If we have the probabilities of $P(A)$ and $P(A∣B)$, how can we calculate the probability of $P(A∣\lnot B)$?
The question statement is as follows:
A disease affects 6.5% of the population. There is however, an inheritance factor. If one’s father has the disease, the probability that the child will get the disease is 0.13. What is the probability a child will get the disease when the father does not have the disease?
How do we do this?
You can do the following \begin{align*} P(A)&=P(A|B)P(B)+P(A|\lnot B)P(\lnot B)\\ &=P(A|B)P(B)+P(A|\lnot B)(1-P(B)) \end{align*} So that \begin{align*} P(A|\lnot B) &= \frac{P(A)-P(A|B)P(B)}{1-P(B)} \end{align*} The problem is that from $P(A)$ and $P(A|B)$ only we cannot determine $P(B)$.