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Statistik topic6 events and probability


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Statistik topic6 events and probability

  1. 1. Topic 6 Events and Probability LEARNING OUTCOMES By the end of this topic, you should be able to 1. explain the concept of event and its occurrence, 2. describe the relationship between two or more events via set’s operation, 3. calculate probability of a given event; and 4. estimate conditional probability. INTRODUCTIONIn this topic we will discuss about events and their probability measures. Someimportant aspects about events to be considered are the: occurrence of events; relationship with other events; and probability of their occurrence.An event is an occurrence that can be observed and its outcome can be recorded.Examples of events are:(a) attendance in school(b) school holidays(c) return to village(d) return to village during school holidays(e) buying stock when the price is low, etc.An occurrence of an event can be measured in two ways depending on how weaddress it. Let an event be “Ahmad passes the final examination”. One way to
  2. 2. 80 TOPIC 6 EVENTS AND PROBABILITYaddress it is by asking him “what is your chance to pass your final examination?”.Another way to measure the occurrence of the above event is by raising questionsuch as “What is the probability that Ahmad can pass the final examination?”. Thenumerical measure for the first method is in the interval of 0% to 100%(inclusive), whereas for the second method the value is between 0 and 1. We willdiscuss these aspects in this topic. Can you define two events from daily activities and try to address it by the two methods, mentioned above? 6.1 EVENTS AND THEIR OCCURANCESWe can see or observe events occurring everyday or even from multimedia news.Following to the events listed in the introduction section, we may introducefollowing additional events:(f) Die/pass way;(g) Cat having horns;(h) Wife wants to enter university if husband gets promotion;(i) Ahmad attends English for SPM in the year 2005 at School A; and(j) Ahmad attends English for SPM in the year 2005 at School B.Various events mentioned earlier can fall between two extreme categories ofevents i.e. the impossible event on the far left hand such as “cat having horns”and the sure (or eventual to occur) event far on the right hand such as “die/passway”. Events (a), (b), (c), and (i) are ordinary events that can occur on their own.Events (d), (e), and (h) are examples of conditional events. The event (d) can berewritten as going back to village if there is a school holiday comprises of event“going back to village” provided there is another event “school holiday”. Thismean that the event “going back to village” can only occur after the other event“school holiday” has occurred. We call this type of event as Conditional Event.Can you explain in the same manner the event (h)? The events (i) and (j) areexamples of two events that cannot occur at the same time.
  3. 3. TOPIC 6 EVENTS AND PROBABILITY 81 (a) Give an example of one impossible event and one very sure event. (b) Define one conditional event. (c) Give two events that cannot occur at the same time. 6.2 EXPERIMENT AND SAMPLE SPACEIn the context of probability, an experiment can be considered as consisting of asequence of trials where each trial possesses its own possible outcomes. The totaloutcomes of an experiment will then be the combination of trials outcomes. Thisset of total possible outcomes is called sample space. An event is defined assubset of sample space.The following are Examples of experiments:(a) throwing Malaysian Coin once,(b) throwing an ordinary dice once,(c) throwing Malaysian Coin twice,(d) throwing Malaysian Coin followed by throwing an ordinary dice.Throwing Malaysian coin is called a trial whose possible outcome is “G, whichstands for Parliament Picture”, and another possible outcome is “N, which standsfor the number on the coin”. Operationally for this trial, if you get ‘G’ on a throw,then you still need at least one more throw to get ‘N’ in order to complete allpossible outcomes of this trial. If it happens that the second throw also obtain “G”(repeat the previous outcome) then discard the throw, and carry on another throwand so forth until you get “N”. Then all possible outcomes of this trial are {G, N}.As a set, this is also equivalent to {N, G} depending how you record the outcomeof each throw.Similarly, throwing an ordinary dice is another example of a trial whose possibleoutcomes are {1, 2, 3, 4, 5, 6}. The numbers 1, 2, 3, 4, 5, 6 are actually thenumber of dots on the respective face of the ordinary dice. So on any singlethrow, you may get either one of these numbers and you need at least six throwsto get all possible outcomes of this trial.In the above examples, (a) and (b) are single trial experiments; (c) is repeatedtrials experiment. The experiment (d) is multiple trials experiment where the
