Let's assume the following:
a) the conditional probability of $B$ given $A$ is 0.8
b) the conditional probability of $B$ given $\text{not }A$ is 0.4
c) the unconditional probability of $B$ is 0.5
What is the probability of $A$?
EDIT
I'm really struggling with this. Here is what I'm thinking and could really use some help.
$$P(B) - P(B|\text{not }A) \cdot P(\text{not }A) = P(B|A) \cdot P(A) \implies P(\text{not }A)=1-P(A)$$
Is this correct?
The law of total probability states $P(A)=\displaystyle \sum_{j=1}^k P(B_j)P(A|B_j)$ with $B_1,...,B_k$ a partition of the sample space.
Here, $A$ and $A^c$ form a partition of the sample space, and also A and B are switched around from the statement of the law. We have $P(B)=P(B|A)P(A)+P(B|A^c)P(A^c)$
$.5=.8P(A)+.4(1-P(A))$
$P(A)=.25$