I try to use Bayes Theorem to calculate the probability of $P(A|B)$. I have $P(A)$ in column1, $P(B|A)$ in colmn2, $P(B)$ in column 3. I get the following:
my calculations were:
$$P(B/A) = 0.8\times A$$ $$P(B) = (Bx*0,55)+((1-Bx)*(0,55))$$ $$P(A/B) = (Ax*Bx)/Cx$$
The probability gets above 1. What am I doing wrong?
Well, then your data is clearly wrong!!
Bayes Theorem or not, you simply cannot have $$P\left(B\,|\,A \right) \times P\left(A\right) > P\left(B\right) $$ but this "impossibility" is exactly what happens in your last two rows!!
The problem is your data! For example, the last row shows that $\mathbb{P}(A) = 1$ and $\mathbb{P}(B|A) = 0.8$. If $\mathbb{P}(A) = 1$ means $A$ is equivalent to all possiblities of event world. Hence, $\mathbb{P}(B|A)$ couldn't be anything except 1.
I agree with @OmG that the table is wrong. However, $\mathbb{P}(B\mid A)$ must be equal to $\mathbb{P}(B)$ and not $1$.
Intuitively: as @OmG said, $\mathbb{P}(A)=1$ means that under $\mathbb{P}$ the event $A$ is not informative therefore you will not update the probability of $B$ after conditioning on $A$ and thus $\mathbb{P}(B\mid A)$ must be equal to $\mathbb{P}(B)$.
Formally: by definition of the conditional probability of an event, $$\mathbb{P}(B\mid A)=\frac{\mathbb{P}(B \cap A)}{\mathbb{P}(A)}=\mathbb{P}(B),$$ since $\mathbb{P}(A)=1$. Thus, $\mathbb{P}(B \cap A)=\mathbb{P}(B)$. It contradicts your table.
