PART IV : PROBABILITY - Binghamton Universitypeople.math.binghamton.edu/jbrennan/home/S13MAT148/Chapter13.pdf · blood types for a huge representative group of Australian people.

Chapter 13 & 14 - ProbabilityPART IV : PROBABILITY

Dr. Joseph Brennan

Math 148, BU

Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 1 / 91

Why Should We Learn Probability Theory?


Let’s Make a Deal!


Uncertainty

Usually the results of a study, observational or experimental, areuncertain: if we repeat a study, we will not get exactly the same results.

Example 1 (A coin). You toss a coin. Will it land heads or tails?

Example 2 (A die). A regular die, a D6, is a cube with six faces:

What number will a die show when it is rolled once?


Uncertainty

Example 3 (Box with marbles) A box contains 7 marbles: 2 red and 5green. Each marble has an equal chance to be selected. One marble is tobe drawn at random from the box. What color will it be?

Example 4 (Box with tickets) A box contains 10 tickets labeled 1through 10. What will be the number of a randomly selected ticket?

Example 5 (Height of students) Seven students will be selected atrandom from the Math 148 class list and their heights will be measured.What will be the average height? Will it change if we choose differentseven students?


Probability

What is the chance or probability that we will make the same conclusionsevery time when we replicate a study? What is the chance or probabilitythat the histogram will change?

Knowledge of probability theory will help us answer these questions.

NOTE: In this part, the words chance and probability will have the same

meaning.

chance = probability


Inferential Statistics & Probability Theory

In Parts II and III we used random samples to collect evidence and makeinferences about population.

Sample mean, x , estimates unknown population mean µ.

Sample standard deviation, s, estimates unknown population standarddeviation σ.

Regression equation is a mathematical model which approximates thetrue relationship between x and y .

In this part we will think in the opposite direction; we will reason from aknown population to randomly selected samples.

Population PopulationSample Sample

Probability Theory Inferential Statistics


Outcomes and Sample Space

Probability theory deals with studies where the outcomes are not knownfor sure in advance. Usually, there are many possible outcomes for a study,we just do not know which particular outcome we will observe.

Sample Space: The set of all possible outcomes of a study. Thesample space of a study is denoted by S .

Every repetition of a study, or a trial, produces a single outcome. Usuallyan outcome is computed from the values of the response variables.

In Example 5 (Height of students) the outcome is the average height, y ,

which is computed from the values of the response variable (y a student’s

height).

Conclusions from a study are based on its outcome.


Sample Spaces

Example 1 (A coin)

When you toss a coin, there are only two possible outcomes: a head or atail. Then the sample space is

S ={ head, tail} or simply S ={H, T}.

Example 2 (A die)

The sample space is S = {1, 2, 3, 4, 5, 6}.

Example 3 (Box with marbles)

We can draw either a red marble or a green one: S={R,G}.

Example 4 (Box with tickets) S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.


Sample Spaces

Example 5 (Height of students) Of all our current examples, example 5is the only continuous quantitative variable. The outcome of this studyis y , an average height. For any sample of 7 students the average heightcould be any number between 40 inches and 100 inches. Then the samplespace is the interval:

S = [40, 100].

Example 6 (Children)For every randomly chosen family the number of boys and girls is recorded.

The outcomes in the sample space are pairs of the form

(number of boys, number of girls)


Sample Space

Example 6 (Two dice)Two dice are thrown. The sample space contains 36 outcomes as shownbelow:

Mathematically, the sample space can be written as

S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), · · · , (6, 6)} .


Probability of Outcomes

Probability: The theoretical probability of an outcome is theproportion (or percent) of times an outcome occurs in the sample space.

Every outcome has a probability.

If an experiment is repeated a large number of times, one would expectthe ratio of occurrences of an outcome to number of experiments to beapproximately the probability of that outcome.

Notation: We will denote the probability of an outcome as

P(outcome),

where P(·) stands for probability.


Example 1 (A coin)

A single coin flip has a simple sample space: S = {H,T}. There are onlytwo potential outcomes: {H}, {T}.

P(H) =number of H in sample space

size of sample space=

1

2

P(T ) =number of H in sample space

size of sample space=

1

2

Is it correct to assume that after flipping 10 coins, 5 would have come upheads?

While imprisoned by the Germans during World War II, the SouthAfrican statistician John Kerrich tossed a coin 10, 000times. Headswas the outcome 5,067 times!

We have 67 more heads then expected. The difference between theobserved percentage and the anticipated result is known as error.


Example 1 (A coin)

The graph shows the percentage of heads minus 50% versus the numberof trials in Kerrich’s experiment.


Example 1 (A coin)

From the graph we can see that the error approaches 0 as the number oftosses increases. After 10,000 tosses there are 67 extra heads, so the erroris:

67

10, 000· 100% = 0.67%.

