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Events An event is technically defined to be any subset of the sample space. 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 space S = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} then the event A is A = {7, 8, 9, 10, 11, 12} 1 With dice, if we represent sample space as S = {(1, 1), (1, 2), (1, 3), ...., (6, 6)}. then the subset that defines the event B is B = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}. 2 Defining Probabilities In the case of equally likely outcomes, the probability of the event E is defined as number of outcomes in event E . P (E) = number of outcomes in sample space S For example, in the case of the event B defined above, P (B) = 6 1 = . 36 6 UNC football example: no reason to assume that all 12 possible outcomes are equally likely. However if we assigned them probabilities p0, p1, ..., p12 (by whatever means) we can write case that P (A) = p7 + p8 + p9 + p10 + p11 + p12. 3 The moral: We can build up probabilities for complicated events by starting with probabilities for elementary events and then manipulating them according to the rules of probability. However, we still need to think carefully about those probabilities for elementary events. 4 Complement If A is an event, then the complement of A, written Ac, means all the possible outcomes that are not in A. For example, if A is the event “UNC wins at least 7 football games”, then Ac is the event “UNC wins between 0 and 6 football games”. 5 We can represent this by a Venn diagram, as follows: S A c A Figure 1. Venn diagram for a single event A and its complement Ac . 6 The Law of Complementary Events states that P (Ac) = 1 − P (A). Example. In the game involving two throws of a die, if A is the event “the total is 10 or greater”, then Ac is the event “the total 1 , so P (Ac ) = 5 . is 9 or smaller”. We know P (A) = 6 6 7 Disjoint events, Intersection and Union Two events A and B are said to be disjoint if they cannot both occur. This is represented by the following Venn diagram: S A B Figure 2. Venn diagram for two disjoint events A and B. 8 For example, in two throws of a die, if A is the event “the total is 10 or larger” and B is the event “the total is 3 or smaller”, it is clear that A and B cannot both be true, so they are disjoint events. With any two events A and B, we define the intersection of A and B, also written “A and B”, to be the event that A and B both occur. With any two events A and B, we define the union of A and B, also written “A or B”, to be the event that at least one of A and B occurs. Note that in common English, if we say “A or B”, that’s often taken as excluding the possibility of both A and B occurring. In the language of probability, “A or B” always includes the possibility that both A and B might occur — unless they are disjoint, in which case it is impossible. 9 So another definition of disjoint events is: two events A and B are disjoint if the intersection “A and B” is an impossible event. The law of addition for disjoint events states that: If two events A and B are disjoint, then P (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 or 1 and it is easy to smaller”. We have already seen that P (A) = 6 1 . Therefore see by similar reasoning that P (B) = 12 1 1 1 P (A or B) = P (A) + P (B) = + = . 6 12 4 We could also figure this out directly, by noting that the event “A or B” consists of 9 outcomes of the sample space ((1,1),(1,2),(2,1) 9 = 1. plus the 6 outcomes that comprise B) so the probability is 36 4 10 The Law of Addition for Non-disjoint Events S A B Figure 3. Venn diagram for two disjoint events A and B. 11 In this case the Law of Addition reads P (A or B) = P (A) + P (B) − P (A and B). Example. In a certain university, 52% of all students take a statistics class, 23% take a computing course, and 7% take both. What percentage of students take at least one of computing or statistics? 12 For a randomly chosen student let A be the event “the student takes statistics”, and let B be the event “the student takes computing”. The Venn diagram to represent this situation is: S A B Figure 4. Venn diagram for this problem. 13 Applying 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 Statistics or Computing. 14