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Chapter 3 Probability3.1 Terminology3.2 Assign Probability3.3 Compound Events3.4 Conditional Probability3.5 Rules of Computing Probabilities3.6 Random Sampling

Homework:3, 7, 17, 23, 27, 29, 32, 35, 37, 43, 50, 55, 57, 61

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Section 3.1: Some terminology of probabilities

We discussed how to understand/describe the information contained in a sample in Chapter II. However, we want to make inference based on the information contained in a sample as well. We will discuss the concept of probability in this chapter because probability plays an important role in inference making process. Let start our discussion with one example.

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<Example 3.1>: (Basic) Suppose there is a bag of M&M chocolate with six different coating colors -- 300 brown, 250 red, 200 yellow, 150 orange, 100 green, and 100 tan. Suppose one piece is drawn at random and the coating color is recorded.

(a) Is this a random experiment?(b) How many sample points in this random experiment?(c) What is the sample space of this random experiment?(d) What is the probability of drawing a yellow M&M

chocolate?(e) What is the event of drawing a piece of M&M with your

favorite colors if your favorite colors are yellow and green?

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Section 3.2: Assign probability to a sample point and to an event

There are two approaches, the relative frequency approach and subjective approach, to assign a probability to a sample point. Both approaches need to follow two basic rules. The first rule is that the probability for each sample point must lie between 0 and 1. The second rule is that the probabilities of all the sample points within a sample space must sum to 1. We use the subjective approach to assign probability in Example 3.2 and 3.4 because the frequency tables are unavailable, and use the relative frequency approach to assign probability in Example 3.1 and 3.3 because the frequency tables are available. Usually, we use relative frequency approach to assign probability if the frequency table is available.

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We can employ the following five steps to compute the probability of an event.

(1) Define the experiment.(2) List all the sample points.(3) Assign probability to each sample point.(4) Find out all the sample points in this event.(5). Sum the probabilities of these sample points.

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<Example 3.2>: (Basic) Two fair coins are randomly tossed, and their up faces are recorded.

(a) Is this a random experiment?(b) Write down the sample space.(c) Assign probabilities to each sample points.(d) Event A is at least one head. Event B is at least one tail.

Compute the probability of event A and event B.

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<Example 3.3>: (Basic) Suppose that we repeat toss a pair of fair coins ten

thousand times. Table 3.1 is the frequency table of this random experiment. Table 3.1 Simple Event TT TH HT HHFrequencies 1977 2855 2423 2745

(a) Assign probabilities to all simple events.(b) Event A is at least one head. Event B is at least one tail.

Compute the probabilities of event A and event B.

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<Example 3.4>: (Basic) Two fair dice are tossed, and the up face on each

die is recorded.(a) Write down the sample space.(b) Assign probabilities to all the sample points.(c)A = {sum of two numbers is equal to 7.}

B = {sum of two numbers is greater than or equal to 8.} C = {3 on the up face of at least one die.}Find the probabilities of events A, B, and C.

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Section 3.3 Compound Events, Complementary Events, and Mutually Exclusiveness

A compound event is a composition of two or more events. A compound event can be formed through either union or intersection operation. The union of two events A and B denoted by the symbol AB consists all the sample points that belong to A or B or both. The intersection of two events A and B denoted by the symbol AB consists all the sample points belonging to both A and B.

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<Example 3.5>: (Basic) Continuation of Example 3.4.Event D = {Sum of two numbers is equal to 10}Event E ={Difference of two numbers is equal to 2}Event F = {Two even numbers}Event G = {Two odd number}Event H = {Sum of two numbers is larger than or equal to

10}(a) Find the probabilities of events D, E, F, G, and H.(b) Find the probability of D E.(c) Find the probability of D E.

