PROBABILITYTHE LIKELIHOOD

THAT SOMETHING WILL HAPPEN

1

2

SWEET STATISTICSWHERE DID ALL THESE COLORFUL CANDIES COME FROM?

Do you know the number of different colors? Do you know that the idea for “M&M’s” Plain

Chocolate Candies was born in the backdrop of the Spanish Civil War?

On a trip to Spain, Forrest Mars Sr. encountered soldiers who were eating pellets of chocolate that were encased in a hard sugary coating to prevent them from melting.

Mr. Mars was inspired by this concept and returned home and invented the recipe for “M&M’s” Plain Chocolate Candies.

3

M&M COLORS BY COUNT

Color Count

Brown 91

Yellow 112

Red 102

Blue 151

Orange 137

Green 99

Total 692

4

M&M COLORS BY PERCENTAGES

Color Percent

Brown 13.2

Yellow 16.2

Red 14.7

Blue 21.8

Orange 19.8

Green 14.3

Total 100.0

5

Percentages behave very much like probability numbers, but the question being asked in probability is quite different.

We are treating the information as sample data and describing the results found.

If we think in terms of probability, we turn the orientation around and treat the complete set of 692 M&M’s as the complete list of possibilities and ask questions about the likeliness of certain events when one M&M is randomly selected from the entire collection of 692.

6

For example, suppose we dump all 692 M&M’s into a large bowl and thoroughly mix them.

Question: If one M&M is selected at random from

the bowl, what is the probability that it will be orange?

Selected randomly means each M&M has the same chance of being selected.

The probability of selecting an orange M&M is 137/692, or 0.198 because there are 137 orange M&M’s in the bowl.

7

PERCENTAGES AND PROBABILITY NUMBERS ARE “THE SAME, BUT DIFFERENT.”

THE NUMBERS HAVE THE SAME VALUE AND BEHAVE WITH THE SAME PROPERTIES; HOWEVER, THE ORIENTATION OF THE SITUATION AND THE QUESTIONS ASKED ARE DIFFERENT.

8

OBJECTIVES To define probability, experiment, random

experiment, sample space, event, simple event, mutually exclusive events and tree diagram

To specify/enumerate the events and sample space in a random experiment

To draw a tree diagram To calculate the probability that a given

event will occur To use the fundamental principles of

counting To find the number of permutations/

combinations of n elements taken r at a time

9

EXPERIMENT The process by which an observation (or

measurement) is obtained

10

RANDOM EXPERIMENT

THE OUTCOME IS NOT NECESSARILY THE SAME WHEN YOU REPEAT THE EXPERIMENT

If you toss a coin, there are 2 possible outcomes: Head or tail

If you roll a die, there are 6 possible outcomes: 1, 2, 3, 4, 5, or 6

11

SAMPLE SPACE THE SET OF ALL POSSIBLE

OUTCOMES OF AN EXPERIMENT

or the set of all simple events

12

EVENT

A collection of one or more simple events

or any subset of the sample space

13

SIMPLE EVENT An event that cannot be decomposed

14

MUTUALLY EXCLUSIVE EVENTS Two events are mutually exclusive if,

when one event occurs, the other cannot, and vice versa

15

SHORT STORY- MILK IN A CUP OF TEA

At a summer afternoon tea party, a guest proclaimed that the tea tastes different depending on whether the milk is poured into the tea or the tea is poured into the milk.

The claim was met with much ridicule. After much bantering, one man

proposed a scientific way to test her hypothesis.

The milk and tea were combined in both manners, then offer her one of each, two at a time in random order, for identification.

16

SHORT STORY-CONT’D… Others quickly joined and assisted with

the testing. She correctly identified 10 in a row. What do you think? Could she tell the difference?