It is already very close to 0! If we continue Kerrich’s experiment, we canexpect the error to become even smaller. We say the error converges to0, which implies that the observed percentage of heads approaches50% which suggests that

P(H) = 0.5

has been calculated correctly.

Question Why is the proportion of heads in Kerrich’s experiment notEXACTLY equal to 0.5?


Example 1 (A coin)

Answer The number of heads is approximately half the number oftosses (5000), but it is off by 67 because of the chance error. Theerror 67 seems to be large in absolute units, but in relative units it isjust 0.67%, which is a very small error.

CONCLUSION: Kerrich’s experiment supports the theoretical

P(H) = 0.5. The observed number of heads may be interpreted as

number of heads = half the number of tosses + chance error,

where the error may appear to be large in absolute terms, but smallrelative to the number of tosses.


Example (Blood Distribution)

Data lifted from Wikipedia.

The following Table shows the distribution of Blood Types of people inAustralia.

O+ A+ B+ AB+ O- A- B- AB- Total

40% 31% 8% 2% 9% 7% 2% 1% 100%

The above distribution was obtained by computing the proportions ofblood types for a huge representative group of Australian people.Therefore, we can say that the numbers in the table are the probabilitiesof different blood types for Australians.


Example (Blood Distribution)

1 What is the probability of AB+ blood for Australians?

Answer: P(AB+) = 0.02

If we randomly sample 1000 Australians, approximately 20 of them areexpected to have AB+ blood type. However, due to chancevariability, we will not observe exactly 20 people with an AB+ bloodtype.

2 In July 2009 the population of Australia was 28,395,716 people.Approximately, how many people in Australia have an A+ bloodtype?

Answer: As the probability of having type A+ blood is 31%, the population,disregarding chance variability, of is

0.31 · 28, 395, 716 = 8, 802, 671.96 ≈ 8, 802, 672 people


Example 2 (A die)

An ace is the face of a die with one spot.

Question What is the probability of getting an ace in a single roll?

Answer:

P(ace) =number of faces with 1

number of faces=

1

6

Indeed, a die has 6 faces and every face has the same chance tobe observed in a single roll. Then

P(1) = P(2) = P(3) = P(4) = P(5) = P(6) =1

6.


Equally Likely Outcomes

Assigning correct probabilities to individual outcomes often requires longobservation of the random phenomenon. In some circumstances, however,we are willing to assume that individual outcomes are equally likelybecause of some balance in the phenomenon.

For the equally likely outcomes:

Probability of a single outcome =1

Number of outcomes in S



Ordinary coins have a physical balance that should make heads andtails equally likely. So the outcomes in Example 1 are equally likely:

P(H) = P(T ) =1

2

A fair die in Example 2 is equally likely to show any of its 6 faces ina single roll.

P(1) = P(2) = P(3) = P(4) = P(5) = P(6) =1

6



In Example 4, every one of the 10 tickets has the same chance to bedrawn:

P(1) = P(2) = P(3) = · · · = P(8) = P(9) = P(10) =1

10

In Example 6 (two dice), there are 36 equally likely outcomes. Theprobability of each outcome is 1

36 .

In Example 3, the outcomes of drawing a red marble and drawing agreen marble are not equally likely. There are more green marbles inthe box, so the probability of selecting a green marble is greater.What is this probability?


Computing Probabilities of Outcomes

RULE: In problems involving drawing at random, the probability ofgetting an object of a particular type in a single draw is equal toproportion of objects of this type in the population.


The population is the box with marbles. The proportion of red marbles is2/7, the proportion of green marbles is 5/7. This implies that in a singledraw

P(R) =2

7P(G ) =

5

7

Notice that P(R) + P(G ) = 1. We will explore this phenomenon later inthis chapter.


Events

We will give the definition of an event in words and mathematically.

Definition of an event in words: An event is some statement aboutthe random phenomenon. Events are usually denoted by capitalletters as A, B, etc.

Example of Events:(a) A = At least one head in two tosses of a coin,(b) B = Sum of the numbers on the dice is five,(c) C = A die rolled once shows an even number.

Mathematical definition of an event: An event is a set ofoutcomes from the sample space S .


Examples of Events

(a) A = At least one head in two tosses of a coin.The sample space for this experiment is S = {HH,HT ,TH,TT}.The event

A = {HH,HT ,TH}

(b) B =Sum of the numbers on the dice is five.The sample space for this experiment contains 36 equally likelyoutcomes which are discussed in Example 2. The event B is

B = { , , , }

(c) C = A die rolled once shows an even number.The sample space is S = {1, 2, 3, 4, 5, 6}. The event C is

C = { , , }


Occurrence of an Event

We say that an event has occurred if ANY of the outcomes thatconstitute it occur.

For instance, in Example (c) if we roll a die and it shows 4, we saythat the event C has occurred. If the die shows 3, the event A doesnot occur because 3 is an odd number, so the event C does notcontain this outcome.