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<Solutions>:(a)Event D = {(4,6), (5,5), (6,4)}

Event E = {(1,3), (2,4), (3,5), (4,6), (6,4), (5,3), (4,2), (3,1)}Event F = {(2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), (6,6)}Event G = {(1,1), (1,3), (1,5), (3,1), (3,3), (3,5), (5,1), (5,3), (5,5)}Event H = {(4,6), (5,5), (6,4), (5,6), (6,5), (6,6)}

P(D) = 3/36 = 1/12;P(E) = 8/36 = 2/9;P(F) = 9/36 = 1/4; P(G) = 9/36 = 1/4; P(H) = 6/36 = 1/6.

(b) Event D E = {(4,6), (6,4)} and P(D E) = 2/36 = 1/18.(c) Event D E = {(1,3), (2,4), (3,5), (4,6), (5,5), (6,4), (5,3),

(4,2), (3,1)} and P(D

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The complement of a event A, denoted by Ac, is the event that A does not occur, i.e. Ac consists all the sample points not belonging to event A. The concept of complementary event is very important in computing the event probability because that in many probability problems calculating the probability of the complement of the event of interest is easier than calculating the probability of the event itself.

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<Example 3.6> (Basic) Continuation of Example 3.4

(a) Is Event F a complementary event of Event G?(b) Find the probability of Fc , Gc , and Hc.

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Event A and Event B are mutually exclusive if A B contains no sample points, that is, A and B have no sample points in common. A and Ac has no sample points in common, i.e. A and Ac are mutually exclusive.

<Example 3.7> (Basic) Continuation of Example 3.4(a) Are events F and G mutually exclusive?

(b) Find P(F G).<Solutions>:(a) Events F and g have no sample points in common, i.e.

they are mutually exclusive.

(b) P(F G) = 0.

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Note: (a) The sum of the probabilities of complementary

events equals one; that is, P(A) + P(Ac) = 1.(b) Event A and event Ac have no sample points in

common.(c) Event A and event Ac are mutually exclusive.(d) P(A) + P(Ac) = 1. This means the Event A and

the event Ac cover the entire sample space.(e) Venn Diagram is a good graphical tool for

understanding the concept of compound events, complementary events, and mutually exclusiveness.

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Section 3.4: Conditional Probability (Intermediate)

The event probabilities we have been discussing based on subjective approach or relative frequency approach are unconditional probabilities because no special conditions are assumed when we compute these probabilities. However, we sometimes we have additional information that might alter the probability of a given event. A probability that reflects such additional knowledge is called conditional probability of the event. The probability that event A occurs given that the event B occurs is called the conditional probability of A given B and can be denoted by the symbol P(A|B), where P(A|B) = P(AB)/P(B) if P(B) > 0.

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<Example 3.8> (Basic)Consider the experiment of tossing a fair coin twice and recording the up face on each toss. The following event are defined:A: {the first toss is a head}B: {the second toss is a head}

Find P(A), P(B), P(B|A), and P(BA).<solutions>:Sample space = {HH, HT, TH, TT}

A = {HH, HT}; B = {HH, TH}; BA = {HH}

P(A) = 2/4 = ½; P(B) = 2/4 = ½; P(BA) = ¼;

P(B|A) = P(BA)/P(A) = (1/4)/(1/2) = ½.

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<Example 3.9>: (Basic) Suppose that the sample space is {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, Event A is {2, 5, 6, 7, 0}, and Event B is {0, 4, 7, 8}. Find P(A), P(B), P(A B), P(A|B), P(B|A), P(A B), and P(A’).

<Solutions>:A B = {0, 2, 4, 5, 6, 7, 8}P(A) = 5/10 = 0.5;P(B) = 4/10 = 0.4;P(A B) = 2/10 = 0.2;P(A|B) = P(AP(B) = 0.2/0.4 = 0.5;P(B|A) = P(BPP(A B) = 7 / 10 = 0.7;P(A’) = 1 - P(A) = 1 - 0.5 = 0.5.

