Conditional probability and calculation if $P(F|E)=3P(F|E^c)$
$P(E)=0.3$
$P(F)=0.4$
Then $P(E|F)=??$
How do i use the conditional probability equation that has been given in the question
 A: From the first equality you wrote, 
$$P(F|E) = \frac{P(F\cap E)}{P(E)} = \frac{P(F\cap E)}{.3} = 3P(F|E^{c}) = 3\frac{P(F\cap E^{c})}{P(E^{c})} = 3\frac{P(F\cap E^{c})}{.7} $$
We can rearrange some terms to get 
$$P(F\cap E) = \frac{3*.3}{.7} P(F\cap E^{c}) = \frac{9}{7}P(F\cap E^{c})$$
Now, since $P(F) = P(F\cap E) + P(F\cap E^{c})$, substitute the above equality into this one to get
$$\frac{2}{5} = .4 = P(F) = P(F\cap E) + P(F\cap E^{c}) = \frac{9}{7} P(F\cap E^{c}) + P(F\cap E^{c}) = \frac{16}{7}P(F\cap E^{c})$$
Therefore $P(F\cap E^{c}) = \frac{\frac{2}{5}}{\frac{16}{7}} = \frac{14}{80} = \frac{7}{40}$. Now
$$P(F\cap E) = \frac{2}{5} - \frac{7}{40} = \frac{9}{40}$$
By definition, $P(E|F) = \frac{P(E\cap F)}{P(F)}$, which will be $\frac{\frac{9}{40}}{\frac{2}{5}} = \frac{45}{80} = \frac{9}{16}$
A: $P(E)=0.3$
$P(E^c)=1-0.3=0.7$
$P(F)=0.4$
$P(F|E)=3P(F|E^c)$
$P(F\cap E) = 3 \frac{P(F\cap E^c)}{P(E^c)}$
$P(F\cap E) = 3 \frac{P(F)−P(F\cap E)}{P(E^c)}$
Because $P(F\cap E^c)=P(F)−P(F\cap E)$
$P(F\cap E).P(E^c) = 3P(F)− 3P(F\cap E)$
$0.7P(F\cap E) + 3P(F\cap E) = 3×0.4$
$3.7P(F\cap E) = 1.2$
$P(F\cap E)=\frac{1.2}{3.7}$
$P(F\cap E)=0.324$
Hope you now solve for $P(E|F)$
