department of mathematics and statisticsaoyet/stat2550/slides/probnotes.pdf · for instance, if i...

DEPARTMENT OF MATHEMATICS AND STATISTICS

Memorial University of Newfoundland St. John's, NewfoundlandCANADA A1C 5S7 ph. (709) 737-8075 fax (709) 737-3010

Alwell Julius Oyet, Phd email: [email protected]

STATISTICS FOR PHYSICAL SCIENCES - LECTURENOTES

2. PROBABILITY

There are several actions in life that have several possible outcomes. For instance, if I take a drug I

can get better, or worse or it can lead to other complications. If I take an exam, my grade will be A,

or B or C or D or E or F . However before I take the exam., I cannot say for sure what the outcome

of the examination will be. Although at the back of my mind or mentally, I may have a vague idea

of the chance of making each of the grades. For instance, I may feel that I can never fail the course.

So, in numerical terms, I am actually saying that my chance of failing is 0%. I may sense that I have

a very good chance of making an A in the course. So, I assign a chance of 70% to my making an

A. The remaining 30% is then distributed amongst the other grades depending on the information I

have about the course. After the exam has been taken and before the result is published I may then

be able to revise the values assigned to my chances of making each of the grades. These updated

values may then be more accurate than the previous values because I now have more information

about the exam. Once the result is release, the grade I made will now be the observed value of my

action.

These values which represent the chance of making a particular grade are what we will refer to as

probability. In quality control departments, the decision to accept or reject a batch of raw materials

is usually based on the probability that the batch contains an unacceptable number of defective

items. If that probability is very high based on a technique called acceptance sampling, the batch

is rejected and accepted otherwise. So, probability is a measure of the degree of con¯dence we have

in the occurence of an outcome or group of outcomes of an action. It is a measure of uncertainty.

Under the topic of probability, we will study random and uncertain events and attempt to assign

speci¯c numbers to the chance that they will occur.

The di±culty in being certain of the outcome of an action or experiment arises from the fact that

the outcomes occur in a completely random manner and there are several factors we cannot control

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that can in°uence the outcome of an action. Sometimes, some particular outcomes of an action

or experiment may have very signi¯cant impact on our lifes, positive or negative. For instance, in

August 2000 I said to an American Statistician whom I met at a conference that I will never °y

American Airlines because of my assessment of the very high risk of an attack against the airline.

For me, the chance of an attack against AA compared to Air Canada or British Airways was so

high that I was not willing to take the risk. He °ew Korean Airline while I °ew Air Canada and

Japan Airline. The September 11 attacks had very signi¯cant impact on everybody. If there was a

person like me travelling on that date, the choice of airline based on the chance of an attack (risk

assessment) may have saved their life. My assessment was based on information on the actions of the

US government around the world and the nature of its relationship with groups around the world.

I am certain that most of us are taking this course today after an assessment of our chance of passing

this course. You are here because you feel you have a fair or a very good chance of passing. As

you gather more information about the course, twice a week, you also mentally review your chance

of passing the course. If at some point you feel that your chances are not good you will then take

action (e.g. spend more time on it or move to drop the course) to correct the problem.

Thus based on probability (chance) we are able to make informed decisions on a daily basis whether

we recognise it or not. It therefore seems appropriate to start our statistics course with a study of the

concept of probability. Some ¯nd this concept very di±cult to follow because we are called upon to

use our imagination for risk assessment and because of the language that is normally used to discuss

the subject. In this course, we will lay the foundation necessary for understanding the subject of

statistics and its application in the real world.

We begin our study by establishing notations and de¯ning terms that will be needed.

x2.1 Sample Spaces and Events

De¯nition (Experiment): An Experiment is any action or process that generates observations

(This de¯nition is su±cient for our purposes in this class).

Experiment: Prof. delivers lecture.

Outcomes: \Boring" Lecture, \Interesting" Lecture.

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Experiment: Wait for metrobus.

Outcomes: Waiting time(in mins.): 0, 0.5, 0.58, 1, 2, 3,¢ ¢ ¢

As you can see, we cannot list all the outcomes of certain experiments. What I want to know is the

chance of waiting for 0, 1, 2, 3, ¢ ¢ ¢ mins. so that I can plan my trip and not wait for too long - evenif I can't list the entire outcomes. In the ¯rst part of this course, we will concentrate on studying

techniques for calculating these probabilities.

De¯nition (Sample Space): The set of all possible outcomes of an experiment, denoted by S, iscalled the sample space of that experiment.

Example 1.2: If a new type-D °ashlight battery has a voltage that is outside certain limits, that

battery is characterized as a failure (F); if the battery has a voltage within the prescribed limits, it

is a success (S). Suppose an experiment consists of testing each battery as it comes o® an assembly

line until we ¯rst observe a success. What are the possible outcomes of this experiment ?

Solution: First battery examined can be a success - S;

First, a failure; Second, a success - FS;

First, a failure; Second, a failure; Third, a success - FFS; and so on.

Therefore S = fS, FS, FFS, FFFS,¢ ¢ ¢g.

Example 1.3: PROBLEM 2, PAGE 57

Suppose that vehicles taking a particular freeway exit can turn right (R), turn left (L), or go straight

(S). Consider observing the direction for each of three successive vehicles.

(a) List the possible outcomes of this experiment.

(b) List all outcomes with the property that all three vehicles take di®erent directions.

(c) List all outcomes with the property that exactly two vehicles go in the same direction.

Solution: (a) Using a tree diagram, it is easy to ¯nd that

S = f(R,R,R), (R,R,L), (R,R,S), (R,L,R), (R,L,L), (R,L,S), (R,S,R), (R,S,L), (R,S,S),(L,R,R), (L,R,L), (L,R,S), (L,L,R), (L,L,L), (L,L,S), (L,S,R), (L,S,L), (L,S,S),

(S,R,R), (S,R,L), (S,R,S), (S,L,R), (S,L,L), (S,L,S), (S,S,R), (S,S,L), (S,S,S)g.

