The Venn diagram explicitly tells us that the universe consists of 60 objects, and
[tex]P(D\cap E\cap F)=\dfrac1{60}[/tex]
[tex]P(D\cap E)=\dfrac{1+6}{60}=\dfrac7{60}[/tex]
[tex]P(D\cap F)=\dfrac{1+5}{60}=\dfrac1{10}[/tex]
[tex]P(E\cap F)=\dfrac{1+7}{60}=\dfrac2{15}[/tex]
[tex]P(D)=\dfrac{13+6+5+1}{60}=\dfrac5{12}[/tex]
[tex]P(E)=\dfrac{4+6+7+1}{60}=\dfrac3{10}[/tex]
[tex]P(F)=\dfrac{21+5+7+1}{60}=\dfrac{17}{30}[/tex]
[tex]P(D\cup E\cup F)=1-\dfrac3{60}=\dfrac{19}{20}[/tex]
By definition of conditional probability, we have
[tex]P(D\mid F)=\dfrac{P(D\cap F)}{P(F)}=\dfrac{\frac1{10}}{\frac{17}{30}}=\dfrac3{17}[/tex]
[tex]P(E\mid D)=\dfrac{\frac7{60}}{\frac5{12}}=\dfrac7{25}[/tex] (only this one is correct)
[tex]P(D\mid E)=\dfrac{\frac7{60}}{\frac3{10}}=\dfrac7{18}[/tex]
[tex]P(F\mid E)=\dfrac{\frac2{15}}{\frac3{10}}=\dfrac49[/tex]
[tex]P(E\mid F)=\dfrac{\frac2{15}}{\frac{17}{30}}=\dfrac4{17}[/tex]
