EventsAn event is technically defined to be any subset of the samplespace. Usually events are denoted by capital letters, A, B, etc.Two possible events are• A: UNC wins at least seven football games, and• B: In two throws of a die, the total is at least 10.Need to represent an event as a subset of the sample space.With UNC football games, if we define sample spaceS = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}then the event A isA = {7, 8, 9, 10, 11, 12}1With dice, if we represent sample space asS = {(1, 1), (1, 2), (1, 3), ...., (6, 6)}.then the subset that defines the event B isB = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}.2Defining ProbabilitiesIn the case of equally likely outcomes, the probability of the eventE is defined asP (E) =number of outcomes in event Enumber of outcomes in sample space S.For example, in the case of the event B defined above,P (B) =636=16.UNC football example: no reason to assume that all 12 possibleoutcomes are equally likely. However if we assigned them prob-abilities p0, p1, ..., p12(by whatever means) we can write casethatP (A) = p7+ p8+ p9+ p10+ p11+ p12.3The moral: We can build up probabilities for complicated eventsby starting with probabilities for elementary events and then ma-nipulating them according to the rules of probability. However,we still need to think carefully about those probabilities for ele-mentary events.4ComplementIf A is an event, then the complement of A, written Ac, meansall the possible outcomes that are not in A.For example, if A is the event “UNC wins at least 7 footballgames”, then Acis the event “UNC wins between 0 and 6 footballgames”.5We can represent this by a Venn diagram, as follows: AAcSFigure 1. Venn diagram for a single event A and its complementAc.6The Law of Complementary Events states thatP (Ac) = 1 − P (A).Example. In the game involving two throws of a die, if A is theevent “the total is 10 or greater”, then Acis the event “the totalis 9 or smaller”. We know P (A) =16, so P (Ac) =56.7Disjoint events, Intersection and UnionTwo events A and B are said to be disjoint if they cannot bothoccur. This is represented by the following Venn diagram: A BSFigure 2. Venn diagram for two disjoint events A and B.8For example, in two throws of a die, if A is the event “the totalis 10 or larger” and B is the event “the total is 3 or smaller”, itis clear that A and B cannot both be true, so they are disjointevents.With any two events A and B, we define the intersection of Aand B, also written “A and B”, to be the event that A and Bboth occur. With any two events A and B, we define the unionof A and B, also written “A or B”, to be the event that at leastone of A and B occurs.Note that in common English, if we say “A or B”, that’s oftentaken as excluding the possibility of both A and B occurring.In the language of probability, “A or B” always includes thepossibility that both A and B might occur — unless they aredisjoint, in which case it is impossible.9So another definition of disjoint events is: two events A and Bare disjoint if the intersection “A and B” is an impossible event.The law of addition for disjoint events states that: If two eventsA and B are disjoint, thenP (A or B) = P (A) + P (B).Example. Consider the toss of two dice where A is the event“the total is 10 or larger” and B is the event “the total is 3 orsmaller”. We have already seen that P (A) =16and it is easy tosee by similar reasoning that P (B) =112. ThereforeP (A or B) = P (A) + P (B) =16+112=14.We could also figure this out directly, by noting that the event “Aor B” consists of 9 outcomes of the sample space ((1,1),(1,2),(2,1)plus the 6 outcomes that comprise B) so the probability is936=14.10The Law of Addition for Non-disjoint Events A BSFigure 3. Venn diagram for two disjoint events A and B.11In this case the Law of Addition readsP (A or B) = P (A) + P (B) − P (A and B).Example. In a certain university, 52% of all students take astatistics class, 23% take a computing course, and 7% take both.What percentage of students take at least one of computing orstatistics?12For a randomly chosen student let A be the event “the studenttakes statistics”, and let B be the event “the student takescomputing”. The Venn diagram to represent this situation is: A BSFigure 4. Venn diagram for this problem.13Applying the Law of Addition,P (A or B) = 0.52 + 0.23 − 0.07 = 0.68.In other words, 68% of students take at least one of Statisticsor