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(b) A = f(R,L,S), (R,S,L), (L,R,S), (L,S,R), (S,R,L), (S,L,R)g

(c) B = f(R,R,L), (R,R,S), (R,L,R), (R,L,L), (R,S,R), (R,S,S), (L,R,R), (L,R,L), (L,L,R),(L,L,S), (L,S,L), (L,S,S), (S,R,R), (S,R,S), (S,L,L), (S,L,S), (S,S,R), (S,S,L)g

Example 1.3 clearly illustrates the fact that sometimes, we are not just interested in the individual

outcomes of an experiment but rather in outcomes with a certain desirable property. For instance,

the property of interest in part (b) is that the vehicles turn in diferent directions. Notice that A andB are both subsets of the sample space S. When this happens, we say that A and B are events ofthe experiment. Events can be represented graphically in what is called a Venn diagram.

De¯nition: An event is any collection or subset of outcomes contained in the sample space of an

experiment.

Types of Events

(a) Simple Events: Is any event containing only one outcome.

(b) Compound Events: Is any event containing more than one outcome.


Each of a sample of four home mortgages is classi¯ed as ¯xed rate (F) or variable rate (V).

(a) What are the 16 outcomes in S ?(b) Which outcomes are in the event E1 that exactly three of the selected mortgages are ¯xed rates ?(c) Which outcomes are in the event E2 that all four mortgages are of the same type ?(d) Which outcomes are in the event E3 that at most one of the four is a variable rate mortgage ?

Solution: Let VFFV represent ¯rst mortgage is variable rate; second is ¯xed rate; third is ¯xed and

fourth is variable rate. Then,

(a) S = fFFFF, FFFV, FFVV, FVVV, VVVV, VVVF, VVFF, VFFF, VFFV, VFVV, VVFV,FFVF, FVVF, FVFV, VFVF, FVFFg(b) E1 = fFFFV, VFFF, FFVF, FVFFg(c) E2 = fFFFF, VVVVg(d) E3 = fFFFF, FFFV, VFFF, FFVF, FVFFg.

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Suppose, we perform the experiment in Example 1.5 by taking one sample of four home mortgages

and ¯nd that the ¯rst was a variable rate mortgage (V); the second, third and fourth were ¯xed

(F) (i.e VFFF). Since the sample VFFF belongs to E1 and E3, we say that events E1 and E3 haveoccurred. When an experiment is performed, a particular event E is said to have occurred if theresulting outcome is an element of E .

Question: Can two simple events occur simultaneously ?


A college library has ¯ve copies of a certain text on reserve. Two copies (1 and 2) are ¯rst printings,

and the other three (3, 4 and 5) are second printings. A student examines these books in random

order, stopping only when a second printing has been selected. One possible outcome is 5, and

another is 213.

(a) List the outcomes in S.(b) Let A denote the event that exactly one book must be examined. What outcomes are in A ?

(c) Let B be the event that book 1 is not examined. What outcomes are in B ?

Solution:

(a) S = f3, 4, 5, 13, 14, 15, 23, 24, 25, 123, 124, 125, 213, 214, 215g(b) A = f3, 4, 5g(c) B = f5, 15, 25, 125, 215g(d) C = f3, 4, 5, 23, 24, 25g

Some Relations From Set Theory: Observe that both the sample space of an experiment and

events are sets. So, we adopt some operations of set theory in studying events.

Union: The union of A and B denoted by A [ B is the event consisting of all outcomes that areeither in A or in B or in both A and B (without repeating any outcomes).

Example: Consider the experiment described in Problem 6, Page 57. Here, A [ B= f3,4,5,15,25,125,215g;B [ C = f3,4,5,15,23,24,25,125,215g.

Intersection: The intersection of A and B denoted by A \ C is the event consisting of all outcomesthat are in both A and B.

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Example: For the experiment in Problem 6, Page 57, A \ B = f5g; B \ C = f5g and A \ C =f3,4,5g.

Compliment: The compliment of the event A denoted by A0 is the event consisting of all outcomes

in S that are not in A.

Example: For the experiment in Problem 6, Page 57, C0 = f13,14,15,123,124,125,213,214,215g

Example 1.7: Refer to PROBLEM 2, PAGE 57.

Solution:

B0 = f(R,R,R), (R,L,S), (R,S,L), (L,R,S), (L,L,L), (L,S,R), (S,R,L), (S,L,R), (S,S,S)g.

A [ B = f(R,R,L), (R,R,S), (R,L,R), (R,L,L), (R,L,S), (R,S,R), (R,S,L), (R,S,S),(L,R,R), (L,R,L), (L,R,S), (L,L,R), (L,L,S), (L,S,R), (L,S,L), (L,S,S),

(S,R,R), (S,R,L), (S,R,S), (S,L,R), (S,L,L), (S,L,S), (S,S,R), (S,S,L)g.

A \ B = f g or ; - the empty set.

The events A and B are said to be mutually exclusive or disjoint because they have no commonoutcomes or their intersection is empty. Note that these operations apply to more than two events.

x2.2 Axioms and Properties of Probability

So far, we have only studied how to construct sample spaces of experiments and events. Beginning

from this section, we will study how to assign a number P (A), to an event A. This number P (A)

measures the chance that A will occur. First, we state some basic properties, called axioms, which

all assigned numbers P (A) must satisfy, for consistency.

Axiom 1: For any event A, P (A) ¸ 0. That is, negative probabilities are not allowed.

Axiom 2: P (S) = 1. That is, the sample space is an event that must occur when the experimentis performed.

