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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:

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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?

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  • $\begingroup$ What is $Bx$, $Ax$, and $Cx$? $\endgroup$ – OmG Nov 15 '19 at 13:47
  • $\begingroup$ The columns a=column1, b column2, etc.. x indicates the element per row. $\endgroup$ – havaey Nov 15 '19 at 13:54
  • $\begingroup$ Then your data in the last two rows is obviously wrong! It's simply impossible for P(B | A) * P(A) > P(B) $\endgroup$ – ManRow Nov 28 '19 at 11:45
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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.

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  • $\begingroup$ That explains why things go wrong for the last row but not why they go wrong in the second to last row, rtight? $\endgroup$ – havaey Nov 15 '19 at 13:58
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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!!

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