  4. 4. 82 TOPIC 6 EVENTS AND PROBABILITYfirst trial is “throwing a coin” follow by “throwing an ordinary dice” as the secondtrial.Sample space (denoted by S) of single trial experiment is the same as the possibleoutcomes of the trial. Thus the sample space for experiment (a) is S = {G, N},and for experiment (b) is S = {1, 2, 3, 4, 5, 6}. Table 6.1 bellow shows howpossible outcomes are recorded according to the result of each throw. Normal rulein a set theory says that order of elements of a set is not important, therefore S1S2 S3 S = {1, 2, 3, 4, 5, 6}. Table 6.1 Recorder Results Recorded Per Throw Sample SpaceAhmad (8 throws), 2 repeated 6, 2, 4, 3, 5, 1 S1 = {6, 2, 4, 3, 5, 1}Yap (10 throws),4 repeated 4, 2, 1, 3, 5, 6 S2 = {4, 2, 1, 3, 5, 6}Samy (6 throws), no repeated 6, 5, 1, 3, 4, 2 S3 = {6, 5, 1, 3, 4, 2}A tree diagram is commonly used to obtain sample space of repeated or multipletrials experiment. We have the following diagram for Experiment (b): Figure 6.1The followings are steps of drawing a tree diagram:(a) List all possible outcomes of each trial involved
  5. 5. TOPIC 6 EVENTS AND PROBABILITY 83(b) Visualise the whole operation of the experiment which comprises of first, second, third and so on of trials involved(c) Start by drawing the first branch which represents an outcome of the first trial, it is immediately followed by a branch representing an outcome of the second trial, third trial and so on.(d) Then go back to the original point and start to draw a branch for another outcome of the first trial, follow through for the rest of the trials involved.To draw Figure 6.1, we know that only two branches for the first trial and thesecond trial. Let us draw the Branch ‘G’ of the first trial, and immediately drawBranch ‘G’, and Branch ‘N’ of the second trial. Now for step (d), go back to theorigin and draw Branch ‘N’ representing the only other outcome of the first trial,it is followed immediately by Branch ‘G’ and Branch ‘N’ of the second trial.Connection of first trial and branch of the second trial will produce combinationof first trial outcomes and the second trial outcomes. Thus, we have a totalcollection of combined outcomes GN, GG, NG, and NN. Therefore, the samplespace of this experiment is S = {GG, GN, NG, NN}. You can draw a tree diagramfor experiment (d) where the first trial which is “throwing coin” has two branches,one branch for ‘G’ and one for ‘N’. The second trial which is “throwing ordinarydice” has six branches, one for each outcome 1, 2, 3, 4, 5, 6. Thus the samplespace for Experiment (d) is S = {G1, G2, G3, G4, G5, G6, N1, N2, N3, N4, N5,N6}. 1. If we change the sequence of trials in Experiment (d), draw the suitable tree diagram and obtain the sample space. 2. Refering to the sample space of Experiment (c), can you interpret the outcome NG. Do you think outcome GN = NG? Explain why?
  6. 6. 84 TOPIC 6 EVENTS AND PROBABILITYExample 6.1A fair Malaysian coin is tossed three consecutive times. Obtain the sample spaceof this experiment.SolutionThis experiment is of the type repeated trials where the trial is “tossing a fairMalaysian coin”. Actually it is an extension of Experiment (c) given above. Wecan extend easily Figure 6.1 to accommodate outcomes of the third toss coin. Thepossible sample space will be: S = {GNG, GNN, GGG, GGN, NGN, NGG, NNN, NNG}.As has been explained earlier, sample space S is just a universal set of all possibleoutcomes of an experiment. As a set, the order of its elements is immaterial.Referring to the above sample space S, one can have different arrangement ofelements for the same S; probably he may record NNG first followed by the otherelements. Can you draw the tree diagram to suit this outcome?However, for each element in S, the order of items is very important becauseitems are representing outcomes of trials in that order. For instance, the elementGNG with items G, N and G has to be in that order because the first G is theoutcome from the first throw, followed by item N as the outcome of the secondthrow and finally followed by item G the outcome of the third throw. Thuselement GNG is not equivalent to element GGN in that contact. ACTIVITY 6.1 (a) A special fair dice consists of four faces numbered 1,2,3,4 is thrown. Write down the sample space. (b) Now the dice in (a) is thrown, followed by tossing a fair Malaysian coin. What is the category of this experiment? Write down its sample space.