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Axiom 3: If A1; A2; ¢ ¢ ¢ ; An is a ¯nite collection of mutually exclusive events, then

P (A1 [A2 [ ¢ ¢ ¢ [An) = P (A1) + P (A2) + ¢ ¢ ¢+ P (An) =nXi=1

P (Ai):

If A1; A2; An; ¢ ¢ ¢ is an in¯nite collection of mutually exclusive events, then

P (A1 [ A2 [An [ ¢ ¢ ¢) = P (A1) + P (A2) + P (A3) + ¢ ¢ ¢ =1Xi=1

P (Ai):

Properties of Probability

For any event A,

P (A) = 1¡ P (A0):If A and B are mutually exclusive events, then

P (A \B) = 0:

For any two events, A and B,

P (A [B) = P (A) + P (B)¡ P (A \B):

Examples

1. Problem 13, Pg 65

A computer consulting ¯rm presently has bids out on three projects. Let Ai = fawardedprojectig, for i = 1; 2; 3, and suppose that P (A1) = 0:22, P (A2) = 0:25, P (A3) = 0:28,P (A1 \A2) = 0:11, P (A1 \A3) = 0:05, P (A2 \A3) = 0:07, P (A1 \A2 \A3) = 0:01. Computethe probability of each event.

(a) A1 [A2 (the probability that at least one of the ¯rst two projects was awarded).(b) The probability that neither of the ¯rst two projects was awarded.

(c) The probability that at least one of the three projects was awarded.

Solution

(a) P (A1 [A2) = P (A1) + P (A2)¡ P (A1 \A2) = .22 + .25 - .11 = .36.(b) P (A01 \ A02) = 1¡ P (A1 [ A2) = 1 - 0.36 = 0.64.(c) P (A1 [ A2 [ A3) = P (A1) + P (A2) + P (A3) ¡ P (A1 \ A2) ¡ P (A1 \ A3) ¡ P (A2 \ A3) +P (A1 \A2 \ A3) = .22 + .25 + .28 - .11 - .05 - .07 + .01 = 0.53This example is an illustration of the fact that the properties of probability which we stated

for two events can be extended to more than two events.

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2. Problem 15, Page 65

Consider the type of clothes dryer (gas or electric) purchased by each of ¯ve di®erent customers

at a certain store.

(a) If the probability that at most one of these purchases an electric dryer is 0.087, what is the

probability that at least two purchase an electric dryer ?

(b) If P (all ¯ve purchase gas) = 0.0768 and P (all ¯ve purchase electric) = 0.0102, what is the

probability that at least one of each type is purchased ?

Solution

(a) Let A = fat most one of ¯ve purchases an electric dryerg; B = A0 = fat least two purchasean electric dryerg. Then, P(A) = P(0 or 1 customer will purchase a dryer) = 0.087. Theprobability of interest is

P(B) = P(A0) = P(at least two will purchase an electric dryer)

= P(2 or 3 or 4 or 5 will purchase a dryer).

The experiment here is noting the number of customers that buy clothes dryer at the store. So,

the sample space S = f0,1,2,3,4,5g. Observe that S = A [ B = f0; 1g [ f2; 3; 4; 5g = A [ A0.Now,

P (S) = P (A) + P (A0) = 1: So;P (B) = P (A0) = 1¡ P (A) = 1¡ 0:087 = 0:913:

(b) At least one of each type is purchased means that all ¯ve will not purchase a gas dryer and

all ¯ve will not purchase an electric dryer. Now, if A = fall ¯ve purchase gasg and B = fall¯ve purchase electricg, then the required probability isP (A0 \B0) = 1¡ P (A [B) = 1¡ P (A)¡ P (B) = 1 - 0.087 = 0.913.

3. Problem

A family that owns two cars is selected at random. Let A1 = fthe older car is Canadiang andA2 = fthe newer car is Canadiang. If P(A1) = 0.7, P(A2) = 0.5 and P(A1\A2) = 0.4, computethe following:

a. P(A1 [A2) (the probability that at least one car is Candian)b. The probability that neither car is Canadian.

c. The probability that exactly one of the two cars is Canadian.

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Solution

a. Using the Addition Law,

P (A1 [A2) = P (A1) + P (A2)¡ P (A1 \A2)= 0:7 + 0:5¡ 0:4 = 0:8:

b. P(neither car is Canadian) = P(not A1 and not A2) = P(A01 \A02).

Since no formula exist for this probability, we use a venn diagram to develop one. Based on

the venn diagram,

P (A01 \ A02) = 1¡ P (A1 [ A2)= 1¡ 0:8 = 0:2:

c. P(exactly one car is Canadian) = P[(A1 and not A2) or (A2 and not A1)]. Based on a venn

diagram, we have that

P [(A1 \A02) [ (A01 \A2)] = P (A1 [ A2)¡ P (A1 \ A2)= 0:8¡ 0:4 = 0:4:

4. Problem

A construction ¯rm is working on two di®erent projects. Let A be the event that the ¯rst one is

completed by the contract date and de¯ne B analogously for the second project. If P(A1 [A2)= 0.9 and P(A1 \ A2) = 0.5, what is the probability that exactly one project is completed bythe contract date ?

Solution

The probability of interest is

P(exactly one project is completed) = P[(A1 and not A2) or (A2 and not A1)].

