math: add more probability rules and slight refactor
This commit is contained in:
parent
01699e04bf
commit
61555fc51c
@ -174,11 +174,20 @@ $$P(A)=\frac{n(A)}{n(U)}$$
|
||||
As event $A$ must be a subset of all possible outcomes $U$, where $1$ indicates that the event always happens and $0$ the opposite:
|
||||
$$0\leq P(A)\leq 1$$
|
||||
|
||||
The complement of **event A** is the probability that it does not happen. It is written as $A^c$, $A'$, or $\pu{not } A$.
|
||||
The **complement** of event $A$ is the probability that it does not happen. It is written as $A^c$, $A'$, or $\pu{not } A$.
|
||||
$$P(A')=1-P(A)$$
|
||||
|
||||
Events $A$ and $B$ are disjoint if no outcomes between them are common and can never happen simultaneously. As such the probability of one of the events happening is equal to their sum.
|
||||
$$P(A\cup B)=P(A)+P(B)$$
|
||||
Events $A$ and $B$ are **disjoint** or mutually exclusive if no outcomes between them are common and can never happen simultaneously. As such the probability of one of the events happening is equal to their sum.
|
||||
$$
|
||||
P(A\cup B)=P(A)+P(B) \\
|
||||
P(A\cap B)=0
|
||||
$$
|
||||
|
||||
Events $A$ and $B$ are **exhaustive** if their union includes all possible outcomes in the sample space: $A\cup B=U$.
|
||||
$$P(A\cup B)=1$$
|
||||
|
||||
The **principle of inclusion and exclusion** forms a general rule for the union between two sets:
|
||||
$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$
|
||||
|
||||
## Resources
|
||||
|
||||
|
Loading…
Reference in New Issue
Block a user