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Bayes Theorem: $$Pr(A|B) = \frac{Pr(A)*Pr(B|A)}{Pr(B)} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(0)$$

Furthermore: $$Pr(A) = Pr(A|B)*Pr(B) + Pr(A|

eg B)*Pr(|

eg B) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)$$

Scenario:

Suppose we are not given that an event $B$ has occured. Assuming we are given a base Probability of $A$: $(Pr(A_1)$ (this is our prior probability), subsequently, we are now given a $Pr(A|B_1)$, a $Pr(B_1)$, a $Pr(A|

eg B_1)$, and a $Pr(

eg B_1)$ as additional information. Assuming, that I now want to calculate a $Pr(A_2)$ that is the posterior probability in light of the new information.

From $(1)$: $$Pr(A_2) = Pr(A|B_1)\cdot Pr(B_1) + Pr(A|

eg B_1)\cdot Pr(

eg B_1)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)$$

From $(0)$: $Pr(A|B_1) = \frac{Pr(A_1)*Pr(B_1|A)}{Pr(B_1)} \,\,:\,\, Pr(A|

eg B_1) = \frac{Pr(A_1)*Pr(

eg B_1|A)}{Pr(

eg B_1)} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(3)$$

Rewriting $(2)$ in light of $(3)$ $$Pr(A_2) = \frac{Pr(A_1)*Pr(B_1|A)}{Pr(B_1)} \times Pr(B_1) + \frac{Pr(A_1)*Pr(

eg B_1|A)}{Pr(

eg B_1)} \times Pr(

eg B_1)$$

Which gives us: $$Pr(A_2) = Pr(A_1)*Pr(B_1|A) + Pr(A_1)*Pr(

eg B_1|A)$$ Simplifying to: $$Pr(A_2) = Pr(A_1)\cdot\left(Pr(B_1|A) +Pr(

eg B_1|A)\right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(4)$$

Generalising from $(4)$: $$Pr(A_{l+1}) = Pr(A_l)\cdot\left(Pr(B_l|A) +Pr(

eg B_l|A)\right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(5)$$

My Question: