Sample Space, Events and Probability Discussion.

Sample Space, Events and Probability Sample Space and Events There are lots of phenomena in nature, like tossing a coin or tossing a die, whose outcomes cannot be predicted with certainty in advance, but the set of all the possible outcomes is known.Sample Space, Events and Probability Discussion.These are what we call random phenomena or random experiments. Probability theory is concerned with such random phenomena or random experiments.

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Consider a random experiment. The set of all the possible outcomes is called the sample space of the experiment and is usually denoted by S. Any subset E of the sample space S is called an event. Here are some examples.Example 1Tossing a coin. The sample space is S={H, T}. E={H}is an event.Example 2Tossing a die. The sample space isS={1,2,3,4,5,6}.E={2,4,6}is an event,which can be described in words as ”the number is even”. Example 3Tossing a coin twice. The sample space isS={HH, HT, T H, T T}.E={HH, HT}is an event, which can be described in words as ”the first toss results in a Heads.Example 4Tossing a die twice. The sample space isS={(i, j) :i, j= 1,2, . . . ,6}, which contains36 elements. Sample Space, Events and Probability Discussion.”The sum of the results of the two toss is equal to 10” is an event.Example 5Choosing a point from the interval(0,1). The sample space is S= (0,1).E=(1/3,1/2)is an event.Example 6Measuring the lifetime of a light bulb. The sample space is S= [0,∞). E= [90,∞)is an event.Example 7Keeping on tossing a coin until one gets a Heads. The sample space of this experiment is S={H, T H, T T H, T T T H, . . .}.E={H, T H}is an event.Suppose that E is an event. We say that the event E”occurs” if the outcome of the experiment is contained in E.Since events are simply subsets of the sample space, we can talk about various set theoretic operations on events. In the following,E,F,G,Ei, = 1,2, . . .are events.E∪F denotes the union ofEandF.E∩F denotes the intersection of E and F.E c stands for the complement of E, that isE=SE.E⊂F means that E is a subset of F. IfE∩F=∅,we say that E and F are disjoint.Similarly, we can talk about the union and intersection of more than two events:∪ni=1Ei,∪∞i=1Ei,∩ni=1Ei,∩∞i=1Ei.1

Now we recall some properties of set theoretic operations:Commutativity:E∪F=F∪E, E∩F=F∩E.Associativity:(E∪F)∪G=E∪(F∪G),(E∩F)∩G=E∩(F∩G). Distributivity:(E∪F)∩G= (E∩G)∪(F∩G),(E∩F)∪G= (E∪G)∩(F∪G).De Morgan’s law:(∪ni=1Ei)c=∩ni=1Eci,(∩ni=1Ei)c=∪ni=1Eci,(∪∞i=1Ei)c=∩∞i=1Eci,(∩∞i=1Ei)c=∪∞i=1Eci.Axioms of Probability Consider an experiment with sample space S. A real-valued function P on the space of all events of the experiment is called a probability measure if(i) for all events E, 0≤P(E)≤1;(ii)P(S) = 1;(iii) for any sequence of eventsE1, E2, . . .which are mutually disjoint P(∪∞n=1En) =∞∑n=1P(En). Sample Space, Events and Probability Discussion.For any event E, we refer to P(E) as the probability of E.Here are some examples.Example 8.Tossing a fair coin. In this case, the probability measure is given byP(H) =P(T) =12.If the coin is not fair, the probability measure will be different.Example 9Tossing a fair die. In this case, the probability measure is given by P(1) =P(2) =···=P(6) =16. If the die is not fair, the probability measure will be different.Example 10Tossing a fair coin twice. In this case, the probability measure is given by P(HH) =P(HT) =P(T H) =P(T T) =14.Example 11Tossing a fair die twice. In this case, the probability measure is given by P((i, j)) =136, i, j= 1, . . . ,6.2

Example 12Tossing a point from{1, . . . , n}at random, that is each point is equally likely to be chosen. In this case, the probability measure is given by P(i) =1n, i= 1, . . . , n.Example 13Choosing a point from the interval(a, b)at random, that is each point is equally likely to be chosen. In this case, the probability measure is given byP((c, d)) =d−cb−a,for all interval(c, d)⊂(a, b).Sample Space, Events and Probability Discussion. Example 14Measuring the lifetime of a light bulb. Depending on the manufacturer, the probability measure will be different. One possible probability measure is given byP(E) =∫Ee−tdt,for anyE⊂[0,∞). Example 15Keeping on tossing a fair coin until one gets a Heads. In this case, the probability measure is given by P(H) =12,P(T H) =14,P(T T H) =18, . . ..The following result list some properties of probability measures.Theorem 16Suppose thatPis a probability measure. Then it satisfies the following properties.(i)P(∅) = 0.(ii) For any n≥2, ifE1, . . . , En are disjoint events, then P(∪ni=1Ei) =∑ni=1P(Ei).(iii) IfE⊂F, thenP(E)≤P(F).(iv) For any eventE,P(Ec) = 1−P(E).(v) For any events EandF,P(E∪F) =P(E) +P(F)−P(E∩F).(vi) For any n≥2and any eventsE1, . . . , En,P(∪ni=1Ei) =n∑i=1P(Ei)−∑i1<i2P(Ei1∩Ei2)+···+(−1)k+1∑i1<···<ik P(∩kj=1Eij)+···+(−1)n+1P(∩ni=1Ei). The last formula is called the inclusion-exclusion formula.Proof.We are only going to prove (i), (ii), (v). (vi) follows from (v) and the induction.To prove (i), takeE1=S,E2=E3=···=∅. ThenE1, E2, . . .is a sequence of disjoint events,so we have1 =P(S) =∞∑n=1P(En) = 1 +P(∅) +P(∅) +···. Consequently we have P(∅) = 0.3

To prove (ii), takeEn+1=En+2=···=∅, thenE1, E2, . . .is a sequence of disjoint events, so we have P(∪ni=1Ei) =P(∪∞i=1Ei) =∞∑i=1P(Ei) =n∑i=1P(Ei).To prove (v), notice that E= (E∩F)∪(E∩Fc),F= (E∩F)∪(F∩Ec) and E∪F=(E∩F)∪(E∩Fc)∪(F∩Ec). Hence P(E) =P(E∩F) +P(E∩Fc),P(F) =P(E∩F) +P(F∩Ec),P(E∪F) =P(E∩F) +P(E∩Fc) +P(F∩Ec).(v) follows immediately from the three identities above.2Example 17A fair die is tossed twice. Find the probability that the sum of the two results is even.Solution. Let E be the event that the sum of the two results is even. Fori= 2,4,···,12, let Ei be the event that the sum of the two results is i. ThenE2, E4,···, E12are disjoint and/E/is the/union ofE2, E4,···, E12. One can easily find the probability of each/Ei, adding them up, we get/P(E) =12.Example 18A fair die is tossed 100 times. Find the probability that there is at least one 5.Solution. Let E be the event that there is at least one 5. Then/Ecis the event that there is no5 and/P(Ec) = (5/6)100. Thus P(E) = 1−(5/6)100.Example 19Suppose that E and Fare two events. If we know the probabilities of/E,Fand E∩F, we can find the probability of any set theoretic combination of E and F. For instance, ifP(E) =1/2,P(F) =1/3andP(E∩F) =1/4, then P(E∪F) =P(E) +P(F)−P(E∩F) =1/2+1/3−1/4andP(E∩Fc) =P(E)−P(E∩F) =1/3−1/4. Sample Space, Events and Probability Discussion.