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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
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\[P(E \cup F)=P(F)+P(E)-P(E and F)\]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,\[P(E|F)=\frac{P(E)+P(F)-P(E \cup F)}{P(F)}=\frac{0.3+0.1-0.4}{0.1}=0\]
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