Since there is no formula for calculating this probability, we will use a venn diagram of the

sample space and events to develop a formula. Based on the venn diagram, the probability of

interest is

P [(A1 \A02) [ (A01 \A2)] = P (A1 [ A2)¡ P (A1 \ A2)= 0:9¡ 0:5 = 0:4:

5. Problem

A video store sells two di®erent brands of VCRs, each of which comes with either two heads

or four heads. The accompanying table gives the percentages of recent purchasers buying each

type of VCR:

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Number of Heads

Brand 2 4M 25% 16%Q 32% 27%

Suppose a recent purchaser is randomly selected and both the brand and the number of heads

are determined.

a. What are the four simple events ?

b. What is the probability that the selected purchaser bought brand Q, with two heads ?

c. What is the probability that the selected purchaser bought brand M ?

Solution

a. S = fM2;M4; Q2; Q4g.b. P(purchaser bought brand Q, with two heads) = P(Q2) = 0.32.

c. P(purchaser bought brand M) = P(M2 or M4)

P (M2 or M4) = P (M2 [M4)= P (M2) + P (M4)

= 0:25 + 0:16 = 0:41:

Note the strategy we adopt in solving these problems

(1) Write down the information provided in the problem.

(2) Write down the probability you are asked to calculate.

(3) Obtain a formula based on known formulae and a venn diagram for computing (2).

(4) Use the formula to calculate the required probability.

This approach may not work all the time but it is very useful - going from the known to the unknown.

The Relative Frequency Interpretation of Probability

Consider the experiment of tossing a balanced two-sided coin under identical and independent con-

ditions. Independent, in the sense that the outcomes of previous trials does not in°uence, in any

way, the outcome of current or future trials. Let A be the event that the outcome in a particular

toss is the tail. If we toss the coin repeatedly n number of times, the outcome will be tails sometimes

and heads at other times. Let n(A) be the number of times the tail turns up. Then the ratio n(A)n-

the proportion of tails in n trials - is called the relative frequency of the event A. As the number of

trials n becomes larger and larger, the relative frequency n(A)nwill tend to stabilize to a particular

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value called the limiting relative frequency of the event A. It is this limiting relative frequency that

we interpret as the probability of the event A, P (A). In practice, this limiting relative frequency of

an event may not be known. Thus, we will have to assign probabilities based on our beliefs about

the limiting relative frequency of events under study.

Examples

1. Experiment: toss a fair two-sided coin

Limiting relative frequency interpretation suggests assigning P(H) = 0.5 and P(T ) = 0.5. That

is, if the number of times the experiment is performed is large enough, the limiting relative

frequency of head and tail is 50%.

2. Experiment: toss a fair six-sided die

Limiting relative frequency interpretation suggests assigning P (1) = P (2) = ¢ ¢ ¢ = P (6) = 16.

That is, if the number of times the experiment is performed is large enough, the limiting relative

frequency of 1, 2,¢ ¢ ¢, 6 is 1=6 for each of the six possible outcomes.

3. Problem 16, page 65

An individual is presented with three di®erent glasses of cola, labeled C, D, and P. He is asked

to taste all three and then list them in order of preference. Suppose that the same cola has

actually been put into all three glasses.

(a) What are the simple events in this ranking experiment, and what probability would you

assign to each one ?

(b) What is the probability that C is ranked ¯rst ?

(c) What is the probability that C is ranked ¯rst and D is ranked last ?

Solution

(a) Using a tree diagram, it is easy to ¯nd that the sample space of the experiment is

fCDP,CPD,DCP,DPC,PCD,PDCg. P (Ei) = 1=6, (i = 1; : : : ; 6), where Ei is any simple eventin the sample space.

(b) Let A = fC is ranked ¯rstg = fCDP,CPDg. Then P (A) = 2=6.(c) Let B = fC is ranked ¯rst and D is ranked lastg = fCPDg. Then P (B) = 1=6.

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Equally Likely Outcomes

Consider an experiment with N possible outcomes. That is, S = fO1; O2; : : : ; ONg. Let Ei, (i =1; 2; ¢ ¢ ¢ ;N) be the ith simple event (i:e event with only one outcome) of the experiment. That is,E1 = fO1g, E2 = fO2g, and so on. If

P (Ei) =1

N; i = 1; 2; ¢ ¢ ¢ ; N;

then each outcome of the experiment is said to be equally likely. That is, the outcomes are equally

likely if they have equal probability of occurence.

Now, let A be a compound event of an experiment with N equally likely outcomes. Let N(A) be the

number of outcomes contained in A. Then,

P (A) =N(A)

N:

For instance, if A = fO3; O6; O8g, then

P (A) =N(A)

N=3

N:

Examples


Human visual inspection of solder joint on printed circuit boards can be very subjective. Con-

sequently, even highly trained inspectors can disagree on the disposition of a particular joint. In

one batch of 10,000 joints, inspector A found 724 that were judged defective, inspector B found

751 such joints, and 1159 of the joints were judged defective by at least one of the inspectors.

Suppose that one of the 10,000 joints is randomly selected.

(a) What is the probability that the selected joint was judged defective by neither of the two

inspectors ?

(b) What is the probability that the selected joint was judged to be defective by inspector B

but not by inspector A ?

Solution

Let A = fjudged defective by inspector Ag and B = fjudged defective by inspector Bg. Now,N = 10000;

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N(A) = number of joints judged defective by A = 724;

N(B) = number of joints judged defective by B = 751;

N(A [B) = number of joints judged defective by at least one of the two inspectors = 1159.(a) The required probability is P (A0 \B0) = 1¡ P (A [B).Now, P (A [B) = N(A[B)

N(S) = 0:1159. Therefore, P (A0 \B0) = 1¡ 0:1159 = 0:8841.

(b) The required probability is P (A0 \ B) = P (B) ¡ P (A \ B) = P (B) ¡ [P (A) + P (B) ¡P (A [B)] = P (A [B)¡ P (A) = 0.1159 - 0.0724 = 0.0435.

