Condintional Probaility: Bayes' Theorem vs. Kolmogorov's definition I always seem to get stuck on probability problems where I can't identify whether or not I should be using Bayes' theorem, or Kolmogorov's definition. 
That is, for a given problem, how can I tell whether I should solve it using this definition:
$$
P(A|B) = \frac{P(AB)}{P(B)}
$$
Or Bayes':
$$
P(A|B) = \frac{P(B|A)P(A)}{P(B)}
$$
I know that the two are equal to each other - it's just that some solutions immediately start with Bayes' and others start with Kolmogorov's.
 A: I'll give examples of each and then say what I think is different about them.
Bayes:
Suppose 1 in 1000 people has a certain disease, and we have a test for the disease which always notices the disease if present but which has a 1% false positive rate. What's the chance someone has the disease given that the test comes back positive?
Answer: $$\begin{align*}
P(\text{ill}|\text{positive test})&=\frac{P(\text{positive test}|\text{ill}).P(\text{ill})}{P(\text{positive test}|\text{ill}).P(\text{ill})+P(\text{positive test}|\text{well}).P(\text{well})}\\
&=\frac{1.\frac{1}{1000}}{1.\frac{1}{1000}+\frac{1}{100}.\frac{999}{1000}}=\frac{100}{1099}
\end{align*}$$
Kolmogorov:
What's the probability that a 6 sided die comes up on a prime number given that it comes up on an odd number?
Answer: $P(\text{odd})=3/6$, $P(\text{odd and prime})=2/6$ so $P(\text{prime}|\text{odd})=\frac{2/6}{3/6}=2/3$.
In the Bayes example the two propositions can naturally be thought of as hypothesis and data. The hypothesis is whether or not the person is ill, and the data is the test. The important thing is that if we knew which hypothesis was correct, we would know just how likely each bit of data was. That is, we know the conditional probabilities $P(\text{positive test}|\text{ill})$ and $P(\text{positive test}|\text{well})$. Then we use Bayes theorem to update our probability for the hypothesis when we get the data.
In the Kolmogorov example the things we care about don't obviously split up as hypothesis and data. The proposition "X is prime" is approximately of the same type as "X is odd". We aren't told the conditional probabilities for one given the other. But we can easily work out the (non-conditional) probabilities for various outcomes just by counting the number of die faces where that outcome is satisfied. Note how the right hand side of Kolmogorov's formula doesn't have any conditional probabilities in it. What Kolmogorov does is tell you the proportion of the die's faces which are prime out of those which are odd.
A: If you accept the first definition, then $P(B|A)=\frac{P(AB)}{P(A)}$ and so $P(AB)=P(B|A)P(A)$. 
