Given 7 women and 10 men;
a) the top 3 are all men:
[tex]\begin{gathered} ways\text{ to choose 3 men out of 10 men is:} \\ 10C_3=\frac{10!}{(10-3)!3!} \\ \Rightarrow\frac{10!}{7!3!}=\frac{10\times9\times8\times7!}{7!\times3\times2\times1} \\ \Rightarrow\frac{10\times9\times8}{3\times2\times1}=120 \\ \text{ways to choose 3 men from 17 people(10men +7women) is:} \\ 17C_3=\frac{17!}{(17-3)!3!} \\ \Rightarrow\frac{17!}{14!\times3!}=\frac{17\times16\times15\times14!}{14!\times3\times2\times1} \\ \Rightarrow\frac{17\times16\times15}{3\times2\times1}=680 \end{gathered}[/tex]Therefore, the probability that the top 3 are all men is:
[tex]P_{all\text{ men}}=\frac{120}{680}=0.1765[/tex]b) the top 3 are all women:
[tex]\begin{gathered} \text{ways to choose 3 women from 7 women is:} \\ 7C_3=35 \\ \text{ways to choose 3 women from 17 people is:} \\ 17C_3=680 \end{gathered}[/tex]Therefore, the probability that the top 3 are all women is:
[tex]P_{\text{all women}}=\frac{35}{680}=0.0515[/tex]c) 2 men and 1 woman;
[tex]\begin{gathered} ways\text{ to choose 2 men out of 10 men is:} \\ 10C_2=45 \\ \text{ways to choose 1 woman from 7 women is:} \\ 7C_1=7 \\ \text{Thus, ways to choose 2 men and 1 woman }=45\times7=315 \end{gathered}[/tex]Therefore, the probability that the top 3 finishers are 2 men and 1 woman is:
[tex]P=\frac{315}{680}=0.4632[/tex]d) 1 man and 2 women;
[tex]\begin{gathered} \text{ways to choose 1 man from 10 men is;} \\ 10C_1=10 \\ \text{ways to choose 2 women from 7 women is:} \\ 7C_2=21 \\ \text{Thus, ways to choose 1 man and 2 women is 10}\times21=210 \end{gathered}[/tex]Therefore, the probability that the top 3 finishers are 1 man and 2 women is:
[tex]P=\frac{210}{680}=0.3088[/tex]