Let A be the event of landing at least one heads when tossing a coin2 times, as in Example (a). If we toss and get {H,T} or {T ,H} or{H,H} then we’ll say that A has occurred. Only when we have twotails appearing does A not occur.


Probabilities of Events

The probability of an event A is denoted by P(A).

In all our examples, except for Example 6, the sample space consisted offinitely many outcomes which we could list. In this case the sample spaceis called finite. An event is a collection of outcomes which support it.

Theorem 1 (Probability of an event in a finite sample space)The probability of any event is the sum of probabilities of theoutcomes making up the event.

In the special case that the outcomes in the sample space are equallylikely, the probability of an event E is computed as

P(E ) =# of outcomes supporting E

Totat # of outcomes in S. (1)


Probabilities of Events

(a) A = At least one head in two tosses of a coin. There are 4 equallylikely outcomes in this experiment. The probability of every outcome

is1

4. Then

P(A)= P(HH) + P(HT ) + P(TH)=1

4+

1

4+

1

4=

3

4.

(b) B = Sum of the numbers on the dice is five. There are 36 equallylikely outcomes. Out of them, 4 outcomes support the event B.

P(B) =4

36=

1

9.

(c) C = A die rolled once shows an even number. There are 6 equallylikely outcomes, 3 of them support C .

P(C ) =3

6=

1

2.


Example 3 (Box with Marbles)

A box contains 7 marbles, 2 red and 5 green. A marble is drawn atrandom, its color is recorded, and the marble is put back in the box. Then,the next marble is drawn at random. This example is of sampling withreplacement, which we will define later.

Find the probability that 2 red marbles are drawn.

Solution: The sample space for this experiment is:

S = {RR,RG ,GR,GG}

The outcomes are not equally likely since the red and green marbles haveunequal chances to be selected. Intuitively, the RR is the least likelyoutcome. This outcome can be described:

A = First two randomly chosen marbles are red


Example 3 (Box with Marbles)

We can use the following two events to describe event A:

A1 = The first selected marble is redA2 = The second selected marble is red

Then it follows that A = A1 and A2. Note that if we replace the marbleback to the box, a red marble has the same chance to be selected at everydraw:

P(A1) = P(A2) =2

7

We have calculated P(A1) and P(A2), but how can we use this tocalculate P(A1 and A2)?


Special Events

We have the ability to calculate the probability of simple events; thoseresulting from one repetition of an experiment with equally likelyoutcomes. To tackle more complex problems, we must define some specialevents.

Some special events:

Case 1: Sure-to-happen or certain event.

Case 2: Impossible event.

Case 3: Opposite event.

Case 4: Mutually exclusive or disjoint events.

Case 5: Independent events.


Certain Event

Certain Event: An event which is guaranteed to happen at everyrepetition of the experiment.

The event A = A head or a tail in a single toss of a coin. In everytoss a coin lands up either heads or tails. There are no otherpossibilities. So A is a certain event.

Notice that A = {H, T} which coincides with the sample space S forthis experiment.

The event B = rolling a die once will show a number less than 7, is acertain event.

Notice that B = {1, 2, 3, 4, 5, 6} is the whole sample space.

A certain event is equal to the sample space.


Impossible Event

Impossible Event: An event which never can occur.

The event A = a coin lands up neither heads nor tails is an impossibleevent.

In all the coin-related experiment we usually assume that the coinmay not lay on the edge. Eliminating this possibility makes event Aimpossible.

The event B = the number that shows up when a die is rolled once is−2 is an impossible event.

Mathematically an impossible event is written as ∅, the empty set.


Opposite Event

Opposite Event: An event B is opposite to A if it happenswhenever A does not happen.

The event opposite to event A contains all the outcomes from thesample space S which do not belong to A.

The opposite event for event A is called not A event.

An opposite event for A is denoted as A.

A and A split the sample space into two parts. (Also called a partition)

Quite often, events are plotted on a graph as certain areas in thesample space. Such graphs are called the Venn diagrams.


Opposite Event on a Venn diagram

The Venn diagram shows the opposite event to the event A as the shadedregion.


Opposite Events: Examples

(a) Event A = At least one head in two tosses of a coin. The oppositeevent

A = It is NOT true that at least one H in two coin tosses

= No heads in two tosses of a coin

= {TT}

The probability is given by

P(A) = P(TT ) =1

4



(b) Event B = Sum of five when rolling two dice. The opposite event:

B = Sum of the numbers on the dice is NOT equal to five.

B = {(1,1), (1,2), (1,3), (1,5),(1,6), (2,1), (2,2), (2,4),(2,5), (2,6), (3,1), (3,3), (3,4), (3,5), (3,6), (4,2),(4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4),(5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}.

There are 32 equally likely outcomes in B. The sample space for thisexperiment contains 36 equally likely outcomes. This implies

P(B) =32

36=

8

9



(c) Event C = A die rolled once shows an even number.