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<Example 3.10>: (Intermediate)Five hundred married men and women were asked if they would marry their current spouses if they were given a chance to do it over again. Their responses are recorded in the following : Table 3.2

Yes NoMale 125 175Female 55 145

(a). The following events are defined:Event A: Want to marry their current spouses againEvent B: Female; Event C: MaleEvent D: Do not want to marry to their current spouses again; Find P(A), P(A|B), P(B|A), and P(A B)

(b). Are the events B and D mutually exclusive? Explain.

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<Solutions>: (a) P(A) = (125+55)/500 = 0.36;

P(B) = (55+145)/500 = 0.4;P(C ) = (125+175)/500 = 0.6;P(D) = (175+145)/500 = 0.64;P(A

P(AP0.11/0.4 = 0.275;P(B|A) = P(B P(A) = 0.11/0.36 = 11/36;P(A B) = (125+55+145)/500 = 0.665.

(b) P(B D) = 145 / 500 = 0.29. Thus, events B and D are not mutually exclusive because there are many unhappy wives who do not want to marry their current spouses again.

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Section 3.5: Rules of Computing the Probabilities

Both additive rule and multiplicative rule are very useful when we want to compute the probabilities of compound events. The additivity rule of probability is used to find the probability of union of two events and the multiplicative rule of probability is used to find the probability of intersection of two events. Let A and B be two events, the additivity rule of probability is P(A B) = P(A) + P(B) - P(A B) and the multiplicative rule of probability is P(A B) = P(A|B)*P(B) = P(B|A)*P(A).

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We say two events A and B are independent events if the probability of the occurrence of one event (say A) does not alter the probability that another event (say B) has occurred. P(A|B) = P(A) and P(B|A) = P(B) when A and B are independent events. If events A and B are independent, the multiplicative rule of probability becomes to P(AB) = P(A)*P(B) = P(B)*P(A). We say events A and B dependent events if A and B are not independent.

Note:(1) P(A B) = P(A) + P(B) if events A and B are mutually

exclusive because P(AB) = 0.

(2) P(AB) = P(A) * P(B) if events A and B are independent events.

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<Example 3.11>: (Basic)When an American marriage ends in divorce, it is most likely that the women is the partner who is dissatisfied. A study of National Center for Health Statistics reported the following table:

1975 1986By wife 67.2% 61.5%By husband 29.4% 32.6%Jointly 3.4% 5.9%Suppose that two women are interviewed, one of whom was divorced in 1975, and the other in 1986.

(a) What is the probability both were filed by themselves?(b) What is the probability of the first was filed by herself and

the second was filed by her exit-husband?

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<solutions>:(a) Let A: {divorced in 1975 and filed by herself}

B: {divorced in 1986 and filed by herself}C: {divorced in 1986 and filed by her exit-

husband} then P(A) = 0.672, P(B) = 0.615, and P(C) = 0.326. Since A and B are independent events,

P(AB)=P(A)*P(B)=0.672*0.615=0.41328.

(b) Since A and C are independent events, P(AC) = P(A)*P(C ) = 0.672*0.326 = 0.219072.

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<Example 3.12>: (Intermediate)A smoke detector system uses two devices, P and

Q. If smoke is present, the probability that it will be detected by device P is 0.95: by device Q is 0.98: by both devices is 0.94. Considering the following events:A: {The smoke will be detected by device P}B: {The smoke will be detective by device Q}

(a) Find the probability of smoke will be detected by at least one device.

(b) Find the probability that the smoke will not be detected.

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<Solutions>:

(a) P(A) = 0.95, P(B) = 0.98, and P(AB) = 0.94 are given. Applying the additivity rule of probability, we have P(AB) = P(A) + P(B) - P(AB) = 0.95 + 0.98 - 0.94 = 0.99.

(b) Smoke will not be detected is a complement event of the event that smoke will be detected by at least one device. Applying the probability of complement, we have P(smoke will not be detected) = 1 - P(AB) = 1 - 0.99 = 0.01.