17

TREE DIAGRAM A drawing that schematically represents

the various possible outcomes of an experiment

18

EXAMPLE 1 Experiment:

Toss a coin and observe the face that appears

Events:Event A: observe a headEvent B: observe a tail

19

EXAMPLE 2 Experiment:

Roll a die and observe the number that appears on the upper face

Events: Event A: Observe an even number Event B: Observe a number greater than 3 Event E1: Observe a 1 Event E2: Observe a 2 Event E3: Observe a 3 Event E4: Observe a 4 Event E5: Observe a 5 Event E6: Observe a 6

20

NOTE When event E1 occurs, event E2 cannot

occurThese two events are mutually exclusive

Events A and B in example 1 are mutually exclusive

Events A and B in example 2 are not mutually exclusive

21

EXAMPLE 3 Experiment

Toss a single coin and observe the result Events:

Event E1 : Observe a head (H)Event E2 : Observe a tail (T)

Tree diagram:

Sample space, S = {E1, E2}

or S = {H, T}

H

T

“Root”

Toss

22

EXAMPLE 4 Experiment:

Toss a single coin twice and observe the result

Events:Event E1: Observe a head in the 1st toss and

a head in the 2nd tossEvent E2: Observe a head in the 1st toss and

a tail in the 2nd tossEvent E3: Observe a tail in the 1st toss and a

head in the 2nd tossEvent E4: Observe a tail in the 1st toss and a

tail in the 2nd toss23

CONTINUATION… Experiment:

Toss a single coin two times and observe the result

Tree diagram:

“Root”

H

T

T

T

H

H

Toss 1

Toss 2

24

SAMPLE SPACE, S

TTTHHTHHS

or

EEEES

,,,

,,, 4321

Number of possible outcomes: 4

25

EXAMPLE 5 Experiment:

Toss a single coin three times and observe the result

Tree diagram:

H

H

H

T

T

T

H

H

T

T

H

H

T

T

Toss 1

Toss 2

Toss 3

“Root”

26

SAMPLE SPACE, S

Number of possible outcomes: 8

S = {E1, E2, E3, E4, E5, E6, E7, E8}

or S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

27

EXAMPLE 6 Experiment:

Roll two distinct dice. Each die is numbered 1 to 6.

Number of possible outcomes: 36 mn rule can also be used (to be discussed

later, p.62)11 E1 21 E7 31 E13 41 E19 51 E25 61 E31

12 E2 22 E8 32 E14 42 E20 52 E26 62 E32

13 E3 23 E9 33 E15 43 E21 53 E27 63 E33

14 E4 24 E10 34 E16 44 E22 54 E28 64 E34

15 E5 25 E11 35 E17 45 E23 55 E29 65 E35

16 E6 26 E12 36 E18 46 E24 56 E30 66 E36

28

EXAMPLE 7 Experiment

Toss a die and observe the number that appears on the upper face.

Events: Event A: Observe an odd number Event B: Observe a number less than 4 Event E1 : Observe a 1 Event E2 : Observe a 2 Event E3 : Observe a 3 Event E4 : Observe a 4 Event E5 : Observe a 5 Event E6 : Observe a 6

29

EXAMPLE 8 Experiment

Record a person’s blood type. Events:

Event E1 : Blood type AEvent E2 : Blood type BEvent E3 : Blood type ABEvent E4 : Blood type O

Sample space, S = {E1, E2, E3 ,E4}

30

EXAMPLE 9 Experiment

A medical technician records a person’s blood type and Rh factor. List the simple events in the experiment.

Events: Event E1 : Blood type A+ Event E5 : Blood type

AB+ Event E2 : Blood type A- Event E6 : Blood type

AB- Event E3 : Blood type B+ Event E7 : Blood type

O+ Event E4 : Blood type B- Event E8 : Blood type

O-

Sample space, S = {E1, E2, E3, E4,E5,E6,E7,E8}

S = {A+, A-, B+, B-, AB+, AB-, O+, O-}

31

SOLUTION Tree diagram

A

B

AB

O

A+

_

AB-

O-

B+

AB+

O+

B-

+

+

+

+

_

_

_Bloodtype

Rh factorOutcome

32

A-

PROBABILITY OF AN EVENT THE RELATIVE FREQUENCY IN WHICH AN

EVENT CAN BE EXPECTED TO OCCUR

33

CALCULATING PROBABILITIES USING SIMPLE EVENTS The probability of an event A is a

measure of our belief that the event A will occur.

