In our example of a coin flip, where S={H,T}, we can list out all possible events (all subsets of S):
- {H}
- {T}
- {H,T}
- ∅ (null set/null event)
We can do the same with our die roll example, S={1,2,3,4,5,6}:
- ∅
- {1},{2},{3}...{6}
- {1,2},{1,3}...{1,6},{2,3}...{6,6}
- ...
- {1,2,3,4,5,6}
All of these events can be assigned probability values, as we previously discussed.
The key thing to remember about probability values is that they represents frequency over the long run – in other words, it’s the average amount of times that an event will occur if an experiment is repeated many times.
What is the probability that…
- …event A doesn’t happen?
- …both event A and event B happen?
- …either event A or B happens?
These questions represent new events that are generated from information about other events.
Definition 1. A′ or AC is called the complement of A – the event where A doesn’t happen.
Definition 2. A∩B is called the intersection of A and B – the event where both A and B happen.
Definition 3. A∪B is called the union of A and B – the event where either A or B happens.
Thus our three questions can be represented by P(A′), P(A∩B), and P(A∪B).
Because these operations give new events, we can chain multiple of them together: P(A∩B∩C) is the probability that A, B, and C all occur.
Definition. When A∩B=∅, A and B are said to be mutually exclusive (or disjoint) events. This implies that it is not possible for A and B to both occur in a given experiment.
- P(A)≥0 for all events A.
- P(S)=1 for all sample spaces S.
- If A1,A2,... is an infinite sequence of disjoint events, then P(A1∪A2∪...)=∑i=1∞P(Ai).
We can derive all other properties of probability from these three axioms.
- P(∅)=0.
- If A1,A2...An is a finite sequence of disjoint events, then P(A1∪A2∪...∪An)=∑i=1nP(Ai). (Finite version of 3rd axiom)
- P(A)+P(A′)=1.
- A∩A′=∅
- A∪A′=S
- P(A)≤1.
- P(A∪B)=P(A)+P(B)−P(A∩B). (You need to subtract away the probability of events in A AND B so they don’t get double counted)
- P(A∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)−P(B∩C)−P(A∩C)+P(A∩B∩C). (Same reasoning as above)