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<Example 3.13>: (Advance)In a study involving a manufacturing process, 15% of all parts tested were defective, and 40% of all parts were produced by machine I. If a part was produced by machine I there is a 20% chance that it is defective.Let A = {the part is produced by machine I}

B = {the part is defective}.(a) Find P(A) and P(B).(b) Find P(B|A).(c) Find P(A B).(d) If a part is found defective, what is the probability that it

came from machine I?(e). Are A and B independent events? Explain.

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<Solutions>:(a) P(A) = 40% = 0.4 and P(B) = 15% = 0.15.(b) P(B|A) = 0.20. Note: (1) The condition is the part was produced by machine I, i.e., event

A. (2) P(B|A) is the conditional probability that the defective part was produced by machine I. (3) Thus, P(B|A) is given in the problem and P(B|A) = 0.20.

(c) P(A B) = P(B|A) * P(A) = 0.20 * 0.4 = 0.08.(d) P(A|B) = P(A B) / P(B) = 0.08 / 0.15 = 8/15.Note: (1) The condition now is the part was found defective. (2) P(A|B)

is the conditional probability that the part produced by machine is defective. (3) Thus, P(A|B) = 8/15.

(e) P(A) * P(B) = 0.4 * 0.15 = 0.06 and P(A B) = 0.08. Thus, P(A B) P(A) * P(B) and A and B are not independent events.

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Section 3.6 Random SamplingHow a sample is selected from a population is of vital importance in statistical inference. Although there are many sampling procedures can be used to obtain a useful sample in statistical inference, we will only discuss the most frequently and simplest form of sampling procedure, simple random sampling procedure, in this semester. A sample of size n is called a simple random sample (or random sample) if every set of n elements in the population has equal chance of being selected. Usually, we rely on random number generator that are built into most statistical software to generate the random sample. We say a sample is biased if not all set of n elements has equal chance of being selected. Usually, we want to avoid to use a biased sample in statistical inference.

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Collection of Definitions: Random Experiment A random experiment is a

process that can be repeated under the same conditions. And each outcome of this process can not be predicted in advance without uncertainty.

Sample Point A sample point is the most basic outcomes from a random experiment.

Sample Space Sample space is the collection of all possible sample points from a random experiment.

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Event An event is a specific collection of sample points.

Probability of a sample point The probability of a sample point is a number between 0 and 1 that reflects the chance that the outcome will occur when the experiment is performed.

Union The union of two events A and B is the event that either A or B or both occur in a single trail of the experiment. We denote the union of A and B by the symbol A B.

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Intersection The intersection of A and B is the event that both A and B occur n a single trail of the experiment. We denote the intersection of A and B by the symbol A B.

Complementary Events The complement of an event A is the event that A does not occur. This means the complement event of A consists all sample points that are not in event A. We denote the complement of event by the symbol A’ or .

Mutually Exclusive Events Events A and B are mutually exclusive events if event A and event B have no sample points in common.

AC

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Conditional Probability The probability that event A occurs given that the

condition of event B occurs is called the conditional probability of A given B. The conditional probability of event A given event B is denoted by the symbol P(A|B), and P(A|B) = P(AB)/P(B) if P(B) > 0.

Additive Rule of Probability The probability of the union of events A and B, denoted by

P(AB), equals P(A) + P(B) - P(AB). In particular, P(A B) = P(A) + P(B) if events A and B are mutually exclusive.

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Multiplicative Rule of Probability The probability of the intersection of events A and B,

denoted by P(AB), equals to P(A)*P(B|A) or equals to P(B) * P(A|B). P(AB) = P(A) * P(B) if events A and B are independent.

Independent and Dependent Events A and B are independent events if the occurrence of

B does not alter the probability that A has occurred. That is P(A|B) = P(A) and P(B|A) = P(B). Events A and B are dependent if they are not independent,