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 $P(A∪B)$, $P(A|B)$, $P(A|B∪C)$.

  • 1
    $\begingroup$ Look up "Bayes theorem." $\endgroup$ May 21, 2020 at 21:50
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    $\begingroup$ $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. Other relations: you need to specify start. $\endgroup$ May 21, 2020 at 21:56
  • $\begingroup$ Please see How to ask a good question $\endgroup$
    – amWhy
    May 21, 2020 at 21:58

1 Answer 1


$P(A\cap B)=P(A|B)P(B)=P(B|A)P(A)$. 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: $P(A|B\cup C)=\frac{P(B\cup C|A)P(A)}{P(B\cup C)}$.


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