  7. 7. TOPIC 6 EVENTS AND PROBABILITY 856.3 EVENT AND ITS REPRESENTATIONThere are two ways of representing event either by Venn diagram or Algebraicset. For clarification purpose, both representations can be put together. Sinceevent is a subset of the sample space S, therefore the sample space S will play therole of the universal set. This means that any event A, must be defined relative toits sample space S. Thus, if S belongs to the experiment of throwing ordinarydice, then event B = {G} is meaningless or is not defined.Figure 6.2 below shows how Venn diagram is very useful to clarify further thealgebraic set of an event. This figure depicts the sample space S = {1, 2, 3, 4, 5,6} and its defined events A = {1, 2, 3}, B = {4, 5, 6}, and C = {1, 3, 5}. You canobserve that there are some overlapping between C and A, as well as between Cand B. However, A and B are not overlapping. We will discuss about theserelationships between any two events later on. Figure 6.2: Venn diagram of events A, B, C, and SExample 6.2Refering to the sample space of Example 6.1, define some examples of events.SolutionThe sample space is restated here as: S = {GGG, GGN, GNG, GNN, NGG, NGN, NNG, NNN}
  8. 8. 86 TOPIC 6 EVENTS AND PROBABILITYLet us define the following events: A : Throw any two consecutive pictures B : Throw at least two pictures C : Obtain two pictures for the last two throws D : Obtain same items for the three throws E : Obtain alternate items in the three throws.For the above sample space, it is better to represent the events using algebraic setsas follows: A = {GGN, NGG} B = {GGG, GGN, GNG, NGG} C = {GGG, NGG}. D = {GGG, NNN}. E = {GNG, NGN}.Example 6.3Let S be the sample space of experiment throwing ordinary dice, once. Definesome events from this sample space.SolutionThe sample space is given by S = {1, 2, 3, 4, 5, 6}.Let us define the following events: A : Throws numbers less than 4 B : Throws numbers more than 3 C : Throw odd numbers D : Throw even numbers E : Throws numbers more than 4 F : Throws numbers less than 2 G : Throws any numbers H : Throws no numbers.
  9. 9. TOPIC 6 EVENTS AND PROBABILITY 87We can use algebraic set to clarify the elements of each event as follows: A = {1, 2, 3} B = {4, 5, 6} C = {1, 3, 5} D = {2, 4, 6} E = {5, 6} F = {1} G = {1, 2, 3, 4, 5, 6} H = , empty set.For further clarification, Venn diagram as in Figure 6.3 below can also be drawnas follows so that we can see any relationships between them. Figure 6.3: Venn diagram for A, B, ...., H of experiment throwing ordinary dice ACTIVITY 6.2 Refer to the experiment in Exercise 6.1, write down elements of the following events: A: obtain numbers less than 4; B: obtain numbers 2 and above; C: obtain odd numbers; and D: obtain even numbers.
  10. 10. 88 TOPIC 6 EVENTS AND PROBABILITY6.3.1 Mutually Exclusive Events(a) Involving Two Events Any two events defined from the same sample space is said to be mutually exclusive if both of them cannot occur at the same time. It means that their algebraic sets do not overlap, or their overlapping is equal to null set. Let us go back to Example 6.3, we have A and B as mutually exclusive events; likewise events C and D.(b) Involving Three Events Three events P, Q and R are said to be mutually exclusives if and only if the three pairs P and Q, P and R, and Q and R are all mutually exclusives. Refering to sample space of throwing ordinary dice, let P = {1,3}, Q = {2, 4}, and R = {5, 6} where P Q = P R = Q R = , empty set. Therefore events P, Q and R are three mutually exclusive events.6.3.2 Independent EventsTwo events P and Q are said to be independent of each other if and only if theoccurrence of P does not affect the occurrence of Q and vice versa.6.3.3 Complement EventTwo events P and Q defined from the same sample space are said to becomplementing each other if their union completes or forms the sample space i.e.P Q = S. Then we may write Q PC which mean Q is a complement of P.Similarly, we can write P QC. Thus we have equations P PC = S and alsoQ QC = S. As an example, Events A and B in Example 6.3 are complementingeach other since A B = {1, 3, 5, 2, 4, 6} = S.6.3.4 Simple EventSimple event is an event which possess one element of the sample space.Algebraicaly it means singleton set. For S = {1, 2, 3, 4, 5, 6}, we have six simpleevents which are {1}, {2}, {3}, {4}, {5}, {6}. For S = {GG, GN, NG, NN}, wehave four simple events which are {GG}, {GN}, {NG}, {NN}. We can observethat the union of all simple events is equal to the sample space S.
