How do I find $p(E|F')$?

Question

Assume ' is equal to not or complement here.

Alright, you are given the following information:

p(E)= 1/3
p(F)=1/2
p(E|F)=2/5

You are asked to find p(F|E).

Bayes theorem is:

p(F|E)=p(E|F)p(F) divided by P(E|F)p(F)+p(E|F')p(F')

I know that p(E|F')= 1-p(E'|F') How do I find p(E|F')?

I know F'=F here.

Answer 1

$P(F\mid E) = \dfrac{P(E\mid F)\cdot P(F)}{P(E)} = \dfrac{\dfrac{2}{5}\cdot \dfrac{1}{2}}{\dfrac{1}{3}} = \dfrac{3}{5}$

Answer 2

Finding $\Pr(E|F')$, as shown by the answer of ah-huh-moment, not necessary in order to find $\Pr(F|E)$. That is the best possible answer to your question. But we show how to compute $\Pr(E|F')$ anyway, in case we need it for something else.

Note that $\Pr(E\cap F)=\Pr(E|F)\Pr(F)=\frac{1}{5}$.

But $\Pr(E)=\Pr(E\cap F)+\Pr(E\cap F')$. Thus $\Pr(E\cap F')=\frac{1}{3}-\frac{1}{5}=\frac{2}{15}$.

Thus $\Pr(E|F')=\frac{\Pr(E\cap F')}{\Pr(F')}=\frac{2/15}{1/2}$.

Remark: In my experience, students do better with elementary conditional probability problems if they stick to the basic definition $\Pr(A|B)=\frac{\Pr(A\cap B)}{\Pr(B)}$ than if they attempt to use Bayes' Formula.