(This is a section in the notes here.)
We want to calculate probabilities for different events. Events are sets of outcomes, and we recall that there are various ways of combining sets. The current section is a bit abstract but will become more useful for concrete calculations later.
Operations on Events.
Definition [Union] For events , the union of and is the set of outcomes that are in or . The union is written We can say “ or ” as well as “ union ”.
Definition [Intersection] For events , the intersections of and is the set of outcomes that are in both and . The union is written We can say “ and ” as well as “ intersection ”.
Definition [Complement] For event , the complement of is the set of outcomes in the sample space that are not in . We denote the complement of with1 We can say “not ”.
Definition [Relative Complement] For events , the relative complement of relative to is the set of outcomes not in that are in . The relative complement of relative to is written We can say “A not B”.
Warning! Later we will define to denote the probability of given . The line in is straight and whereas line in in the relative complement is not a straight. So be warned that, in general,
Here are a few facts about sets
Lemma 1. For events
i) ii) iii) iv) (De Morgan’s Laws) v) If then .
The proof of each statement above is quite straightforward. For each statement, we can draw a Venn diagram (i.e. we draw 1,2, or 3 intersecting circles and shade the appropriate areas to check the result.) For instance, the first statement in Lemma 1iii) can be seen in Figure 1, below.
From i) and ii) we see that it does not matter what order we apply unions and intersections. For that reason, when we apply a sequence of unions or a sequence of intersections we can unambiguously write
The statement iii) is analogous to the statement that .
In De Morgan’s Laws, iv), we can think of the complement of union being intersection. I.e. and . 2 Thus
Probability Rules for Operations on Sets
We can now start to think about what these operations on events imply for the probabilities of those events. Here are a sequence of lemma to this effect. (You can skip the proofs on first reading.)
Lemma 2. If and have not outcome in common, i.e. , then3
Lemma 2 can be extended to a countable set of sets with . So The above equality is actually an axiom of probability.
Lemma 3. If then
Proof. Using Lemma 1ii) and v), Using Lemma 2 and the fact probabilities are non-negative
Suppose I have a statement like “I have a red car” this implies “I have a car”. So the probability “I have a car” is more than the probability “I have a red car”. The above shows that this inequality holds for any two statements where one implies the other.
(An intuitive proof of this lemma: draw a Venn diagram with events and . Notice if we colour in and then colour in then we end up colouring twice. So if we want to count that region once, which we do for the event , we need to subtract once.)
Proof. Note that where as Subtracting from gives Rearranging the above expression gives the result.
It is often easy to calculate from and (see later discussion on independence). The above lemma gives a way to access from .
Proof. while . Also since probabilities sum to one (see Definition [probdef]). So Rearranging gives the required result.
Sometimes the number of outcomes in is large. So we may need to sum the probability of a lot of different outcomes to get . However, it may be that only has a few outcomes. So it may be simple to calculate . The above lemma then gives us a way to get .