  11. 11. TOPIC 6 EVENTS AND PROBABILITY 896.3.5 Impossible EventRelative to a given sample space S, if P does not contain any single element of S,then P is said to be an impossible event. For example, let S = {1, 2, 3, 4, 5, 6},events P = {0}, and Q = {7} are impossible events. Similarly, if S = {GG, GN,NG, NN}, then event R = {GNG} is an impossible event.6.3.6 Equiprobable EventsA random experiment is an experiment whereby we cannot predict its outcomesbefore the experiment is completed. Take the experiment of throwing ordinarydice; we cannot determine which number will come out after each trial. This is anexample of random experiment, and we have therefore each simple event of thesample space S is equally probable event. In simple words, each element of S willhave equal chance to occur.6.3.7 Occurrence of EventAn event, A defined from S is said to occur if and only if an outcome of a trial isan element of A. The example below will explain the meaning of occurrence of adefined event.Example 6.4Let us refer back to Example 6.3 regarding to throwing of ordinary dice once andalso the defined events there. Since there are six possible numbers as possibleoutcomes, therefore the throw or trial has to proceed untill all numbers haveappeared. Outcome of each throw has to be recorded accordingly. Table 6.2shows the recorded oucome of each throw as well as occurrence of the definedevent pertaining to the recorded outcome. Table 6.2 The Throw Outcome of the Throw Events said to be Occurred 1 6 B, D, E, G 2 5 A, D, G 3 4 B, D, G 4 3 A, C, G 5 2 C, B, E, G 6 1 A, C, F, G
  12. 12. 90 TOPIC 6 EVENTS AND PROBABILITYWe can draw some conclusions from the table:(i) The event G seems to occur at all throws which means G is a sure event. Since G always occurs at all situations, then G is actually the sample space of the experiment.(ii) On the other hand, the event H does not occur at any throw to indicate that H is impossible event.(iii) Events B, D and E can occur at the first throw i.e. the throw that produce ‘6’. However B and D both can occur at the third throw which produce outcome ‘4’. Algebraically we can say that B D E = {6}, and B D = {4}. Since their overlapping is not empty set, therefore they are non-exclusive events.(iv) Events A and C seem to occur together at two different situations i.e. when obtaining ‘3’ at the fourth throw and when getting ‘1’ at the sixth throw. Thus, algebraically, we have A C = {3,1}.(v) From the above table we can also find some mutually exclusive events which cannot occur at the same time or on the same throw. Events A and D; C and D; D and F; A and E; B and F are pairs of mutually exclusive events. ACTIVITY 6.6 Consider the experiment tossing a Malaysian coin followed by throwing a fair ordinary die. Obtain its sample space, S, then (i) State the elements of the following events: (a) P : Obtain G on coin and even numbers on the die, G represents picture on the coin; (b) Q : Obtain prime numbers on the die; (c) R : Obtain G on coin and odd numbers on die. (d) P Q (e) Q R (f) Only event Q to occur. (ii) Amongst P, Q and R determine pairs of mutual exclusive events.
  13. 13. TOPIC 6 EVENTS AND PROBABILITY 916.3.8 Probability of an EventLet E be an event defined from a given equiprobable sample space S, theprobabality of occurrence of E is given by: Number of elements in E n( E ) Pr( E ) Number of elements in S n( S ) Formula 6.1Propositions:(a) If E is an empty set then n(E) = 0, and therefore Pr(E) = 0. Thus, any impossible event will have zero probability to occur.(b) If E is sure event then E S, and n(E) = n(S), and therefore Pr(E) = 1. Thus for any event which is sure to occur will have probability value 1.(c) If E is any event occuring between the above two extreme events, then the Pr(E) is lying in the interval 0<Pr(E)<1.Thus, any event which is close to the extreme left (i.e.impossible situation), itsprobability value is fairly close to zero; on the other hand, if event E is close to theextreme right (i.e.sure event), then its probability value is fairly close to 1. Ingeneral, any event E will have its probability lying in the interval of 0 Pr (E) 1. Formula 6.2Example 6.5Refer back to Example 6.3 and find the probability of the occurrence of events A,B, C, E, G, H.SolutionThe sample space S of the throwing an ordinary die is equiprobable which meansthat all simple events in S has equal probability to occur. Thus we can useFormula 6.1 to find the required probabilities as follows:
  14. 14. 92 TOPIC 6 EVENTS AND PROBABILITYPr(A) = n(A)/n(S) = 3/6 = 0.5. Pr(B) = n(B)/n(S) = 3/6 = 0.5. Pr(C) = 0.5. Pr(E) =2/6 = 1/3. Pr(G) = 6/6 = 1, and Pr(H) = 0/6 = 0.We conclude the followings: Probability of each event is lying in the interval between 0 and 1, H is an impossible event, G is a sure event, Pr(A) + Pr(B) = 1, this proves that A and B are complement events whereby A B = S. ACTIVITY 6.4 A box contains 5 red balls and 3 blue balls. A ball is randomly taken out from the box and its colour recorded before returning back to the box. Find the probability of obtaining a: (a) red ball; (b) blue ball. 6.4 COMPOUND EVENTSWe have mentioned earlier that an event must be defined from a given samplespace. If we have two defined events from the same sample space, then by usingset operations we can construct new events which will be categorised ascompound events. Consider three defined events P, Q, R, and then we have threeset operations used to develop compound events as follows:(a) Union operation, “ ” (i) Union of two events. B compound event A = P Q. A will contain all elements in P, and elements in Q, as well as common elements in them. It is the event that P occurs, or Q occurs, (or both). For example, if P = {1,2,3,4}, Q = {3, 4,5}, then A = P Q = {1,2,3,4,5}. (ii) Union of three events, A compound event B = P Q R which will contain elements in P, in Q and in R, as well as elements in common to them. It is the event that P occurs or Q occurs or R occurs; or P and Q,