2. If a standard deck of 52 cards (spades, hearts, diamonds and clubs; with ace, 2,....,10, jack,

queen, and king in each suit) is well mixed before a single card is drawn, all 52 outcomes are

equally likely. De¯ne events A = fheartg, B = fblack cardg, C = fjack, queen, or kingg andD = fselected card is at most a 5g and use elementary counting along with the axioms andproperties to compute the following probabilities.

a. P(A), P(B), P(C), and P(D).

b. P(A [B) and P(A [ C).c. P(A [B [ C [D).Solution

a. Since the outcomes are equally likely with N = 52,

N(A) = number of hearts = 13,

N(B) = number of black cards = 26,

N(C) = number of jacks, queens and kings (4 each) = 12,

N(D) = number of cards between ace and 5 = 20, then

P (A) =N(A)

N=13

52=1

4; P (B) =

N(B)

N=26

52=1

2;

P (C) =N(C)

N=12

52=3

13; P (D) =

N(D)

N=20

52=5

13:

b. From part (a) and the addition law,

P (A [B) = P (A) + P (B); since P (A \B) = 0;=

1

4+1

2=3

4:

P (A [ C) = P (A) + P (C)¡ P (A \ C);where P (A \ C) = P (jack, queen and king of hearts) = 3

52;

Therefore, P (A [ C) = 1352+12

52¡ 352=22

52:

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Alternatively, note that N(A[C) = number of hearts, jacks, queens and kings in suit withoutrepetition. = N(A) + N(C) - number of jack, queen and king of hearts = 13 + 12 - 3 = 22.

Therefore,

P (A [ C) = N(A [ C)N

=22

52:

c. First, we simplify the expression P (A [ B [ C [D) to the form we can recognize. To dothis, we let E = A [B and F = C [D. Then, using the addition law, we have

P (A [B [ C [D) = P (E [ F ) = P (E) + P (F )¡ P (E \ F )= P (A [B) + P (C [D)¡ P [(A [B) \ (C [D)]= P (A) + P (B) + P (C) + P (D)¡ P [(A [B) \ (C [D)]=

13

52+26

52+12

52+20

52¡ P [(A [B) \ (C [D)]

Note that (A [B) = fheart, black cardg; (C [D) = fjack, queen, or king, selected card is atmost a 5g. So that, (A[B)\(C[D) = fjack, queen and king of heart, black jack, black queenand black king, heart and black cards that are at most 5g. So, N [(A [B) \ (C [D)] = 24. Itfollows that

P (A [B [ C [D) = 1352+26

52+12

52+20

52¡ 2452=47

52:

An alternative and simpler approach is to note that the only outcomes not in A [B [ C [Dare the 6, 7, 8, 9 and 10 of diamonds. So, using the fact that P (A) = 1¡ P (A0) we have

P (A [B [ C [D) = 1¡ 552=47

52:

x2.3 Counting Techniques

We have seen that if the N outcomes of an experiment are equally likely, then the probability of a

compound event A is simply

P (A) =N(A)

N;

where N(A) = the number of outcomes in A. That means, we need to be able to count the number of

outcomes in S and in A. Sometimes, this is not possible because of the complexity of the experimentor the outcomes are too many for us to be able to list them before counting. So, it is important for

us to develop or study some rules for counting the number of outcomes in any event of an experiment

without necessarily listing the outcomes. The ¯rst of these rules is the product rule for ordered pairs.

14

METHOD I: Product Rule for Ordered Pairs

De¯nition: By an ordered pair, we mean that if, O1 and O2 are objects, then the pair (O1; O2) is

di®erent from the pair (O2; O1). We will use an example to illustrate this rule before stating it.

Examples:


(a) Beethoven wrote 9 symphonies and Mozart wrote 27 piano concertos. If a university radio

station announcer wishes to play ¯rst a Beethoven symphony and then a Mozart concerto, in

how many ways can this be done ? In other words, what is the number of elements N in the

sample space of this experiment ?

Solution:

This experiment involves playing a Beethoven, then a Mozart concertos. Note that there is an

ordering involved - Beethoven ¯rst before a Mozart. So, (M1; B1) is di®erent from (B1;M1).

In fact (M1; B1) does not qualify to be an outcome of this experiment. The problem is to

determine the number of outcomes in the sample space of the experiment.

Let n1 = 9, n2 = 27. Then, from the tree diagram, this number is

N = n1 £ n2 = 9£ 27 = 243.Note that in this example, the selection is made from two di®erent sets. The ¯rst element must

come from the set of Beethovens and the second from the set of Mozarts. This is an important

component of any problem which requires the application of the product rule in its solution.

We now state the product rule.

Product Rule for Ordered Pairs:

If the ¯rst element or object of an ordered pair can be selected in n1 ways, and for each of

these n1 ways the second element of the pair can be selected in n2 ways, then the total number

of pairs is N = n1 £ n2.

We use part (b) of Problem 31, Page 73 to show that the product rule can be extended to an

ordered sequence consisting of more than two objects (e.g. an ordered sequence of 3 objects).

Such ordered sequences consisting of k objects is called a k-tuple. A 2-tuple is an ordered

15

pair; A 3-tuple an ordered sequence of 3 objects or elements; etc. Suppose, I'm °ying from St.

John's to Edmonton. The sequence of my trip could be (St. John's, Toronto, Edmonton) and

my return trip is (Edmonton, Toronto, St. John's). These two ordered sequences of 3 cities are

not the same although they contain the same cities.

Problem 31, Page 73

(b) The station manager decides that on each successive night (7 days per week), a Beethoven

symphony will be played, followed by a Mozart piano concerto, followed by a Schubert string

quartet (of which there are 15). For roughly how many years could this policy be continued

before exactly the same program would have to be repeated ?

Solution:

N = n1 £ n2 £ n3 = 9£ 27£ 15 = 3645:

Since there are approximately 365 days in a year, the policy can be continued for 3645365

¼ 10yearswithout repeating exactly the same program.