Event C can be described via its outcome space as C = {2, 4, 6}.The sample space for this experiment is S = {1, 2, 3, 4, 5, 6}.Then the event C will contain all the outcomes from S that Cdoesn’t contain. In particular,

C = {1, 3, 5}

In words, C = A die rolled once shows an odd number. Theprobability P(C ) = 1

2 .

Rule: For an event A,

P(A) = 1− P(A).



A box contains 7 marbles: 2 red and 5 green. Two marbles are chosenwith replacement. We found the sample space S = {RR,RG ,GR,GG}.We reconsider the event

A =two red marbles in two draws with replacement = {RR}.

What is the opposite event?

It follows that A = {RG ,GR,GG} can be expressed as:

A = NOT exactly two red marbles in two draws

= At most one red marble in two draws

= Not all the marbles in two draws were red

= At least one green marble in two draws

P(A) =3

4= 1− 1

4= 1− P(A)


Mutually Exclusive Events

Mutually Exclusive Events: Two events A and B are mutuallyexclusive if they both cannot happen at the same time.

Examples of mutually exclusive events:

Events A = An odd numbered face showing and B = The 2 faceshowing are mutual events in a single die role.

Events A = The coin lands up heads and B = The coin lands up tailsare mutually exclusive events in a single coin flip experiment.

Events A and B are mutually exclusive if they do not have commonoutcomes. We say A and B do not overlap or do not intersect. For thisreason, mutually exclusive events are also called disjoint events.


Mutually Exclusive Events

A Venn diagram depicting two mutually exclusive events A and B in thesample space S .

Single outcomes of an experiment which make up the sample spaceare always mutually exclusive.

An event and its opposite are mutually exclusive: A and A aremutually exclusive.


Example (Two dice)

A pair of dice is rolled once. Consider the following events:

A = Sum of numbers on the dice is 11.B = Both dice show an even number.C = Both dice show the same number.

Are the events A and B mutually exclusive?

Are the events B and C mutually exclusive?

Let’s express the events as sets of outcomes and see if they have commonoutcomes.

A = { ; }B = { ; ; ; ; ; ; ; ; }C = { ; ; ; ; ; }.


Example (Two dice)

A = { ; }B = { ; ; ; ; ; ; ; ; }C = { ; ; ; ; ; }.

Events A and B are mutually exclusive since there are no commonoutcomes.

Events B and C are not mutually exclusive since they have threecommon outcomes.

What are the probabilities of events A, B, and C?


Intersection of Events

If A and B are not mutually exclusive events, they have some commonoutcomes. The set of common outcomes is an event which is called theintersection of events A and B. We will denote the intersection of eventsA and B by

A and B

Example:B = { ; ; ; ; ; ; ; ; }C = { ; ; ; ; ; }.

Denote by D the intersection of events B and C . Then

B and C = D = { ; ; }.

D = The face showing on each die are the same and even.


Intersection of Events

When we connect two events with AND, we are interested in theirintersection.

The intersection of two mutually exclusive events is the impossibleevent. If A and B are mutually exclusive, then

A and B = ∅


Example

Suppose you toss a fair coin twice. You are counting heads, so two eventsof interest are

A = First toss is a head,B = Second toss is a head.

What is the probability of the intersection P(A and B)?

The events A and B are both disjoint, they occur together when bothtosses give heads. Indeed, the sample space in this experiment is

S = {HH, HT , TH, TT}

All the outcomes are equally likely. The events are

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

The intersection is A and B = {HH}.

P(A and B) =1

4


Independent Events

Independent Events: Two events A and B are independent ofeach other if knowing that one event occurs does not change theprobability that the other event occurs.

Tosses of a fair coin are independent events. Further, it isequally likely that each side show after a toss.

Whether there are two heads in a row or twenty, the chance of gettinga head next time is still 0.5. It is the memoryless property of thecoin.

A fair die also has the memoryless property, i.e., the rolls areindependent. For instance, the probability of an ace at every roll is 1

6no matter how many aces has appeared in the previous rolls.

We will assume that child births are independent events. Child birthsare independent events.

The actual probability of a boy is slightly higher than 0.5, but we will

assume it to be 0.5.


Independent Events

Consider two events arising from rolling two dice:

A = Getting a 6 on the first roll.B = The sum of the numbers seen on the first and second rolls is 11

Are A and B independent events?Solution 1: The event B consists of the following outcomes:

B = { ; }.

The probability of event B is P(B) = 236 = 1

18 .

Suppose event A has happened. Then the possible outcomes for thesecond trial are 1, 2, 3, 4, 5, 6. Event B will happen if the secondoutcome was 5. The probability of this is 1

6 which is larger than 118 .

Knowing that event A has happened changes the probability that event Bwill happen. Hence, A and B are not independent events.


Independent and Disjoint Events

If events A and B are independent, they can not be mutually exclusive.Independent events always have common outcomes (intersect).

The opposite statement is also true: Mutually exclusive events are notindependent (dependent).

