A group of seven people consisting of two single men, three single women and one married couple are arranged to be seated in a row.

1. Find the total number of unrestricted seating arrangements.

Total number of ways to seat 7 people in a row = $7! = 5040$.

2. Find the number of arrangements where the married couple is seated together.

Treat the married couple as a single unit.
There are now $6$ units (5 individuals + 1 couple unit) to arrange in the row, which can be done in $6!$ ways.
Within their unit, the couple can be arranged in $2!$ ways.
Number of ways the couple is together = $6! \times 2! = 1440$.

3. Calculate the required probability.

$\text{Probability} = \frac{1440}{5040} = \frac{2}{7}$

1. Find the number of ways all 4 women are together AND the married couple is together.

For all 4 women (3 single women and the wife) to be together while the married couple is also together, the husband must sit next to his wife but cannot be positioned between any of the women.
Thus, the wife must be at either end of the block of 4 women, with the husband immediately beside her on the outside.
Possible configurations for this 5-person block: (Husband)(Wife)(3 Single Women) or (3 Single Women)(Wife)(Husband).
Internal arrangements for the 3 single women $= 3! = 6$.
So there are $2 \times 6 = 12$ ways to arrange this 5-person block internally.
Treating this block as 1 unit alongside the 2 single men gives 3 units, which can be arranged in $3! = 6$ ways.
Total ways for both conditions to be met $= 12 \times 6 = 72$.

2. Calculate the conditional probability.

The given condition that 'the married couple is seated together' defines the reduced sample space. Treating the married couple as 1 unit alongside the 5 other individuals gives 6 units to arrange.
Total ways the married couple is seated together $= 6! \times 2! = 1440$.

$P(\text{Women together} \mid \text{Couple together}) = \frac{72}{1440} = \frac{1}{20}$

1. Identify the total number of men and women and establish the required alternating pattern.

The group consists of 3 men (2 single, 1 married) and 4 women (3 single, 1 married).
For 3 men and 4 women to be seated alternately in a row of 7, the arrangement must start and end with a woman:
$\text{W - M - W - M - W - M - W}$

2. Calculate the number of ways to seat the men and women alternately.

The number of ways to arrange the 4 women in their positions is $4!$.
The number of ways to arrange the 3 men in their positions is $3!$.
Total number of alternating arrangements $= 4! \times 3! = 24 \times 6 = 144$.

3. Find the probability by dividing by the total unrestricted arrangements.

The total number of unrestricted arrangements for the 7 people is $7! = 5040$.
$\text{Probability} = \frac{144}{5040} = \frac{1}{35}$

1. Find the total number of ways to arrange the 7 people at a round table.

Total number of ways $= (7 - 1)! = 6! = 720$

2. Find the number of ways to select and arrange 2 men adjacent to the particular woman.

There are 3 men in total (2 single men $+ 1$ married man).
Number of ways to choose 2 men to sit next to the particular woman $= \binom{3}{2} = 3$
Number of ways to arrange these 2 men on either side of her $= 2! = 2$
Total ways to form this group of three $= 3 \times 2 = 6$

3. Find the number of favourable arrangements by treating the group as a single unit, and calculate the final probability.

Treating the group of 3 (man-woman-man) as 1 unit, there are 4 other people remaining, making a total of $1 + 4 = 5$ units to arrange in a circle.
Number of ways to arrange these 5 units $= (5 - 1)! = 4! = 24$
Total number of favourable arrangements $= 6 \times 24 = 144$
$\text{Probability} = \frac{144}{720} = \frac{1}{5}$