Solving Bayes' Theorem

Question

I am provided $P(A|B) = 0.980$, $P(B) = 0.0005$, $P(A'|B') = 0.987$ and I calculated $P(B') = 0.9995$. I am asked to calculate the overall probability of $P(A)$. I tried to solve it by rearranging the Bayes' Theorem formula but I was left with two unknowns e.g. $P(B|A)P(A) = 0.00049$

Answer 1

You have what you need.

$$P(A)=P((A\cap B) \cup (A\cap B')) = P(A\cap B) + P(A\cap B')$$

$$= P(A|B)P(B)+P(A|B')P(B')$$

and use

$P(A|B')=1-P(A'|B')$

Answer 2

Bayes theorem is basically: $$P(A\cap B)=P(A|B)P(B)=P(B|A)P(A)$$ The total probability of A, $P(A)$ is going to be the probability of A, given B, multiplied by the probability of B i.e. $P(A|B)P(B)$ plus the same but if B is not true, which translates to: $$P(A)=P(A|B)P(B)+P(A|B')P(B')$$ Now it is clear that: $$P(A|B')=1-P(A'|B')$$ and: $$P(B')=1-P(B)$$ so combining it all we get: $$P(A)=P(A|B)P(B)+[1-P(A'|B')][1-P(B)]$$ and these are all values you are given, so you are done :)