  15. 15. TOPIC 6 EVENTS AND PROBABILITY 93 P and R, or Q and R (or all of them) ocuurs. For example, if P = {1,2,3,4}, Q = {3,4,5}, and R = {4,5,6}, then B = P Q R = {1,2,3,4,5,6}. (b) Overlapping operation, “ ” (i) The compound event C = Q P contains elements in common in them. It is the event that occurs that if Q occurs and R occurs. For example, if P = {1,2,3}, Q = {3, 4,5}, then C = P Q = {3}. (ii) The compound event D = P Q R contains elements in common to three of them. This means the event that occurs when all three of them occur at the same time. For example, if P = {1,2,3,4}, Q = {3, 4,5}, and R = {4,5,6}, then D = P Q R = {4}. (c) To conclude, event B in (a) (ii) will contain events P Q R = {4}, P Q ={3,4}, P R = {4}, Q R = {4,5}, P alone = {1,2}, Q alone = R alone = {6}. 6.4.1 Probability of Compound Event We have the following rules to obtain the probability of compound events. (a) Compound by union operation, ‘ ’ Let P, Q and R be any three events defined from a given sample space S, then (i) Pr( P Q) Pr( P) Pr(Q) Pr( P Q) Formula 6.3 (ii) If P and Q are mutual exclusive events, then Pr( P Q) Pr( P) Pr(Q) Formula 6.3(a) (iii)Pr( P Q R) P( P) Pr(Q) Pr( R) Pr( P Q) Pr( P R) Pr(Q R) Pr( P Q R) Formula 6.4
  16. 16. 94 TOPIC 6 EVENTS AND PROBABILITY (iv) If P, Q and R mutual exclusive events, then Pr( P Q R) Pr( P) Pr(Q) Pr( R) Formula 6.4(a)(b) Compound events by ‘ ’ Let P, and Q be any two events defined from a given sample space S, then: (i) n( P Q ) Pr( P Q) n( S ) Formula 6.5 (ii) Pr( P Q) Pr( P ) Pr(Q P) , where Q P is conditional event, see later. Formula 6.5(a) (iii) If P, and Q are independent events, then Pr ( P Q) Pr( P) Pr(Q) Formula 6.5(b)(c) Complement events If Qc is the complement of event Q and they are defined from the same S, then: Qc Q S, c Pr(Q ) Pr(Q) Pr( S ) 1; c Pr(Q ) 1 Pr(Q) Formula 6.6
  17. 17. TOPIC 6 EVENTS AND PROBABILITY 95Example 6.6Refer back to the Example 6.3, and find the probability of the followingcompound events:(a) B C,(b) B DSolution(a) From the defined events, we have B C = {5}, and B D = {4,6}, then the Formula 3.1, Formula 3.3 and Formula 3.5 give: n( B ) 3 1 3 1 Pr ( B) , Pr (C ) n( S ) 6 2 6 2 3 1 Pr ( D) 6 2 1 2 1 Pr ( B C ) , and Pr ( B C ) 6 6 3 Then by using Formula 3.3 1 1 1 5 Pr ( B C) 2 2 6 6
  18. 18. 96 TOPIC 6 EVENTS AND PROBABILITY ACTIVITY 6.5 1. A fair Malaysian coin is tossed, and immediately followed by tossing a fair four faces die marked 1,2,3,4. Obtain sample space and events. (a) A: Picture and even numbers appear; B: Number on coin and odd numbers on die; C: Numbers 3 and above appear on die. (b) A C, B C. 2. Which of the events A, B and C are mutually exclusive? 3. Obtain probabilities of A, C and (A C). 6.5 TREE DIAGRAM AND CONDITIONAL PROBABILITYA Tree diagram is usually used to resolve the followings:(a) analyse a given experiment in the context of its operation and presenting the trials in the proper sequence as they should be occurring in the real situation. In other words, tree diagram should represent the actual operation of an experiment.(b) the correct tree diagram plays an important role of getting a correct combination of trials outcomes to generate sample space, especially for repeated trial and multiple trials experiments.(c) shows appropriate probability values at each branch with a pair branch has a total probability equal to 1.(d) depicts outcomes of each trial as well as conditional events.Consider a tree diagram in Figure 6.4 below. We should be able to read andinterpret the diagram and try to imagine the operation aspect of the experimentrepresented by the diagram. We may try the following way:
  19. 19. TOPIC 6 EVENTS AND PROBABILITY 97 Figure 6.4: Example of a tree diagram(i) The diagram shows that the first trial produce outcomes either {M}, or {K} where {M} has probability 3/20 to occur.(ii) After getting {M} it is immediately follow by the second trial which produce either {W}, or {V} where {W} has probability 7/15 to occur.(iii) Similarly, the first trial may produce outcome {K}, and it is immediately followed by second trial which produces either {V} with probability 8/15, or {W}.(iv) From observations (i) and (ii) we can understand that the experiment represented by the diagram is possibly of the category multiple trials.(v) Event {M}, or {K} are called unconditional events, they can occur on their own. However, events {W} and {V} are conditional events. This means that {W} can occur conditionally in two situations: after the outcome {M} and in probability language we term this conditional W as W|M, after the outcome {K} and in probability language we term this conditional W as W|K, We can explain in a similar way for conditional event V. In general we conclude that outcomes of the second trial are conditional events which means they occur after the outcomes of the first trial. Similarly, the outcomes of the third trial are conditional based on the outcomes of the second trial, and so forth.