Explanation: If we know that A and B are mutually exclusive events with

nonzero probabilities P(A) > 0 and P(B) > 0, and we know that event A

has happened, this implies that the probability that event B also happened

is 0.

Example: Toss a coin twice. Two events A = First outcome is a head and B =

Second outcome is a head are independent, but not mutually exclusive.

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

Events A and B are not mutually exclusive since they have a common outcome

{HH}.


Union of Events

Union: The union of events A and B is the event which happens wheneither event A or event B or both happen. The union of two events isexpressed as

A or B

In mathematics, the word or is not inclusive, rather it is exclusive.Inclusive means either or, but not both; as in a restaurant. Exclusivemeans either or both.

In other words, the union of events A and B is the event which happenswhen at least one of events A or B happens.


Union of Events: Venn diagram

Mathematically, the union of two events A and B is found by combining the

outcomes from A and B into a single set A or B.

The following Venn diagram illustrates the union of events A and B.

The shaded region corresponds to A or B.


Example

Find the probability of event

E = at least one 3 in two successive rolls of a die.

Solution: The event E consists of the following outcomes:

; ; ; ; ; ; ; ; ; ;

The sample size for the two dice problem has 36 equally likely outcomes.As a consequence:

P(E ) =11

36.


Example

Observe that event E can be represented as the union of two events:

A = First outcome is 3 = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)}B = Second outcome is 3 = {(1, 3), (2, 3), (3, 3), (4, 3), (5, 3), (6, 3)}

Therefore:

E = At least one 3 in two successive rolls of a die

= First outcome is 3 or second outcome is 3

= A orB

When combining the outcomes from two events into a single event, donot repeat the same outcomes twice. Observe that each of the events Aand B has 6 outcomes, but their union (event E ) has 11 outcomes. This isbecause A and B have one common outcome (3, 3) which is written justonce in E . Notice A and B= {(3, 3)}.


Rules of Probability

The following rules simplify many probability computations:

Rule 1: The probability P(A) of any event A satisfies

0 ≤ P(A) ≤ 1.

In other words, the probability of any event is between 0 and 1.

Rule 2: If S is the sample space for an experiment, then P(S) = 1.The probability of a certain event is 1.

Rule 3: The probability of an impossible event is 0. Hence,P(∅) = 0.

Rule 4: If events A and B are independent, then the probabilitythat they both happen is the product of their probabilities:

P(A and B) = P(A) · P(B)


Rules of Probability

Rule 5: For any events A and B, the probability of their union isequal to the sum of their individual probabilities minus the probabilityof their intersection:

P(A or B) = P(A) + P(B)− P(A and B)

Subtracting the probability of the intersection is needed to avoiddouble counting.

A special case: if A and B are disjoint events, then the probability oftheir union is the sum of individual probabilities:

P(A or B) = P(A) + P(B)

Rule 6: For any event A

P(A) = 1− P(A)


A Few Remarks

Some important notes about the rules of probability:

1. The closer the probability of an event is to 1, the more likely thisevent is to happen.

2. Unlikely events have probabilities close to 0.

3. Outcomes partition the sample space:

Every cell in the above picture is an outcome.

4. Probabilities of all the outcomes add up to 1 because the set of alloutcomes is S and P(S) = 1.


Example

We reconsider the following events arising from rolling a die twice:

E = At least one 3 in two rolls of a dieA = The first outcome is 3B = The second outcome is 3.

It should be clear that E = A or B, or that E is the union of both A andB.

The experiment which consists in throwing 2 fair dice has 36 equally likely

outcomes.

P(A) =6

36=

1

6= P(B)

The intersection of events A and B is { } with probabilityP(A and B) = 1

36 .

P(E ) = P(A or B)= P(A) + P(B)− P(A and B)=1

6+

1

6− 1

36=

11

36


Example

Problem 1. A total of 30% of American males smoke cigarettes, 7%smoke cigars, and 5% smoke both cigars and cigarettes. What percentageof males smoke neither cigars nor cigarettes?


Example (Blood Type - 2) From Moore and McCabe

All human blood can be ABO-typed as one of A, B, O, or AB, but thedistribution of the types varies among groups of people. Here is thedistribution of blood types for a randomly chosen person in the US:

Blood type O A B AB

U.S. probability 0.45 0.40 0.11 ?

(a) What is the probability of type AB blood in the US?

P(AB) = 1− 0.45− 0.40− 0.11 = 0.04

(b) Maria has type B blood. She can safely receive blood transfusionsfrom people with blood types O and B. What is the probability that arandomly chosen American can donate blood to Maria?


Example (Blood Type - 2)

Blood type O A B AB

U.S. probability 0.45 0.40 0.11 0.04

Consider the events:

O = Randomly selected person has O blood typeB = Randomly selected person has B blood type.

Events O and B are disjoint since a person can not have type O and typeB of blood at the same time. We are interested in the probability that arandomly selected person has a blood type of either O or B type. Then

P(O or B)= P(O) + P(B) = 0.45 + 0.11 = 0.56.


