Blood Type Probability I'm given this table: 

I have to work out the numbers. I have only been able to pin down the easiest I guess.
I have $$P(A)=0,70$$ so $$P(\neg A)=0,30$$
the same way I have $$P(B)=0,50$$
thus $$P(\neg B)=0,50$$
as well. What I cannot get are the other numbers on the table, as I don't know how to since blood types aren't mutually exclusive or have the same probability (you can either have A, or B, or both, or none).
Can you please help me on how to figure out the rest of the table?
Okay, since there were missing elements, apparently these are the results I'm supposed to get, I just don't know how and none of the answers seemed to satisfy the results:

I also have this specific formula since the probabilities are non-exclusive and non-equiprobabilistic:
$$P(A ∪ B) = P(A) + P(B) - P(A ∩ B)$$
 A: I show you that this exercise has no unique solution provided that the table is a classic 2-way-contingency table without assuming the probabilities to be independent. At the end you see the solution for the case of independence.
Given the data the entries must satisfy the following:
$$\begin{matrix}
x & \color{blue}{0.7}-x & \color{blue}{0.7} \\
y & \color{green}{0.3}-y & \color{green}{0.3} = 1-\color{blue}{0.7} \\
x+y=\color{orange}{0.5} & \color{orange}{0.5} & 1
 \end{matrix}$$
So, 
$$\mbox{for any }0 \leq y\leq 0.3 \mbox{ set } x=0.5-y$$
you get a possible set of valid table entries.
For example
$$\color{red}{y=0.2} \mbox{ and } \color{red}{x= 0.3} \mbox{ or } y=0.1 \mbox{ and } x= 0.4$$
Conclusion:
So, you need more information. This could be, for example, the independence of the probablities in question.
Then, you would get:
$$\begin{matrix}
\color{orange}{0.5}\cdot\color{blue}{0.7} = 0.35 & 0.35 & \color{blue}{0.7} \\
\color{orange}{0.5}\cdot \color{green}{0.3} = 0.15 & 0.15 & \color{green}{0.3} \\
\color{orange}{0.5} & \color{orange}{0.5} & 1
 \end{matrix}$$
Update: (after OP edited post)
Your specific solution corresponds to one of the examples I gave (highlighted now in $\color{red}{red}$). 
Important is to understand the first part of my answer:
Without any further information there are infinitely many solutions to your problem.
A: The table just corresponds to a Venn diagram.

Since there is nothing else than $A$ and $B$ (and their complements), then just calculate the areas of the various combinations.