  20. 20. 98 TOPIC 6 EVENTS AND PROBABILITY(vi) The combinations of branches will produce the required combinations of outcomes. Thus the events combinations are: {MW}, {MV}, {KV}, {KW}. The sample space of the experiment is S = {MW, MB, KB, KW}.(vii) As for probability, we have the rule that sum of probability of a pair branches is equal to 1. Thus, Pr(M) + Pr(K) = 1, and Pr(K) = 1 – 3/20 = 17/20. Pr(W|M) + Pr(V|M) = 1, and Pr(V|M) = 1 – 7/15 = 8/15. Pr(V|K) + Pr(W|K) = 1, and Pr(W|K) = 1 – 8/15 = 7/15.(viii) So for the whole experiment the probability of all simple events in S are given by Pr(MW) = Pr(M) Pr(W|M) = 3/20 7/15 = 7/100. Pr(MV) = Pr(M) Pr(V|M) = 3/20 8/15 = 6/75. Pr(KV) = Pr(K) Pr(V|K) = 17/20 8/15 = 34/75. Pr(KW) = Pr(K) Pr(W|K) = 17/20 7/15 = 119/300. You can verify that Pr(S) = Pr(MW)+ Pr(MV)+ Pr(KV)+ Pr(KW) = 1.By looking at the above sample space S, we should now understand that the eventMW actually means the event M W. Therefore in probability language weshould write Pr(MW) as Pr(M W) and we have the following relationship: Pr(M W) = Pr(M) x Pr(W|M)From the above relationship we can have the general formula of conditionalprobability for conditional event as follows:Let P and Q are two defined events from the same sample space S, then theprobability of conditional event Q|P is given by: Pr(Q P) Pr(Q P ) Pr( P) Formula 6.7
  21. 21. TOPIC 6 EVENTS AND PROBABILITY 99Since Q P = P Q, therefore we also have Pr(Q P) Pr( P Q ) Pr(Q) Formula 6.7(a)Example 6.7We refer back to Example 6.3, find the conditional probability of the conditionalevent E|B.SolutionFrom Formula 6.7, we have Pr( E B) Pr( E B ) Pr( B)Thus we need to find the compound event E B, and its probability; as well asthe Pr(B). From the given information in Example 6.3, we have B = {4,5,6}, andE = {5, 6}; therefore E B = {5, 6}: Pr(B) = n(B)/n(S) = 3/6 = 0.5; Pr(E B) = n(E B)/n(S) = 2/6 = 1/3. Pr(E B) = (1/3)/(1/2) = 2/3. ACTIVITY 6.6 Given two boxes containing oranges as follows: Box A contains 10 oranges including 4 bad oranges. Box B contains 6 oranges including 1 bad orange. One of the above boxes is selected randomly and an orange is randomly taken out from it. What is the probability that the taken orange is a bad one?
  22. 22. 100 TOPIC 6 EVENTS AND PROBABILITY6.5.1 Bayes TeoremLet us go back to the tree diagram given in Figure 6.4. The sequence of trialsshows that V always happened after M or after K has had happened. This meanswe are talking about conditional event V|M or V|K. Now, supposing V is theoutcome of the experiment, therefore either simple events {MV} or {KV} of thesample space are said to happen or in short X is said to happen where X = {MV,KV}. In this situation, we are actually not sure either V in {MV} or V in {KV}.Thus we can raise several questions on probabilities of conditional events asfollows:(a) What is the probability of event X|M?(b) What is the probability of event X|K?(c) Given ‘V’, what is the probability that it is ‘after’ M? or equivalently to say what is the probability of event M|X to occur?,(d) Given ‘V’, what is the probability that it is ‘after’ K? or equivalently to say what is the probability of event K|X to occur?We can use Formula 6.7 to answer question (a) as follows: Pr( X M ) Pr({MV }) Pr(V M ) Pr( M ) Pr(X M) = Pr(V M ) , Pr( M ) Pr( M ) Pr( M )It is the probability value at the branch MV. The answer for question (b) can beobtained in a similar way. So, it is very straight forward for these two conditionalevents, but it is not so for questions (c) and (d), and we need Teorem Bayes asderived below.Let us partition exhaustively the sample space S into two mutual exclusive eventsM and K. Here the word exhaustive is to mean K M = S. Further, let X be acompound event generated from the same experiment.
