Questioning on bayes's theorem

Question

Let $P(E) = 0.4$ and $P(F) = 0.7$, with $E$ and $F$ independent. How to calculate $P(F \mid (E^c \cup F^c))$?

So:$$\begin{array}{l}\mathrm{P}(F \mid E^c \cup F^c) \\ = \dfrac{\mathrm{P}(F \mid E^c \cup F^c)}{\mathrm{P}(E^c \cup F^c)} \\ = \dfrac{\mathrm{P}(F)\cdot\mathrm{P}(E^c \cup F^c \mid F)}{\mathrm{P}(F)\cdot\mathrm{P}(E^c \cup F^c\mid F)+\mathrm{P}({F^c})\cdot\mathrm{P}(E^c \cup F^c\mid\ {F^c})} \\ = \dfrac{\frac{7}{10}\cdot\frac{}{}}{\frac{7}{10}\cdot\frac{}{}+\frac{3}{10}\cdot\frac{}{}} \\ = \dfrac{}{}\end{array}$$

i know E compliment union F compliment is 0.72 but have no idea how to get rest of conditional probability. am i doing bayes right in the first place?

Answer 1

You are right, $\Pr(E^c\cup F^c)=0.72$.

By the definition of conditional probability, we want $$\frac{\Pr(F\cap (E^c\cup F^c))}{\Pr(E^c\cup F^c)}.$$

The top is just $\Pr(F\cap E^c)$, which by independence is $(0.7)(0.6)$. And you know the bottom.

Answer 2

Since it cannot happen that both $F$ and $F^c$ happen simultaneusly, $P(A\cup F^c|F) = P(E^c|F)$.

Also, as they are independent, $P(E^c|F) = P(E^c)$.

Thus, $$P(F|E^c\cup F^c) = P(F) P(E^c\cup F^c|F) / P(E^c\cup F^c) = P(F) P(E^c) / P(E^c\cup F^c) = P(F) \frac{P(E^c)}{1 - P(E\cap F)} = P(F) \frac{P(E^c)}{1 - P(E)P(F)}$$