Bayes Theorem application confusion The probability of a person NOT having Lyme disease is .99793  When tested for Lyme disease there is a false positive rate of .03 and a false negative rate of .063 .... find the probability that a person has Lyme disease given that the test result is positive.

I think we are looking for $\textbf{P(A|B)}$. I am confused on how to do that. Is Bayes Theorem required?
$$ P(A|B) = \frac{P(B)P(A|B)}{???}$$
Am I correct in saying that since $P(A) = P(A&B) +P(A&B')$, then $$P(A|B) = \frac{P(B)P(A|B)}{P(B)P(A|B) + P(B')P(A|B')}$$   I think I have been thinking about this wrongly. I have to think about the total probability of A happening. Is this correct logic? That is
\textbf{ The way I thought about it was that I had to divide by the total probability that A would happen. And that is the sum of P(A&B) and P(A&B')}
EDIT: PLEASE IGNORE THE $P(A|B) = .03$ and $P(A|B') = .06$ assignment. The image was from my early scratch work. 
 A: Given your conventions ($A \equiv$ lime disease, $A' \equiv$ not lime disease, $B \equiv$ positive test, $B' \equiv$ negative test), you're trying to find $P(A|B)$, as you correctly guessed. 
Bayes Theorem states:
$$
P(A|B) = \frac{P(B|A) P(A)}{P(B)}
$$
and the law of total probability (I forget the official name) states:
$$P(B) = \sum P(B|C_i) P(C_i)$$
where the $\left\{C_i\right\}$ form a set of mutually exclusive elements that span a complete probability space. In our case that is, 
$$
P(B) = P(B|A) P(A) + P(B|A') P(A')
$$
which is applies here, since $A$ and $A'$ describe a set of mutually exclusive scenarios that form the whole space. So the object you want to calculate is:
$$P(A|B) = \frac{P(B|A) P(A)}{P(B|A) P(A) + P(B|A') P(A')}.$$
We are given:
$P(A) = 0.00207$. That is, the probability of having lime disease. Of course this implies that $P(A') = 0.99793.$
Also, the false positive rate is 0.03. That means that, given that a person doesn't have lime disease, the probability of getting a positive test is 0.03. That is, $P(B|A') = 0.03$. 
The probability of getting a negative test, given that the patient does have lime disease is 0.063. That is, $P(B'|A) = 0.063$. Thus, the probability of getting a positive test for a lime disease patient (the only other possibility given the patient has lime disease) is $P(B|A) = 0.937$. 