2. A homeowner doing some remodelling requires the services of both a plumbing contractor and

an electrical contractor. If there are 12 plumbing contractors and 9 electrical contractors avail-

able in the area, in how many ways can the contractors be chosen ?

Solution: Denote the plumbers by P1; ¢ ¢ ¢ ; P12 and the electricians by E1; ¢ ¢ ¢ ; E9. Then weseek the number of pairs of the form (Pi; Ej), (i = 1; ¢ ¢ ¢ ; 12; j = 1; : : : ; 9). From the treediagram, let n1 = 12, n2 = 9, then N = 12£ 9 = 108.

METHOD II: Permutation

Consider a ¯xed set A consisting of 4 distinct objects:

A = fa; b; c; dg

Suppose that we form a 3-tuple by selecting successively from the set A without replacement so that

an element can appear in at most one of the 3 positions. The question is, how many 3-tuples can we

construct from set A ?

16

Using a tree diagram, we see that the ¯rst element can be selected in 4 ways; for each of the 4 ways

in which the ¯rst can be selected, the second element can be selected in 3 ways and the third in 2

ways. So, the total number of 3-tuples possible is 4£ 3£ 2 = 24.

De¯nition: Any ordered sequence of k objects taken from a set of n distinct objects is called a

permutation of size k of the objects. The number of permutations of size k that can be constructed

from the n objects is denoted by Pk;n.

In general, suppose that A contains n distinct objects or elements. How many k-tuples can we con-

struct from the set A ?

As before, 1st element can be chosen in n ways;

2nd element in n¡ (2¡ 1) = n¡ 1 ways;3rd element in n¡ (3¡ 1) = n¡ 2 ways; and so onThe kth element in n¡ (k ¡ 1) ways.Therefore, the total number of k-tuples we can construct from the set A is

Pk;n = n£ (n¡ 1)£ (n¡ 2)£ ¢ ¢ ¢ £ (n¡ k + 1) = n!(n¡ k)!

Examples


A real estate agent is showing homes to a propective buyer. There are ten homes in the desired

price range listed in the area. The buyer has time to visit only three of them.

(a) In how many ways could the three homes be chosen if the order of visiting is considered ?

Solution

(a) The problem here is to select three homes to visit in a set consisting of ten homes H =

fh1; h2; : : : ; h10g without replacement - order being important. Thus, n = 10 and k = 3.Implying that

Pk;n = P3;10 =10!

(10¡ 3)! = 10£ 9£ 8 = 720:

(b) In how many ways could the three homes be chosen if the order is disregarded ?

This question requires that we use a method called combination.

17

METHOD III: Combination

We have ten houses in the set H = fh1; h2; : : : ; h10g. Now we wish to select 3 elements fromH without replacement. If order were to be important, we would have P3;10 = 720 possible

permutations of size 3 of the 10 homes in H. Since order is no longer important, the selection

(h1; h5; h8) ´ (h8; h1; h5):

So, the number of possible selections or arrangements will be smaller. When order is not

important, any k selection or arrangement of objects or elements from a set, say A, consisting

of n objects is called a combination. The number of combinations of size k that can be formed

from n distinct objects denoted byÃnk

!=

n!

k!(n¡ k)! :

Thus the solution to part (b) of the problem is as follows.

Solution:

(b) Here, n = 10 and k = 3Ãnk

!=

n!

k!(n¡ k)! =10!

3!(10¡ 3)! =10£ 9£ 83£ 2£ 1 = 120:

Next we use the idea of combination to answer part (c).

(c) If four of the homes are new and six have previously been occupied and if the three homes

to visit are randomly chosen, what is the probability that all three are new ?

Solution:

(c) In this problem the set of homes has been split into two disjoint sets A = fnew homesg; B= fused homesg. If E = fthe three selected homes are newg, then

P (E) =N(E)

N

where N = the total number of ways of selecting three homes from ten. Now, N(E) = n1£n2,where n1 = total number of ways of selecting three new homes from four; n2 = total number

of ways of selecting zero used homes from six. Thus,

P (E) =N(E)

N=n1 £ n2N

=

Ã43

!£Ã60

!Ã103

! = 4120

=1

30:

18

Alternatively, we can use the concept of permutation to solve the problem

P (E) =N(E)

N=n1 £ n2N

=P3;4 £ P0;6P3;10

=24

720=1

30:

In solving this problem, we have used the product rule and combinations or permutations.


The student Engineering Council at MUN has one student representative from each of the ¯ve

engineering majors (civil, electrical, industrial, materials and mechanical). In how many ways

can both a council president and a vice-president be selected ?

Solution The problem is to ¯ll two positions by selecting from a set consisting of ¯ve depart-

ment representatives without replacement. Thus, n = 5 and k = 2. Implying that

Pk;n = P2;5 =5!

(5¡ 2)! = 5£ 4 = 20:


Shortly after being put into service, some buses manufactured by a certain company have

developed cracks on the underside of the main frame. Suppose a particular city has 20 of these

buses, and cracks have actually appeared in 8 of them.

(a) How many ways are there to select a sample of 5 buses from the 20 for a thorough inspection

?

(b) In how many ways can a sample of 5 buses contain exactly 4 with visible cracks?

(c) If a sample of 5 buses is chosen at random, what is the probability that exactly 4 of the 5

will have visible cracks ?

(d) If buses are selected as in part (c), what is the probability that at least 4 of those selected

will have visible cracks ?

4. A rental car service facility has 10 foreign cars and 15 domestic cars waiting to be serviced on

a particular Saturday morning. Because there are so few mechanics working on Saturday, only

six can be serviced. If the six are chosen at random, what is the probability that

a. 3 of the cars selected are domestic and the other three are foreign ?

b. at least 2 domestic cars are selected ?

