How to find the conditional probability of the following Can you please explain the math to find this? Similar problems, can you leave a link in the comments.

In a Venn diagram:
  $\mathsf (E)$ is $.2$, $\mathsf (F)$ is $.25$, $\mathsf P( E ∩ F)$ is $.2$, and the outside is $.35$
Find:
  
  
*
  
*$\mathsf P(E^\mathsf C \mid F)$ (in fraction form please!)
  
*$\mathsf P(F \mid E ∩ F)$
  
*$\mathsf P(E^\mathsf C \mid E)$
  

 A: The main maths involved is that by definition of conditional probability: $$\mathsf P(A\mid B) = \dfrac{\mathsf P(A\cap B)}{\mathsf P(B)}$$
Other than that, everything is just set algebra to establish what the particular $A, B, A\cap B$ and their measures are in respect to the four given values on your Venn Diagram.
A: You can use the conditional probability formula
$$P(A|B)=\frac{P(A \cap B)}{P(B)}$$
This essentially means we want to know what parts of $A$ are true when $B$ is true i.e. we are restricting the sample space.
Making small Venn diagrams & shading the appropriate areas can help with individual questions.
I also note that you probably meant $P(E)=0.4, P(F)=0.45 \,\&\, P(E\cap F)=0.25$ therefore giving you the values you stated above being shown in the appropriate part of the Venn diagram.


*

*$P(E^c|F)=\frac{P(E^c\cap F)}{P(F)}=\frac{0.25}{0.45}=\frac{5}{9}$

*$P(F|E\cap F) =1$
because we are restricting the sample space to where $E$ & $F$ are both true then $F$ will always be true $\Rightarrow$ the probability will be 1

*$P(E^c|E)=0$
given we are restricting the sample space to the event $E$ then it is impossible for $E^c$ to be true $\Rightarrow$ the probability will be 0.

