(i) The committee must have either 4 men and 2 women or 5 men and 1 woman. The number of ways to choose 4 men from 10 and 2 women from 9 is \(\binom{10}{4} \times \binom{9}{2}\). The number of ways to choose 5 men from 10 and 1 woman from 9 is \(\binom{10}{5} \times \binom{9}{1}\). Therefore, the total number of committees is:

\(\binom{10}{4} \times \binom{9}{2} + \binom{10}{5} \times \binom{9}{1} = 210 \times 36 + 252 \times 9 = 7560 + 2268 = 9828\).

(ii) To find the probability that Albert and Tracey are both on the committee, we first choose 3 more men from the remaining 9 men and 1 more woman from the remaining 8 women. The number of ways to do this is \(\binom{9}{3} \times \binom{8}{1}\). The total number of committees is 9828, so the probability is:

\(\frac{\binom{9}{3} \times \binom{8}{1}}{9828} = \frac{84 \times 8}{9828} = \frac{672}{9828} = 0.0812\).

(iii) To find the number of committees that include either Albert or Tracey but not both, we calculate separately:

- Albert but not Tracey: Choose 3 more men from 9 and 2 women from 8: \(\binom{9}{3} \times \binom{8}{2}\).

- Tracey but not Albert: Choose 4 men from 9 and 1 more woman from 8: \(\binom{9}{4} \times \binom{8}{1}\).

The total number of such committees is:

\(\binom{9}{3} \times \binom{8}{2} + \binom{9}{4} \times \binom{8}{1} = 3360 + 1134 = 4494\).

(iv) The committee consists of 4 men and 2 women. The total number of ways to arrange 6 people is \(6!\). The number of ways to arrange the 4 men and 2 women such that the women are not next to each other is:

- Total arrangements: \(6! = 720\).

- Arrangements with women together: Treat the women as a single unit, so \(5! \times 2! = 240\).

- Arrangements with women not together: \(720 - 240 = 480\).

The probability is:

\(\frac{480}{720} = \frac{2}{3}\).