If an experiment is performed n times, then the relative frequency of a particular occurrence --- say, A --- is

experiment theof srepetition ofnumber n

occurredA event the timesofnumber frequency

n

frequencyfrequency Re

where

lative

34

If an experiment is repeated a large

number of times (i.e., n → infinity) and event A occurs nA times, the relative frequency of the occurrence of A will be approximately equal to the probability of A

35

PROBABILITY OF AN EVENT A

space sample in the

events simple ofnumber n

Aevent in

events simple of

)(

numbern

wheren

nAP

A

A

36

Since P(A) behaves like a relative

frequency, P(A) must be a proportion lying between 0 and 1

P(A) = 0The event A never occurs

P(A) = 1The event A always occurs

The closer P(A) is to 1, the more likely it is that A will occur

37

MUTUALLY EXCLUSIVE SIMPLE EVENTS Requirements for simple-event

probabilitiesEach probability must lie between 0 and 1The sum of the probabilities for all simple

events in S equals 1

38

EXAMPLE Toss two fair coins and record the

outcome. Find the probability of observing exactly one head in the two tosses. Tree diagram:

H

H

H

T

T

T

“Root”

First coin

Second coin

39

SOLUTION (REFER TO THE TREE DIAGRAM -

PREVIOUS SLIDE)

Sample space, S = (E1, E2, E3, E4) S = (HH, HT, TH, TT)

P(A) = P(observe exactly one head) =P(E2) + P(E3) = ¼ + ¼ = 1/2

40

EXAMPLE A candy dish contains one yellow and

two red candies. You close your eyes, choose two candies one at a time from the dish, and record their colors. What is the probability that both candies are red?

41

SOLUTION Tree diagram

P(A) = P(both candies are red)Event A: both candies are redA = {R1R2 , R2R1)

P(A)= 1/6 + 1/6 = 1/3

R2

Y

R1

R2

Y

R1

Y

R1

R2

Root or start

2nd choice

Simple event

probability

R1Y

R1R2

R2R1

R2Y

YR1

YR2

1/6

1/6

1/6

1/6

1/6

1/6

1st choice

42

EXAMPLEM

What is the probability of getting 3 heads in 3 successive tosses of a fair coin?

Solution (refer to the tree diagram-next slide)

P(3H) = 1/8 = 0.125 (read as “probability of getting 3 heads)

43

TREE DIAGRAM

H

H

H

T

T

T

H

H

T

T

H

H

T

T

Toss 1

Toss 2

Toss 3

“Root”

44

EXAMPLEWhat is the probability of getting a face

card in a well-shuffled deck?

Solution:P(face card) = 12/52 = 0.23(read as “probability of getting a face

card)

45

EXAMPLE In throwing a die, six possible events

may occur. In a fair die, six events are equally likely. What is the probability of getting a 5 in tossing a fair die?

Solution P(5) = 1/6 (read as “probability of getting a 5)

46

EXAMPLE What is the probability of getting either

a 1, a 2, or a 3 in a single throw of a die? Solution P(1,2, or 3) = 3/6 = ½ = 0.5

47

EXAMPLE The proportions of blood phenotypes A,

B, AB, and O in the population of all Filipinos are reported as 0.28, 0.22, 0.09, and 0.41, respectively. If a single Filipino is chosen at random from the population, what is the probability that he or she will have either type AB or type O blood?

Solution P(AB or O) = P(AB) + P(O) = 0.09 + 0.41 = 0.50

48

STEPS IN CALCULATING THE PROBABILITY OF AN EVENT 1. List all the simple events in the

sample space 2. Assign an appropriate probability to

each simple event 3. Determine which simple events result

in the event of interest 4. Sum the probabilities of the simple

events that result in the event of interest

49

TWO CONDITIONS THAT MUST BE SATISFIED 1. Include all simple events in the

sample space 2. Assign realistic probabilities to the

simple events

50

EXERCISE The proportions of blood phenotypes A,

B, AB, and O in the population of all Filipinos in the Philippines are reported as 0.41, 0.10, 0.04, and 0.45, respectively. If a single Filipino is chosen randomly from the population, what is the probability that he or she will have either type A or type AB blood?