19

Solution:

a. Again, let A = fforeign carsg; B = fDomestic carsg; E the event f3 selected cars aredomestic; 3 are foreign g. As before,

P (E) =N(E)

N

where N total number of ways of selecting or arranging 6 cars out of 25. Now, N(E) = n1£n2,where n1 = total number of ways of selecting 3 domestic cars from 15; n2 = total number of

ways of selecting 3 foreign cars from 10. Thus,

P (E) =N(E)

N=n1 £ n2N

=

Ã153

!£Ã103

!Ã256

! = 78253

= 0:3083:

b. Note that, P (at least 2 domestic cars are selected) = P (2, 3, 4, 5, or 6 domestic cars are

selected) = 1 - P (0 or 1 domestic car is selected). Now, P (0 or 1 domestic car is selected) =Ã150

!£Ã106

!Ã256

! +Ã151

!£Ã105

!Ã256

! = 3990Ã256

! = 0:0225296:

Therefore, P (at least 2 domestic cars are selected) = 1 - 0.0225296 ¼ 0.9774.

x2.4 Conditional Probability

The probability assigned to various events is based on information on the experiment that we have

when the assignment is made. Later on, we can revise the assigned probability if extra information

becomes available. Lets take for instance, that your friend visits from out of town to search for you

on Campus. His chance of ¯nding you will be very small because of the large area he/she has to

search. Lets assume that during the search, the information that you are attending a STAT 2510

class at that time becomes available. This extra information then improves his/her chance of ¯nding

you.

Let A = f¯nd a friend in MUN g;Let B = fin STAT 2510 classg.Then, P (A) is the unconditional probability of A.

P (AjB) is called the conditional probability of A given that B has occured.

20

With this extra information B, my sample space for the search reduces from the entire university

campus to the two sections of STAT 2510. So, conditioning on a given event has the e®ect of reducing

our sample space from a larger one to a smaller one.

Example

1. Consider the example on purchasers of brands of VCRS. Suppose the number of customers

purchasing di®erent brands can be summarized as follows.

Number of Heads

Brand 2 4M 3 2Q 4 5

Let A = fcustomer purchased brandM g; and B = fcustomer purchased a VCR with 2 headsg.Now,

P (A) =N(A)

N=5

14:

P (AjB) = 37=N(A \B)N(B)

=P (A \B)P (B)

=314714

:

In general, we have the following de¯nition.

De¯nition: For any two events A and B with P (B) > 0, the conditional probability of A given

that B has occurred is de¯ned by

P (AjB) = P (A \B)P (B)

:

Observe, from this de¯nition, that

P (A \B) = P (AjB)£ P (B):

This is called the multiplication rule for P (A \B).

Examples


At a certain gas station, 40% of the customers use regular unleaded gas (A1), 35% use extra

unleaded gas (A2), and 25% use premium unleaded gas (A3). Of those customers using regular

21

gas, only 30% ¯ll their tanks (B). Of those customers using extra gas, 60% ¯ll their tanks,

while of those using premium, 50% ¯ll their tanks.

(a) What is the probability that the next customer will request extra unleaded gas and ¯ll the

tank (A2 \B) ?(b) What is the probability that the next customer ¯lls the tank ?

(c) If the next customer ¯lls the tank, what is the probability that regular gas is requested ?

Extra gas ? Premium gas ?

Solution:

De¯ne A1 = fcustomer uses regular unleaded gasg; A2 = fcustomer uses extra unleaded gasg;A3 = fcustomer uses premium unleaded gasg; B = fcustomer ¯lls tankg. The events A1, A2and A3 are mutually exclusive and collectively exhaustive in the sense that A1 \A2 \A3 = fg(the empty set) and A1 [ A2 [ A3 = S. Now, P (A1) = 0:4; P (A2) = 0:35; P (A3) = 0:25;P (BjA1) = 0:3; P (BjA2) = 0:6; P (BjA3) = 0:5.(a) P (A2 \B) = P (BjA2)£ P (A2) = 0:6£ 0:35 = 0:21:(b)

P (B) = P (A1 \B) + P (A2 \B) + P (A3 \B)= P (BjA1)£ P (A1) + P (BjA2)£ P (A2) + P (BjA3)£ P (A3)= 0:3£ 0:4 + 0:6£ 0:35 + 0:5£ 0:25 = 0:455:

We generalize the solution to part (b). The generalization is called the law of total probability.

Law of Total Probability: Let A1, A2; ¢ ¢ ¢ ; An be a collection of mutually exclusive andexhaustive events. Then, for any other event B

P (B) = P (BjA1)£ P (A1) + P (BjA2)£ P (A2) + ¢ ¢ ¢+ P (BjAn)£ P (An)=

nXi=1

P (BjAi)£ P (Ai):

(c)

P (A1jB) = P (A1 \B)P (B)

=P (BjA1)£ P (A1)

P (BjA1)£ P (A1) + P (BjA2)£ P (A2) + P (BjA3)£ P (A3)=

0:12

0:455= 0:2637:

22

From the solution to part (c), we obtain a general formula or rule called Bayes' Theorem.

Bayes' Theorem: Let A1; A2; ¢ ¢ ¢ ; An be a collection of mutually exclusive and exhaustiveevents with P (Ai) > 0 for i = 1; ¢ ¢ ¢ ; n. Then, for any other event B for which P (B) > 0

P (AkjB) = P (Ak \B)P (B)

=P (BjAk)P (Ak)Pni=1 P (BjAi)P (Ai)

; k = 1; 2; ¢ ¢ ¢ ; n


A company that manufactures video cameras produces a basic model and a deluxe model. Over

the past year, 40% of the cameras sold have been of the basic model. Of those buying the basic

model, 30% purchase an extended warranty, whereas 50% of all deluxe purchasers do so. If

you learn that a randomly selected purchaser has an extended warranty, how likely is it that

he/she has a basic model ?

