# How conditional probability works

I have some understanding of how probabilities work. My question is: Is there a way to calculate the reunion, intersection or conditional of some probabilities without knowing the specific events of each one (eg. tossing a 2, getting an even number, getting an ace, etc.)?

So for example, it's easy to calculate $$P(A∩B)$$ for some independent events since it's $$P(A)\times P(B)$$, but for some non-independent ones? Same for $$P(A∪B)$$, $$P(A|B)$$, $$P(A|B∪C)$$.

• Look up "Bayes theorem." May 21, 2020 at 21:50
• $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. Other relations: you need to specify start. May 21, 2020 at 21:56
• Please see How to ask a good question May 21, 2020 at 21:58

$$P(A\cap B)=P(A|B)P(B)=P(B|A)P(A)$$. If $$P(A|B)=P(A)$$ and $$P(B|A)=P(B)$$, i.e. if $$A$$ and $$B$$ are independent, then $$P(A\cap B)=P(A)P(B)$$.
$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$. If $$P(A\cap B)=0$$, i.e. if $$A$$ and $$B$$ are mutually exclusive, then $$P(A\cup B)=P(A)+P(B)$$.
The probability of $$A$$ given $$B$$ is defined by $$P(A|B)=\frac{P(A\cap B)}{P(B)}$$.
Since $$P(A\cap B)=P(A|B)P(B)$$, the Bayes' rule follows: $$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$$
Replace $$B$$ with $$B\cup C$$ and you get: $$P(A|B\cup C)=\frac{P(B\cup C|A)P(A)}{P(B\cup C)}$$.