conditional probability

Conditional probability

$A,B$ are two events (outcomes) of a random experiment. The probability of an event $A$ given that another event $B$ has occurred is:

$$P(A|B):=\frac{P(A\cap B)}{P(B)},P(B)>0$$

If we know that $B$ has occured, $B$ becomes the new sample space, and we rescale so $B$ has area one ($÷P(B)$). Everything outside of $B$ is discarded, i.e. the only part of $A$ that survives is $P(A\cap B)$:
→
In this case, $A$ becomes a lot more likely, as it occupies a larger fraction of the sample space than it did before.
It’s easy to see geometrically that the reverse is not true (completely different proportions).

$P(A|B)\ne P(B|A)$

Two dice are rolled. $A$ … one dice is a 3. $B$ … the sum of the dice 6. What’s $P(A\mid B)$?

→ $P(A\mid B)=\frac{1}{5}$, since there’s one out of the 5 possible events (given C), where either of the dices is a 3.

What's $P(A|B)$, from first principles?

$$P(A|B):=\frac{P(A\cap B)}{P(B)},P(B)>0$$