Question

Given: p(e) = 0.42, p(f) = 0.51, and p(e ∪ f) = 0.70 part

Answer 1

$\mathbb P(E\cap F)=\mathbb P(E)+\mathbb P(F)-\mathbb P(E\cup F)=0.42+0.51-0.70=0.23$

$\mathbb P(E\mid F)=\dfrac{\mathbb P(E\cap F)}{\mathbb P(F)}=\dfrac{0.23}{0.51}\approx0.45$

$\mathbb P(F\mid E)=\dfrac{\mathbb P(F\cap E)}{\mathbb P(E)}=\dfrac{0.23}{0.42}\approx0.55$

$E$ and $F$ are not independent. Independence would require that $\mathbb P(E\cap F)=\mathbb P(E)\times\mathbb P(F)$, but $0.42\times0.51\approx0.27\neq0.23$.