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Do I have the proper understanding of Bayes theorem?

The difference between conditional probability and Bayes Theorem of P(A|B), is

For Conditional Probability, it is the probability of A given B $$ P(A|B) =\frac{ P(A \cap B)}{P(B)} $$

While for Bayes Theorem, we utilize "A" as a parameter with B unknown $$ P(A|B) = \frac{P(B|A)P(A)}{P(B)} $$

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I don't know what you mean by "utilize 'A' as a parameter with B unknown." In the example you give, it looks like we need to know quite a bit about both A and B in order to apply the theorem.

And the left-hand side of the equation in your Bayes' Theorem example, $P(A\mid B),$ is the exact same "A given B" as in your Conditional Probability example; Bayes' Theorem is a theorem about conditional probability.

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    For bayes theorem, for P(B), can't we apply law of total probability and equate P(B) as the summation of P(A)P(B|A) ? Thus, we wouldn't necessarily need to know event B, while in the conditional probability it is necessary to know event B. – 2017-02-13
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    Yes, we often do that, but knowing $P(B\mid A)$ is something you know about $B$; and in order to apply the theorem you must know enough such facts to add up to $P(B)$--so in effect you already know $P(B)$ (by total probability) before applying Bayes' Theorem. – 2017-02-13
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    We can also apply the law of total probability to compute $P(B)$ for the conditional probability, by the way, although if we are in the process of _defining_ conditional probability when we do this, we would have to use the $P(B\cap A_i)$ form of total probability rather than the $P(B\mid A_i)P(A_i)$ form. – 2017-02-13