If $P(A | B)=\frac{P(A\cap B )}{P(B)}$ and $P(A\cap B )=P(A)P(B)$ , then is not $P(A | B)=P(A)$? According to https://en.wikipedia.org/wiki/Conditional_probability#Kolmogorov_definition:
$$P(A | B)=\frac{P(A\cap B )}{P(B)}$$
and according to https://en.wikipedia.org/wiki/Probability#Independent_events:
$$P(A\cap B )=P(A)P(B)$$ 
Do not the two above just imply
$$P(A | B)=P(A) \quad ?$$ 
Looking at this example with the dice, how would you use the formulas to solve the final example
$$P(A=2 | A+B ≤ 5) ?$$
 A: As several comments point out, you are using a formula for independent events.
And you are proving
$$P(A|B)=P(A)$$
In English this is "The probability for A given B is the same as the overall probability for A."  And that is another way of saying that A is independent of B.
A: $P(A\cap B) = P(A) P(B) \iff$ $A$ and $B$ are independent.
Let $A = \{\text{Get Ace in first trial}\}, B = \{\text{Get an Ace in second trial}\}$.
Consider two scenarios.


*

*You roll a fair die twice. What is the probability of rolling an an Ace (One) twice? Since the rolls are independent, then
$$P(AB) = P(A)P(B) = \frac{1}{6}\cdot\frac{1}{6} = \frac{1}{36}.$$

*Draw two cards from a standard deck without replacement. What is that probability that you draw two Aces. Clearly, the second draw is dependent of the first. Thus
$$P(AB) = P(A)P(B|A) = \frac{4}{52}\cdot\frac{3}{51} = \frac{1}{221}.$$ 
As for the dice problem referenced in the post, refer to table 3.

Since you are told that the outcome space has been reduced to the outcomes such that $A+B\leq 5$, you are only left with the options in dark gray (10). However, only 3 of these cases have $A = 2$. Thus
$$P(A = 2| A+B\leq 5) = \frac{3}{10}.$$
A: $$P(A=2\mid A+B\leq5)=\frac{P(A=2\wedge A+B\leq5)}{P(A+B\leq5)}=\frac{P(A=2\wedge B\leq3)}{P(A+B\leq5)}$$
If $A$ and $B$ are independent then: $$P(A=2\wedge B\leq3)=P(A=2)P(B\leq3)$$
Now get to work, I would say.
A: $P(A \cap B) = P(A) P(B)$ is equivalent to $P(A \mid B) = P(A)$ so long as $P(A \mid B)$ is actually defined, which it isn't when $P(B) = 0$.  Since we could be dealing with events that have probability zero we prefer the more general definition $P(A \cap B) = P(A) P(B)$.
