[tex]\mathbb P(E\cap F)=\mathbb P(E)+\mathbb P(F)-\mathbb P(E\cup F)=0.42+0.51-0.70=0.23[/tex]
[tex]\mathbb P(E\mid F)=\dfrac{\mathbb P(E\cap F)}{\mathbb P(F)}=\dfrac{0.23}{0.51}\approx0.45[/tex]
[tex]\mathbb P(F\mid E)=\dfrac{\mathbb P(F\cap E)}{\mathbb P(E)}=\dfrac{0.23}{0.42}\approx0.55[/tex]
[tex]E[/tex] and [tex]F[/tex] are not independent. Independence would require that [tex]\mathbb P(E\cap F)=\mathbb P(E)\times\mathbb P(F)[/tex], but [tex]0.42\times0.51\approx0.27\neq0.23[/tex].
