## anonymous 5 years ago If P(E) = 0.3, P(F) = 0.1, and P(E u F) = 0.4, what is P(E|F)? Round your answer to 3 decimal places and are E and F independent

$P(E \cup F)=P(F)+P(E)-P(E and F)$But$P(E|F)=\frac{P(E and F)}{P(F)}$ If you solve the first equation for P(E and F), and sub. into the latter, you get,$P(E|F)=\frac{P(E)+P(F)-P(E \cup F)}{P(F)}=\frac{0.3+0.1-0.4}{0.1}=0$