  23. 23. TOPIC 6 EVENTS AND PROBABILITY 101 Figure 6.5As can be seen from Figure 6.5 above, we can have the following facts: X S; X can possibly overlap with M to give (X M) = {MV,KV} M = {MV}; X is also possible to overlap with K to give (X K) ={MV, KV} K = {KV}; We can write X = (X M) (X K) = {MV, KV}; and (X M) (X K) = .As shown above, Pr (X|M) and Pr (X|K) are considerd known or can be calculateddirectly. Then, we have the following Bayes Theorem for two partitions:(i) Pr( M X ) Pr({MV }) Pr(V M ) Pr( M ) Pr (M X) = , Pr( X ) Pr( X ) Pr( X ) Formula 6.8(ii) Pr( K X ) Pr({KV }) Pr(V K ) Pr( K ) Pr (K X) = , Pr( X ) Pr( X ) Pr( X ) Formula 6.8(a)where Pr( x) Pr MV , KV Pr MV Pr KV Pr M Pr V M Pr K Pr V K
  24. 24. 102 TOPIC 6 EVENTS AND PROBABILITYTherefore to use the above theorem, we require all probabilities of the R.H.S(right hand side) of the equality sign of each formula. We can extend the idea ofpartitioning exhaustively the sample space S to three exclusive events M, K andN. Then we have the following Bayes Theorem for three partitions:We can use the same idea and write: S = (X M) (X K) (X N), and X = (X M) (X K) (X N) = {MV, KV, NV}We tend to find probability of conditional events K|X, M|X, N|X by the followingformulas: Pr( M X ) Pr({MV }) Pr(V M ) Pr( M ) (i) Pr (M X) = , Pr( X ) Pr( X ) Pr( X ) Pr( K X ) Pr({KV }) Pr(V K ) Pr( K ) (ii) Pr (K X) = , Pr( X ) Pr( X ) Pr( X ) Pr( N X ) Pr({NV }) Pr(V N ) Pr( N ) (iii) Pr (N X) = , Pr( X ) Pr( X ) Pr( X ) Formula 6.8(b)Where Pr(X) = Pr{MV, KV, NV} = Pr(MV) + Pr(KV) + Pr(NV) = Pr(M) Pr(V|M) + Pr(K) x Pr(V|K) + Pr(N) x Pr(V|N)Example 6.8Let us go back to the tree diagram given in Figure 6.4. Find the probability Pr(M|X), and Pr (K|X), where X = {MV, KV}.
  25. 25. TOPIC 6 EVENTS AND PROBABILITY 103SolutionPr(M) = 3/20; Pr(K) = 17/20; Pr (V M) = 8/15 ; Pr (V K) = 8/15 ; andX = {MV, KV}, andPr(X) = Pr(MV) + Pr(KV) = Pr(M) Pr (V M) + Pr(K) Pr (V K) = (3/20)(8/15) + (17/20)(8/15) = (40/75) = 8/15. From Formula 6.8, we have Pr( M X ) Pr(V M ) Pr( M ) (3 / 20)(8 / 15) Pr (M X) = 3 / 20 , Pr( X ) Pr( X ) 8 / 15 and From Formula 6.8a, we have Pr( K X ) Pr(V K ) Pr( K ) (8 / 15)(17 / 20) Pr (K X) = 17 / 20 . Pr( X ) Pr( X ) 8 / 15 ACTIVITY 6.7 1. Refer to the information given in Exercise 6.6. If it is found that the taken orange is bad, what is the probability the orange is from Box A? 2. Given three coloured jars containing items as follows: Red Jar: Containing 3 gold coins, and 6 silver coins; Green Jar: Containing 4 gold coins, and 2 silver coins; Blue Jar: Containing 2 gold coins, and 8 silver coins; A jar is selected at random from the above three and a coin is taken out randomly. (a) Construct a tree diagram representing this experiment and obtain its sample space; (b) If the selected coin is of gold type, find the probability that this coin is taken from the Red Jar.
