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Elementary Probability by James Bernhard Suppose a coin is equally likely to land heads or tails If we flip this coin twice, there are four possible outcomes: HH, HT, TH, and TT This can be depicted graphically as follows. . . Possible Outcomes for 2 Coin Tosses HT HH TH TT The pie is divided into 4 slices (in no particular order) since there are 4 possible outcomes The slices are equal sizes since each outcome is equally likely Sample Space for 2 Coin Tosses HT HH TH TT The whole pie can be thought of as depicting the collection of all possible outcomes, which is called the sample space Outcomes and collections are outcomes are called events, and can be depicted similarly The Event HH or TH The Event TT HT HH HT HH TH TT TH TT The Event HT, TH or TT The Event HH, HT, TH, or TT HT HH HT HH TH TT TH TT The probability of an event can be thought of as the fraction of the time that the event would occur in the long run over many repetitions of the process Graphically, the probability of an event can be depicted as the fraction of the pie which the event occupies From either point of view, probability is a number between 0 and 1 In addition to being shown by the areas, probabilities of individual outcomes can also be written into the picture explicitly as well. . . Sample Space for 2 Coin Tosses with Probabilities HT (0.25) HH (0.25) TH (0.25) TT (0.25) Since the event that is the whole sample space takes up the whole pie, its probability must always be 1 HT (0.25) HH (0.25) TH (0.25) TT (0.25) In this case, .25 + .25 + .25 + .25 = 1 The complement of an event is the collection of all outcomes not included in that event The Event HH, HT, or TH The Complement of HH, HT, or TH HT (0.25) HH (0.25) HT (0.25) HH (0.25) TH (0.25) TT (0.25) TH (0.25) TT (0.25) The complement of an event is the collection of all outcomes not included in that event The Event HH, HT, or TH The Complement of HH, HT, or TH HT (0.25) HH (0.25) HT (0.25) HH (0.25) TH (0.25) TT (0.25) TH (0.25) TT (0.25) The probability of an event’s complement equals 1 minus the probability of the event Two events are disjoint if they do not overlap in the sample space Disjoint: HT (0.25) HH (0.25) HT (0.25) HH (0.25) TH (0.25) TT (0.25) TH (0.25) TT (0.25) HT (0.25) HH (0.25) HT (0.25) HH (0.25) TH (0.25) TT (0.25) TH (0.25) TT (0.25) Not disjoint: For two disjoint events, the probability of one or the other is the sum of the probabilities of the individual events HT (0.25) HH (0.25) HT (0.25) HH (0.25) TH (0.25) TT (0.25) TH (0.25) TT (0.25) In this case, .5 + .25 = .75, as combining them shows: HT (0.25) HH (0.25) TH (0.25) TT (0.25) In general, if a random process has n equally likely outcomes, each one will have a probability of 1/n (shown here for n = 8, with 1/8 = .125): Event 3 (0.125) Event 2 (0.125) Event 4 (0.125) Event 1 (0.125) Event 5 (0.125) Event 8 (0.125) Event 6 (0.125) Event 7 (0.125) In this case, an event made up of several outcomes will have probability equal to number of outcomes in the event number of outcomes in the sample space Event 3 (0.125) Event 2 (0.125) Event 4 (0.125) Event 1 (0.125) Event 5 (0.125) Event 8 (0.125) Event 6 (0.125) Event 7 (0.125) To keep things simple, let’s return to the example of tossing 2 coins Sample Space for 2 Coin Tosses HT (0.25) HH (0.25) TH (0.25) TT (0.25) Two events in this sample space are: Event A The 2 tosses come up different from each other. Event B The second toss comes up heads. Event A Event B HT (0.25) HH (0.25) HT (0.25) HH (0.25) TH (0.25) TT (0.25) TH (0.25) TT (0.25) Event A Event B HT (0.25) HH (0.25) HT (0.25) HH (0.25) TH (0.25) TT (0.25) TH (0.25) TT (0.25) What is the probability that Events A and B both occur? This can be seen graphically by looking at their intersection Events A and B HT (0.25) HH (0.25) TH (0.25) TT (0.25) In order to compute the probability of both Events A and B occurring, it is useful to understand conditional probability The probability of Event B, given the information that Event A has occurred, is called the conditional probability of B given A, written P(B|A) In the picture, P(B|A) is given by the area of B that lies within A, expressed as a fraction of the area of A, so P(B|A) = P(A and B) Area of the intersection of A and B = Area of A P(A) HT (0.25) HH (0.25) TH (0.25) TT (0.25) In this case, P(B|A) = .25 = .5 .5 In general, if we want to know the probability that both Events A and B will occur, we have: P(A and B) = Area of the intersection of A and B = (Area of A) · P(B|A) = P(A) P(B|A) That is, P(A and B) = P(A) P(B|A) Geometrically, this says that the area of the intersection of A and B equals the area of A times the fraction of B that is in A Now assume that Events A and B are any two events with nonzero probabilities Events A and B are called independent if P(B|A) = P(B), which means geometrically that the fraction of B that is in A equals the fraction of B in the sample space (This turns out to be the same as the condition P(A|B) = P(A)) In other words, events are independent if knowing that one of them is true tells you nothing new about the probability of the other one For independent Events A and B, the formula for the probability that both will happen simplifies to P(A and B) = P(A) P(B) Geometrically, this simplification is due to the fact that the fraction of B that is in A equals the fraction of B that is in the sample space