Example (Blood Type - 3)

The probability of randomly choosing an individual with blood type A is0.43 in the US and 0.22 in China. What is the probability of randomly andindependently choosing two people, one from the US and the other fromChina, that both have blood type A?

Solution: Define the following events:

X = An American has A blood typeY = A Chinese has A blood type

Notice that events X and Y are independent. We are interested in theprobability of the event X and Y . By the multiplication rule forindependent events

P(E and F ) = P(E ) · P(F ) = 0.43 · 0.22 = 0.0946


Extending the Rules of Probability

1 The multiplication rule for a collection of m mutuallyindependent events.

For any collection of events A1, A2, . . . , Am that are mutuallyindependent, the probability of the intersection of all the events isequal to the product of their individual probabilities:

P(A1 and A2 and . . . and Am) = P(A1) · P(A2) · . . . · P(Am).

2 The addition rule for m disjoint events.

If events A1, A2, . . . , Am are disjoint in the sense that they do nothave common outcomes, then

P(A1 or A2 or . . . or Am) = P(A1) + P(A2) + · · ·+ P(Am).


Example (Universal Donors) From Moore and McCabe

People with O-negative blood are universal donors. That is, any patientcan receive a transfusion of O-negative blood. Only 7% of the Americanpopulation have O-negative blood. If 3 people appear at random to giveblood, what is the probability that at least one of them is a universaldonor?

Solution: Let’s code the outcomes of the experiment in the following way:

Y means a randomly selected person has O-negative bloodN means a person does NOT have O-negative blood.

Then the sample space for this experiment is as follows:

S = {NNN, NNY , NYN, YNN, NYY , YNY , YYN, YYY }


Example (Universal Donors)

We are given that P(Y ) = 0.07. Notice that Y and N are oppositeevents! Therefore

P(N) = 1− P(Y ) = 1− 0.07 = 0.93.

We are interested in the probability of the following event:

A = at least one of the three people is a universal donor

= at least one Y in the outcome

= { NNY, NYN, YNN, NYY, YNY, YYN, YYY}

A has all the outcomes from S except {NNN}, which should be theopposite event.

A = None of the three people is a universal donor

= No Y in the outcome

= {NNN}

Clearly it’ll be easier to find the probability of A rather than A!Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 64 / 91

Example (Universal Donors)

S = {NNN,NNY ,NYN,YNN,NYY ,YNY ,YYN,YYY }A = {NNY ,NYN,YNN,NYY ,YNY ,YYN,YYY } A = {NNN}

Recall that each outcome is the result of three independent events.

Therefore, we may use the multiplication rule of mutually independentevents:

P(A) = P(N and N and N) = P(N) · P(N) · P(N) = (0.93)3 = 0.804357.

As A is opposite A we have

P(A) = 1− P(A) = 1− 0.804357 = 0.195643.

Notice that P(A) can be calculated without finding P(A) by usingmultiplication and addition formulas:

P(A) = P(NNY or NYN or YNN or NYY or YNY or YYN or YYY )

= P(N)P(N)P(Y ) + P(N)P(Y )P(N) + P(Y )P(N)P(N) + . . .


Two Sampling Schemes

We will consider two important sampling schemes:

Sampling with Replacement

Sampling without Replacement

We explore these sampling techniques through the following problem:

Problem. A box contains r red and g green marbles. Define the followingevents:

Ri = The marble selected on the ith draw is red.

For example, R1 is the event for which the first drawn marble is red.

We are interested in the probabilities of events R1, R2, R3, . . ..

Since every marble has the same chance to be picked, we see that

P(R1) =r

r + g



What are the probabilities of events R2, R3, . . .? That depends on thesampling scheme:

Sampling with replacement:

Suppose we return the marble to the box after each draw. In thiscase the conditions of experiment do not change since eachsubsequent draw is made from the same population.

1 Subsequent draws are independent.

2 The probability to select a red marble is the same at every draw:

P(Ri ) =r

r + g, i = 1, 2, . . .



Sampling without replacement:

After each draw the drawn marble is not returned to the box.Subsequently, conditions of the experiment change from draw to drawbecause the population of marbles in the box changes after each draw.

1 The probability of selecting a red marble changes with each draw!

2 Subsequent draws are NOT independent.


Example (Box with Marbles)

A box contains 7 marbles - 2 red and 5 green. We have considered theevent A:

A = The first two chosen marbles are red.

We used the following two events to describe event A :

A1 = The first selected marble is red,A2 = The second selected marble is red.

We found that A = A1 and A2

We know A1 but we are not sure how to calculate A2; is the samplingdone with or without replacement?



Sampling with replacement:

Subsequent draws are independent.

P(A1) = P(A2) =2

7

⇒ P(A) = P(A1) · P(A2) =2

7· 2

7=

4

49

Sampling without replacement:

The probability of A1 is still 27 . The probability of A2 depends on the

outcome of the first draw:The probability to select a red marble on the second draw, given thatthe first selected marble was red is 1

6 .The probability to select a red marble given that the first selectedmarble was green is 2

6 .

