Probability of a device having anomaly type A

Question

$A$ = component has anomaly A

$B$ = component has anomaly B.

$P(B)=0.09$

$P(A|B) = 0.5$

$P(A|B^c) = 0.01$

I figured out that the probability of device having both anomalies is equal to $P(B \cap A) = P(B)*P(A|B)=0.045$

The next question is to figure out what $P(A)$ is.

I tried $P(A \cap B) = P(A)*P(B|A)$ but I only know $P(A \cap B)$ in this equation. I dont see how to apply what I know about $P(A|B^c)$

Answer 1

We have $\mathbb{P}(B^c) = 1 - \mathbb{P}(B) = 0.91$ , so you also know $\mathbb{P}(A \cap B^c) = 0.0091$ and $$ \mathbb{P}(A) = \mathbb{P}(A \cap B) + \mathbb{P}(A \cap B^c) = 0.0541 $$