51

SOLUTION The four events, A, B, AB, and O, do not

have equally likely probabilities. Their probabilities are found using the

relative frequency concept P(A) = 0.41 P(B) = 0.10 P(AB) = 0.04 P(O) = 0.45

52

SOLUTION, CONT’D…

The event of interest consists of two simple eventsP(person is either type A or type AB) =P(A) + P(AB)= 0.41 + 0.04 = 0.45

53

EXERCISES (DRAWING A TREE DIAGRAM)

1. Draw a tree diagram that shows the possible results from rolling a single die once.

54

EXERCISES CONT’D…

2. Draw a tree diagram showing the possible methods of transportation that could be used to travel to a resort area. The possible choices are car, bus, train and airplane.

55

EXERCISES CONT’D…

3. (a) Draw a tree diagram that represents the possible results from tossing two coins.

(b) Repeat part (a) considering one coin a nickel and the other a penny.

© Is the list of possibilities for part (b) any different than it was for part (a)?

56

57

EXERCISES CONT’D…

4. Draw a tree diagram representing the tossing of three coins.

58

EXERCISES CONT’D…

5. (a) Draw a tree diagram representing the possible results that could be obtained when two dice are rolled.

(b) How many branch ends does your tree have?

59

60

EXERCISES CONT’D…

6. Students at our college are to be classified as male or female, as graduates of public or private high schools, and by the type of curriculum they are enrolled in, liberal arts or career. Draw a tree diagram that shows all of the various possible classifications.

61

62

EXERCISES CONT’D…

7. There are two scenic routes (A and B) as well as one business route © by which you may travel from your home to a nearby city. You are planning to drive to that city by way of one route and come home by a different route.

(a) Draw a tree diagram representing all of your possible choices for going and returning.

(b) How many different trips could you plan?

© How many of these trips are scenic in both directions?

63

64

COUNTING RULES… One way to determine the required

number of simple events Can be used to count the elements of a

set without actually listing them

65

THE MN RULE Consider an experiment that is

performed in two stages. If the first stage can be accomplished in m ways and the second stage can be accomplished in n ways, then there are mn ways to accomplish the experiment.

66

EXAMPLE Determine how many options are

available if you can order a car in one of three styles and in one of four paint colors.

Style-color combinations (refer to the tree diagram-next slide)

Number of possible options = mn = 3(4) = 12

67

STYLE-COLOR COMBINATIONS Color

1

Style1

2341234

12

34

2

3

68

EXAMPLE Two dice are tossed. How many simple

events are in the sample space S?

First die can fall in one of m = 6 ways Second die can fall in one of n = 6 ways Total number of simple events in S = mn = 6(6) = 36

(Since the experiment involves two stages)

69

EXAMPLE A candy dish contains one yellow and

two red candies. Two candies are selected one at a time from the dish, and their color is recorded. How many simple events are in the sample space S?

(Refer to the tree diagram, page 43) First candy can be chosen in m = 3 ways Second candy can be chosen in n = 2

ways (Since one candy is gone)

Total number of simple events = mn = 3(2) = 6

70

EXTENDED MN RULE If an experiment is performed in k

stages, with n1 ways to accomplish the first stage, n2 ways to accomplish the second stage, …, and nk ways to accomplish the kth stage, then the number of ways to accomplish the experiment is

n1 n2 n3 … nk

71

EXAMPLE How many simple events are in the

sample space when three coins are tossed?

Solution:

Each coin can land in one of two waysTherefore,

The number of simple events = 2(2)(2) = 8

72

EXAMPLE A truck driver can take three routes

from city A to city B, four from city B to city C, and three from city C to city D. If, when traveling from A to D, the driver must drive from A to B to C to D, how many possible A-to-D routes are available?

73

SOLUTION Let

n1 = number of routes from A to B = 3n2 = number of routes from B to C = 4n3 = number of routes from C to D = 3

Total number of ways to construct a complete routeThe total no. of ways to construct a

complete route, taking one subroute from each of the three groups, A to B, B to C, and C to D, is

n1 n2 n3 =3(4)(3) = 36

74

COUNTING RULE FOR PERMUTATIONA PERMUTATION OF SOME OR ALL OF THE ELEMENTS OF A SET IS ANY ARRANGEMENT OF THE ELEMENTS IN A DEFINITE ORDER

The number of permutations of n distinct elements taken r at a time, denoted , is

1 0!