Sampling multiple times without replacement is not independent!The multiplication rule for independent events cannot be used!


Conditional Probability

Conditional Probability: Let A and B be events. If the eventsare not independent, then the occurrence of B alters the probability thatA will occur. The conditional probability of event A given that event Bhas happened is denoted P(A|B).

Example: (Single roll of a fair die)

Consider two events:

A = The die shows 3 B = The die shows odd.

The ordinary, unconditional, probabilities of events A and B are

P(A) =1

6P(B) =

1

2

Suppose we know that event B happened. Then the conditionalprobability of A given B is

P(A|B) =1

3since there are only 3 odd outcomes possible: 1, 3, and 5, all equally likely.



The conditional probability of an event A given an event B is

P(A|B) =P(A and B)

P(B)

A B

S

When we say ”given B”; we are only considering outcomes covered by event B.

Ignore everything outside of B! Then find the probability that an A outcome

lies in B.Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 72 / 91

Conditional Probability: Examples

(a) Two machines, I and II, produce bolts. Five percent of those from Iand ten percent of those from II are defective.

This can be written as

P(defective |machine I) = 0.05 P(defective |machine II) = 0.10.

Notice that we do not know P(defective), nor can we calculate it withthe information supplied.

(b) Suppose a mortality table shows that the probability of dying withinone year for a 25-year-old male is 0.00193.

This can be stated as

P(male dying within one year | age 25) = 0.00193.



Note 1: Knowledge that B has occurred effectively reduces thesample space from S to B. This is sometimes called the reducedsample space. Therefore, when interpreting the area of an event onthe Venn diagram as its probability, P(A|B) is the proportion of thearea of B occupied by A.

Note 2: If events A and B are mutually exclusive, then P(A|B) = 0as P(A and B) = 0.

Note 3. Recall the formula:

P(A|B) =P(A and B)

P(B)

With algebraic manipulation we find:

P(A and B) = P(A|B) · P(B)

This formula is the general multiplication rule.


The General Multiplication Rule

As and is commutative,

P(A|B) · P(B) = P(A and B) = P(A and B) = P(B|A) · P(A)

The order of conditioning may be changed if needed.

Extension of the General Multiplication Rule:

The events A1,A2, . . . ,An are not necessarily independent.

P(A1 and A2 and A3 and A4 and . . .) =

= P(A1)·P(A2|A1)·P(A3|A1 and A2)·P(A4|A1 and A2 and A3)·. . .


Intersection of 3 Events on a Venn Diagram

The intersection of three events A, B, and C has probability

P(A and B and C ) = P(A) · P(B|A) · P(C |A and B)

A B

S

C


Example (Athletes) (From Moore and McCabe)

Only 5% of male high school basketball, baseball, and football players goon to play at the college level. Of these, only 1.7% enter major leagueprofessional sports. About 40% of the athletes who compete in college andthen reach the pros have a career of more than 3 years. Define these events

A = Competes in college,B = Competes professionally,C = Professional career longer than 3 years.

What is the probability that a high school male athlete competes incollege, reaches a professional level and then goes on to have a pro careerof more than 3 years?


Example (Athletes)

We are given that

P(A) = 0.05P(B|A) = 0.017P(C |A and B) = 0.4.

The probability we want is, therefore,

P(A and B and C ) = P(A) · P(B|A) · P(C |A and B)

= 0.05 · 0.017 · 0.4= 0.00034

Interpretation: Only about 3 of every 10,000 high school male athletescan expect to compete in college, reach a professional level and have aprofessional career for more than 3 years.


Conditional Probability and Independence

When events A and B are independent, knowing that event B hasoccurred gives no additional information about the occurrence of event A.This can be expressed as

P(A|B) = P(A)

To check whether two events A and B are independent, you should checkeither equality: P(A|B) = P(A) or P(A andB) = P(A) · P(B).


Example (Degrees) (From Moore and McCabe)

The counts (in thousands) of earned degrees in the US in the 2005-2006academic year, classified by level and by the sex of the degree recipient:

Bachelor’s Master’s Professional Doctorate Total

Female 784 276 39 20 1119Male 559 197 44 25 825

Total 1343 473 83 45 1944

Tables of this type are called cross-tabulation or contingency tables.


Example (Degrees)

(a) If you choose a degree recipient at random, what is the probabilitythat the person that you choose is a woman?

P(W ) =1119 thousands

1944 thousands≈ 0.5756.

(b) What is the probability of choosing a woman, given that the personchosen received a professional degree?

P(W |P) =39

83≈ 0.4699.

(c) Are the events choose a woman and choose a professional degreerecipient independent? Why or why not?

We found in (a) that P(W ) = 0.5756. In (b) we foundP(W |P) = 0.4699. Since P(W ) 6= P(W |P), the events are notindependent.Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 81 / 91

Example (Degrees)

(d) A randomly chosen person is a man. What is the probability that hereceived a bachelor’s degree?

