Answer :
From the given table, we know tha the grand total is 132, then,
[tex]P(\text{Female }\cap Off\text{ campus)=}\frac{24}{132}=\frac{12}{66}=\frac{4}{22}=\frac{2}{11}[/tex]Similarly,
[tex]P(\text{Male }\cap On\text{ campus)=}\frac{52}{132}=\frac{26}{66}=\frac{13}{33}[/tex]On the other hand, we have
[tex]P(\text{offcampus}\cup Male)=P(offcampus)+P(male)-P(\text{offcampus}\cap Male)[/tex]from the given table, it yields
[tex]P(\text{offcampus}\cup Male)=\frac{13+24}{132}+\frac{52+13}{132}-\frac{13}{132}[/tex]which gives
[tex]\begin{gathered} P(\text{offcampus}\cup Male)=\frac{13+24+52+13-13}{132} \\ P(\text{offcampus}\cup Male)=\frac{89}{132} \end{gathered}[/tex]And finally,
[tex]P(on\text{campus}\cup female)=P(oncampus)+P(female)-P(\text{oncampus}\cap female)[/tex]which gives
[tex]\begin{gathered} P(on\text{campus}\cup female)=\frac{52+43}{132}+\frac{43+24}{132}-\frac{43}{132} \\ P(on\text{campus}\cup female)=\frac{52+43+43+24-43}{132} \\ P(on\text{campus}\cup female)=\frac{119}{132} \end{gathered}[/tex]Therefore, the answers are:
[tex]\begin{gathered} P(\text{Female }\cap Off\text{ campus)}=0.182 \\ P(\text{Male }\cap On\text{ campus)=}0.394 \\ P(\text{offcampus}\cup Male)=0.674 \\ P(on\text{campus}\cup female)=0.902 \end{gathered}[/tex]