$P(A|B)$ is the probability of $A$, given that $B$ has already occurred.
This is not the same as $\frac{P(A)}{P(B}$.
In fact $\displaystyle P(A|B) = \frac{P(A \cap B)}{P(B)}$
Given the phrasing about diseases and tests, I am quite confident by $P(A/B)$ they actually mean $P(A|B)$.
As to why conditional probability needs to be used here is an example.
Suppose there are 50 girls and 50 boys in a class. Out of the 50 boys, 30 have black hair and 20 are blonde. Out of the 50 girls, 40 are blonde and 10 have black hair.
Now I pick a random child, and see that the hair is black. What is the probability that it is a girl?
If you take the two events:
The probability that is being asked is the conditional probability $P(A | B)$.
If you try to reason it out, without any probability formulas, you can reason as:
There are 40 children with black hair. Out of them, 10 are girls.
Thus if the child I picked has black hair, the chance that it is a girl is $\frac{10}{40}$.
Basically, because the child you picked has black hair, your space of possibilities has reduced and you compute the probabilities based on that new information.
This is exactly what the formula $\displaystyle P(A|B) = \frac{P(A \cap B)}{P(B)}$ describes mathematically.
Here $P(B) = \frac{40}{100}$ and $P(A \cap B) = \frac{10}{100}$, and $P(A|B) = \frac{10}{40}$.