) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{7}{276}}{\frac{69}{276}} = \frac{7}{69}$