ComplementIf 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 5 footballgames”, then Acis the event “UNC wins less than 5 footballgames”.1We can represent this by a Venn diagram, as follows: AAcSFigure 1. Venn diagram for a single event A and its complementAc.2The 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) =56.3Disjoint 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.4For 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.5So 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=14.6The Law of Addition for Non-disjoint Events A BSFigure 3. Venn diagram for two disjoint events A and B.7In 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?8For 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.9Applying 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 Computing.10Independent EventsTwo events are said to be independent if the outcome of oneof them does not influence the other. For example, in sportingevents, the outcomes of different games are usually consideredindependent even though that may not be true in a completelystrict and literal sense.The multiplication rule for independent events says that if A andB are independent,P (A and B) = P (A) × P (B).11Example: A football pundit states that the probability that UNCwill beat NC State is 0.4, while the probability that UNC willbeat Duke is 0.8. What is the probability that1. UNC wins both games?2. UNC wins at least one game?3. UNC loses both games?12Solution:1. If A is the event “UNC beats State” and B is the event “UNCbeats Duke”, and if we assume these are independent events,then the probability of A and B is 0.4 × 0.8 = 0.32.2. Apply the Law of Addition:P (A or B) = P (A)+P (B)−P (A and B) = 0.4+0.8−0.32 = 0.88.3. Apply the Law of Complementary Events: “UNC loses bothgames” is the complement of “UNC wins at least one game”,so its probability is 1 − 0.88 = 0.12.13Warning:Don’t confuse the notions of “independent events” and “disjointevents”. Independence means that the outcome of one eventdoes not influence the outcome of the other. Disjoint meansthat if one event occurs then the other cannot occur — the veryopposite of independence!14Conditional ProbabilitiesConsider the example (page 218 of text, referring to the Wim-bledon tennis tournament),A: “Federer misses his first serve”B: “Federer misses his second serve”We are told that Federer misses his first serve 36% of the time,and that of all the times he misses his first serve, he also misseshis second serve 6% of the time.What, then, is the probability he has a double fault?Logically, the answer is 6% of 36%, or 0.06×0.36, which is about0.02.15Now let us rephrase this in the language of conditional probability.We are told that the event A occurs 36% of the time, or in otherwords P (A) = 0.36.We are also told that, given that A has occurred, the event Boccurs 6% of the time. This is written in probability notation asP (B | A) = 0.06.The left hand side is read as the probability of B given A. In thisparticular context, it would not make sense to talk about theprobability of B given Ac, though in other contexts, that wouldmake sense (e.g. free throws in basketball).16The law of multiplication for conditional probabilities saysP (A and B) = P (A) × P (B | A).Note that if we just interchange the role of A and B, we also getP (A and B) = P (B) × P (A | B).Finally, if A and B are independent, we get P (A | B) = P (A)and P (B | A) = P (B) — that formalizes what is meant by sayingthat the outcome of one event does not influence the outcomeof the other. But in that case, either of the last two formulasreduces toP (A and B) = P (A) × P (B)as in our earlier formulation of the multiplication rule for inde-pendent events.17Here is another (more complicated) example.Consider the game in which a player tosses a die twice, and wewant to calculate the probability that the total of the two tossesis at least 10. Define the eventsA: The first throw is a 6.B: The first throw is a 5.C: The first throw is a 4.D: The total of the two throws is at least 10.Note that if the first throw is less than 4, it’s impossible for thetotal to be 10