  26. 26. 104 TOPIC 6 EVENTS AND PROBABILITY6.5.2 Probability of Independent EventsAn event P is said to be independent of another event Q if its occurrence does notaffect in any manner the occurrence of event Q and vice versa. In simple term, itmeans that the occurrence of P has nothing to do with the chance of Q to happen.If we try to conditioning Q on P, the chance of Q to happen still remains the same.Thus, algebraically we can write the following:(a) If event P is independent of event Q, then Pr(Q P ) Pr(Q) , or Pr( P Q ) Pr( P ) Formula 6.9 Formula 6.9(a)(b) The multiplication rule of two independent events becomes: Pr( P Q) Pr( P ) Pr(Q P ) Pr( P ) Pr(Q) Formula 6.9(b)(c) The three events P, Q, and R are said to be independent if and only if the following three conditions are true: (i) Pr( P Q) Pr( P ) Pr(Q) , (ii) Pr( P R ) Pr( P) Pr( R) , (iii) Pr(Q R) Pr(Q) Pr( R) . Formula 6.10(d) The multiplication rule of three independent events becomes: Pr( P Q R) Pr( P ) Pr(Q) Pr( R) Formula 6.10(a)
  27. 27. TOPIC 6 EVENTS AND PROBABILITY 105Example 6.9Try to recall Example 6.1, whereby Malaysian coin is tossed three consecutivetimes. Let us define the following events: A: obtain picture on the first toss; B: obtain picture on the second toss; C: obtain two conescutive pictures;(a) find the probabilities of the events A, B, and C;(b) find the probabilities of the compound events A B, A C, B C;(c) determine whether each pair of A&B, A&C, and B&C is independent; and(d) whether the three events A, B, and C are intradependent.SolutionThe sample space of the experiment as well as the the events concerned are asfollows: (i) S = {GGG, GGN, GNG, GNN, NGG, NGN, NNG, NNN} It is an equiprobable sample space meaning its simple event such as {GNG}has equal probability of (1/2)(1/2)(1/2) = 1/8. (ii) A = {GGG, GGN, GNG, GNN}, B ={GGG, GGN, NGG, NGN}, and C = {GGN, NGG}(a) Pr(A) = (1/8) + (1/8)+(1/8)+(1/8) = 1/2; Pr(B) = (1/8) + (1/8)+(1/8)+(1/8) = 1/2; Pr(C) = (1/8) + (1/8) = 1/4(b) A B = { GGG, GGN }, A C = {GGN}, B C = {GGN, NGG } Pr (A B) = (1/8) + (1/8) = 1/4 = (1/2)(1/2) = Pr(A) Pr(B); Pr (A C) = 1/8 = (1/2)(1/4) = Pr(A) Pr(C) ; Pr (B C) = (1/8) + (1/8) = 1/4 (1/2) (1/4) = Pr(B) Pr(C) .(c) A&B are independent because Pr(A B) = Pr(A) Pr(B); A&C are independent because Pr (A C) = Pr (A) Pr(C); B&C are NOT independent because Pr (B C) Pr (B) Pr(C);
  28. 28. 106 TOPIC 6 EVENTS AND PROBABILITY(d) Since all conditions in Formula 3.11 are not fulfilled, therefore A, B, and C are not intradependent.You can enhance your understanding of the above concepts by doing thefollowing exercises. ACTIVITY 6.8 1. Determine the probability of the following events: (a) obtaining G in the experiment of throwing three Malaysian coins together. (b) obtaining red ball in an experiment of taking out randomly a ball from a jar containing 5 blue balls, 4 red balls and 6 green balls. (c) A group of students consists of 12 male students and 24 female students. A 50% of each subgroup by sex is considered under category high students. A student is selected randomly from the group is found to be Male student or category high students. What is the probability of occurance of this event? 2. An experiment is throwing a red dice and a green dice together. (a) obtain its sample space, S. (b) write down the elements of the following events: P: Getting a sum of eleven or more, the numbers appeared on both die; Q: Getting number 5 on green dice; R: Number 5 appears on at least one of the die. (c) from the defined events in (b), obtain the probability of the following conditional events: (i) P|Q. (ii) P|R. 3. Let P and Q be two events defined from the same sample space S. Given that Pr(P) =3/4, Pr(Q) =3/8, and Pr(P Q) =1/4, find the probabilities of the following events: (i) Pc , (ii) Qc, (iii) Pc Q c, (iv) P c Q c, (v) P Q c,
  29. 29. TOPIC 6 EVENTS AND PROBABILITY 107A given experiment can be classified into three categories:(a) Single trial experiment;(b) Repeated trial experiment; and(c) Multiple trial experiment.Once the category is identified, a tree diagram may be used to depict the operationof the experiment and hence determine the sample space S. With the reference ofa given sample space S, events can be defined as a subset of S. The probabilitiesof defined events can normally be calculated. Various types of events are givensuch as mutual exclusive event, impossible event, sure events, as well asconditional events. The compound events are derived from the defined eventsthrough union, intersection and negation operations of set. A Venn diagram oralgebraic set is normally used to represent these events. Their probabilities can befound using additive and multiplicative rules. The formulas of theorem base aregiven for two exhaustive partition events as well as three exhaustive partitionevents.