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I have three tables of information about $A$ and $B$ (gray cells, black font), their row and column marginal totals (black cells white font), and the grand total (white cell black font). The first two tables have information independent from one another, while the cells in third table in the picture are the sums of the two other tables. To their right I have their respective probability tables. I've used Bayes theorem on the first table to get $P(B|A) = 0.1267605634$.

Here is my work

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I have two questions:

  • If I use Bayes theorem on the second table to also compute $P(B|A)$, and if I want to use the information from table 1, would the correct prior be $0.1267605634$ or $0.32$ (i.e. do I simply assign the prior to table 1's posterior)?

  • If I use Bayes theorem on the third table alone (the combination of both tables) to compute $P(B|A)$, should this conditional probability be identical to the conditional $P(B|A)$ I computed for table 2? Why or why not?

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You didn’t need to use Bayes theorem to calculate P(B|A) for table 1 - you could’ve calculated it the same way you did P(A|B).

  • Probably would use Table 1’s P(B|A) and not P(B). Should probably use table 1’s posterior as the prior for Table 2. Even though the formula looks like it should be P(B), but that makes no sense to actually literally use table 1’s P(B).

  • I don’t think the posterior from the aggregate and the sequential posterior would be the same. I would be quite surprised if they are the same. Because getting P(B|A) from the data all at once is different than calculating the posterior from table 1 and using it as the prior for table 2. I imagine any two sets of two tables that have different entries would produce different results than the aggregate posterior probability. One is the aggregate, and the other does a chain of posterior probabilities.

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In this case, the objective is to derive the joint probability P(A and B) from the conditional P(A|B). Consequently, the correct prior term of the Bayesian numerator is P(B) and NOT the conditional P(B|A) because P(A|B) X P(B|A) ≠ P(A and B). Rather, the correct P(B) to use as the prior for table two is the sum of the updated posterior conditional probability, P(B), from table 1: P(B|A) + P(B|~A). As such, in this case, P(B|A)=0.09/0.71 = 0.13, and P(B|~A) = 0.23/0.29 = 0.79. So, given that A OR `A occurs, B has an occurrence probability of .13 + .79 = .92.

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