P(B|M) =559

825≈ 0.6776.

(e) Use the multiplication rule to find the probability of choosing a malebachelor’s recipient. Check your result by finding this probability directlyfrom the table of counts.

P(M and B) = P(B|M) · P(M) =559

825· 825

1944=

559

1944,

which is the probability obtained directly from the table of counts.



Recall that there are 2 red and 5 green marbles in a box. We defined thefollowing events:

A = Two randomly chosen marbles are redA1 = First selected marble is redA2 = Second selected marble is red.

We found A = A1 and A2, and

P(A1) = P(A2) =2

7.

We want to compute the probability of A under sampling withoutreplacement:

P(A) = P(A1 and A2) = P(A2|A1)P(A1) =1

6· 2

7=

1

21.

We previously calculated that the probability of event A under samplingwith replacement is 4

49 . Since 449 >

121 , event A is more likely to occur

under sampling with replacement.Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 83 / 91

Examples

Example: Two birds are selected at random without replacement from acage containing five male and two female finches. What is the probabilitythat both are males?

P(M and M) = P(M) · P(M|M) =5

7· 4

6=

10

21

Example: A large basket of fruit contains 3 oranges, 2 apples and 5bananas. If two fruit are chosen at random without replacement, what isthe probability that one of the selected fruits is an apple and the other oneis an orange?

P(A and O) = P(A) · P(O|A) =2

10· 3

9=

1

15


Examples

Example: A box of 10 items has 2 defective items. Three items areselected by the researcher without replacement.

(a) What is the probability that the researcher will obtain no defectiveitems?

P(D and D and D) = P(D) · P(D|D) · P(D|D and D) =8

10· 7

9· 6

8=

7

15

(b) Given that the researcher finds the first item selected as defective,what is the probability that the researcher will also have the otherdefective item?

P( (D and D) |D) = P(D|D) · P(D|D and D) =8

9· 1

8=

1

9


Example (Roulette)

Example 20 (Roulette), from Moore and McCabe

A roulette wheel has 38 slots, numbered 0, 00, and 1 through 36. Theslots 0 and 00 are colored green, 18 of the others are red, and 18 are black.

The dealer spins the wheel and at the same time rolls a small ball alongthe wheel in the opposite direction. The wheel is carefully balanced sothat the ball is equally likely to land in any slot when the wheel slows.Gamblers can bet on various combinations of numbers and colors.

(a) What is the probability that the ball will land at any one slot?

P(any slot) =1

38


Example (Roulette)

(b) If you bet on red, you will win if the ball lands on a red slot. What isthe probability of winning?

P(red) =# of red slots

38=

18

38=

9

19<

1

2


Example (Roulette)

(c) The slot numbers are laid on a board on which gamblers place theirbets. One column of numbers on the board contains all multiples of 3, i.e.,3, 6, 9, ..., 36. You place a column bet that wins if any of these numberscomes up. What is you probability of winning?

P(Slot # is a multiple of 3) =12

38=

6

19< 13.

In fact, every game in a casino offers options to the gambler that has lessthan a 50% chance of winning. In later chapters we will discuss probabilityand payout.


Example (Lottery) (From Moore and McCabe.)

A state lottery’s Pick 3 game asks players to choose a three-digit number,000 to 999. The state chooses the winning three-digit number at random.You win if the winning number contains the digits in you number, in anyorder.

(a) Your number is 123. What is your probability of winning?

(b) Your number is 112. What is your probability of winning?

Solution: First of all, note that the sample space S for the experiment,which consists in choosing a 3-digit number at random, contains 1000equally likely outcomes:

S = {000, 001, . . . , 999}.


Example (Lottery)

(a) Consider the event A: a chosen number has digits 1, 2, and 3.

A = {123, 132, 213, 231, 312, 321}

Then P(A) = 61000 = 0.006, quite a small chance!

(b) Define the event B: a chosen number has two 1’s and one 2.

B = {112, 121, 211}

Then P(B) = 31000 = 0.003.

You do not want the digits to repeat in your lottery ticket!


Concluding Remarks

Observation 1: and

and means that we are looking at the intersection of events. We needto compute the probability that ALL events in the statement occursimultaneously. Probability computations in this case usually involveusing one of the multiplication rules.

Observation 2: or

or means that we are looking at the union of events. We need tocompute the probability that at least one event in the union occurs.Probability computations in this case usually involve either one of theaddition rules or the rule for opposite events.

Observation 3: at least, at most, not exactly, not equal

Most of the problems of this type reduce to computing probabilities ofthe unions of events. It is often easier to compute the probability ofthe opposite event first.


PART IV : PROBABILITY - Binghamton Universitypeople.math.binghamton.edu/jbrennan/home/S13MAT148/Chapter13.pdf · blood types for a huge representative group of Australian people.

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