Solution: B = fcustomer purchases a basic modelg; D = fcustomer purchases a deluxemodelg; W = fcustomer purchases extended warrantyg. Now, P (B) = 0:4; P (D) = 0:6;P (W jB) = 0:3; P (W jD) = 0:5.

P (BjW ) = P (B \W )P (W )

=P (W jB)£ P (B)

P (W jB)£ P (B) + P (W jD)£ P (D)=

0:12

0:42= 0:2857:


Show that for any three events A, B and C with P (C) > 0, P (A[BjC) = P (AjC)+P (BjC)¡P (A \BjC).Solution:

P (A [BjC) = P [(A [B) \ C]P (C)

=P (A \ C) + P (B \ C)¡ P (A \B \ C)

P (C)

= P (AjC) + P (BjC)¡ P (A \BjC):


Seventy percent of the light aircraft that disappear while in °ight in a certain country are

subsequently discovered. Of the aircraft that are discovered, 60% have an emergency locator,

whereas 90% of the aircraft not discovered do not have such a locator. Suppose a light aircraft

has disappeared.

(a) If it has an emergency locator, what is the probability that it will not be discovered ?

23

(b) If it does not have an emergency locator, what is the probability that it will be discovered

?

Solution:

A = faircraft discoveredg; B = faircraft has emergency locatorg. Now, P (A) = 0:7; P (BjA) =0:6; P (B0jA0) = 0:6.a.

P (A0jB) = 1¡ P (AjB) = 1¡ P (A \B)P (B)

= 1¡ P (BjA)£ P (A)P (BjA)£ P (A) + P (BjA0)£ P (A0)

= 1¡ 0:6£ 0:70:6£ 0:7 + 0:1£ 0:3 = 0:067:

b.

P (AjB0) = P (A \B0)

P (B0)=P (B0jA)£ P (A)

1¡ P (B)=

0:28

0:55= 0:509:

x2.5 Independence

In the presence of extra information, we have seen that the unconditional probability of an event,

say P (A), is sometimes not equal to the conditional probability of A given that B has occurred

P (AjB). There are situations where P (A) = P (AjB). That is, the extra information that B hasoccurred does not a®ect the chance that A will occur. In such cases, we say that the events A and

B are independent. Thus the two events, A and B are said to be independent if

P (AjB) = P (A); P (B) > 0:

They are otherwise said to be dependent. From the multiplication rule, we have that

P (A \B) = P (AjB)£ P (B):

Implying that if A and B are independent

P (A \B) = P (AjB)£ P (B) = P (A)£ P (B):

24

Examples


An executive on a business trip must rent a car in each of two di®erent cities. Let A denote

the event that the executive is o®ered a free upgrade in the ¯rst city and B represent the

analogous event for the second city. Suppose that P (A) = 0:2, P (B) = 0:3, and that A and B

are independent events.

(a) If the executive is not o®ered a free upgrade in the ¯rst city, what is the probability of not

getting a free upgrade in the second city ?

(b) What is the probability that the executive is o®ered a free upgrade in at least one of the

two cities ?

(c) If the executive is o®ered a free upgrade in at least one of the two cities, what is the prob-

ability that such an o®er was made only in the ¯rst city ?

Solution:

Let A = fexecutive is o®ered free upgrade in city 1gand B = fexecutive is o®ered free upgrade in city 1g.Now, P (A) = 0:2 and P (B) = 0:3. Therefore, using independence we have that P (A \ B) =0:2£ 0:3 = 0:06.(a)

P (B0jA0) = P (B0 \A0)

P (A0)=1¡ P (A [B)1¡ P (A) :

Now,

P (A [B) = 0:2 + 0:3¡ 0:06 = 0:44:

Therefore,

P (B0jA0) = 1¡ 0:441¡ 0:2 = 0:7:

(b) From part (a), P (A [B) = 0:2 + 0:3¡ 0:06 = 0:44:(c)

P [(A \B0)j(A [B)] = P [(A \B0) \ (A [B)]

P (A [B) =P (A \B0)P (A [B)

=P (A)¡ P (A \B)

P (A [B) =0:14

0:44¼ 0:318:

25

2. Problem 72 Page 90

Suppose that the proportions of blood phenotypes in a particular population are as follows:

A B AB O

0.42 0.10 0.04 0.44

Assuming that the phenotypes of two randomly selected individuals are independent of one

another, what is the probability that

(a) both phenotypes are O ?

(b) both phenotypes match ?

Solution:

Let A = fPhenotype 1 is Og and B = fPhenotype 2 is Og.(a) P (A \B) = P (A)£ P (B) = 0:44£ 0:44 = 0:1936:(b) P (both phenotypes match) = P (both are A or both are B or both are AB or both are O)

= 0:42£ 0:42 + 0:1£ 0:1 + 0:04£ 0:04 + 0:44£ 0:44 = 0.3816.

3. Problem 78, Page 90.

Solution:

Let Ai = fcomponent i worksg, (i = 1; 2; 3; 4). So that

System works = (A1 or A2) or (A3 and A4):

P (System works) = P [(A1 [ A2) [ (A3 \A4)]= P (A1) + P (A2) + P (A3 \ A4)¡ P (A1 \ A2)¡ P (A1 \A3 \ A4)

¡P (A2 \A3 \A4) + P (A1 \ A2 \ A3 \A4)= 0:9 + 0:9 + 0:92 ¡ 0:92 ¡ 0:93 ¡ 0:93 + 0:94 = 0:9981:

26

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