)(1)2)...(3)(2-1)(n-n(n !

n r 0for !

!

n

where

rn

nPrn

rnP

75

EXAMPLE Three lottery tickets are drawn from a

total of 50. If the tickets will be distributed to each of three employees in the order in which they are drawn, the order will be important. How many simple events are associated with the experiment?

76

SOLUTION Total number of events

600,117)48)(49(50!47

!50350 P

77

EXAMPLE A piece of equipment is composed of

five parts that can be assembled in any order. A test is to be conducted to determine the time necessary for each order of assembly. If each order is to be tested once, how many tests must be conducted?

78

SOLUTION

Total number of tests

120)1)(2)(3)(4(5!0

!555 P

79

EXAMPLE A museum has 7 paintings to hang and

3 vacant locations, each of which will hold one painting. In how many different ways can these 3 locations be filled by the paintings?

Solution:

2101.2.3.4

1.2.3.4.5.6.7

)!37(

!737

P

80

EXAMPLE How many different ways can 8 people

be seated in a row of 5 chairs?

Solution:

6720)!58(

!858

P

81

EXAMPLEM

Four lottery tickets are drawn from a total of 100. If the tickets will be distributed to each of 4 employees in the order in which they are drawn, the order will be important. How many simple events are associated with the experiment?

Solution:

94,109,400

97.98.99.100)!4100(

!1004100

P

82

EXAMPLEM

Suppose you have three books A, B, and C, but you have room for only two on your bookshelf. In how many ways can you select and arrange the two books?

Solution:

61

1.2.3

)!23(

!323

P

83

OPTIONAL In general, the number of

distinguishable permutations P of n elements taken n at a time, with r1 like elements, r2 like elements of another kind, and so on, is

!...!!

!

321 rrr

nPPnn

84

EXAMPLEM

In how many ways can you arrange the letters in the word STATISTICS?

Solution: There are 10 letters: 3 S’s, 3 T’s, 1 A, 2

I’s and 1 C

400,50

!1!2!1!3!3

!10!

54321

rrrrr

nP

85

ASSIGNMENT

Find the number of permutations of the letters in the word MISSISSIPPI

Solution: There are 11 letters: 1 M, 4 I’s, 4 S’s

and 2 P’s

650,34!2!4!4!1

!11

!...!!

!

321

rrr

nP

86

EXAMPLEM

In how many ways can you arrange 3 red flags, 4 blue flags, and 5 yellow flags in a row?

Solution:

720,27!5!4!3

!12P

87

CIRCULAR PERMUTATIONS n distinct objects are arranged in a circle

)!1(11 nPPnn

88

EXAMPLEM

Five people are to be seated in a circular table of 5 chairs. In how many ways can they be arranged?

Solution:

P = (n-1)! = (5-1)! = 24 ways

89

ASSIGNMENT Problem: In how many ways can 6 individuals be

seated in a round table with 6 chairs? Solution: This is a circular permutation problem.

90

120

1.2.3.4.5!5)!16(1616

P

ASSIGNMENT Problem: In how many ways can 6 persons be

seated around a table with 6 chairs if two individuals wanted to be seated side by side?

Solution:Two individuals who wanted to be seated

side by side: considered one person, n = 5 (5-1)! = 4! = 4.3.2.1 = 24Permutation of the two persons treated as

one2! = 2.1 = 2

Therefore, 24(2) = 48 ways 91

COUNTING RULE FOR COMBINATIONS

The number of distinct combinations of n distinct elements taken r at a time denoted

, is

rnn

rnrn

C

rnr

n

r

PC

rnCgeneral, In

n r 0for

)!(!

!

!

rnC

A COMBINATION OF n ELEMENTS OF A SET TAKEN r AT A TIME IS ANY r-ELEMENT SUBSET OF THE GIVEN SET

92

RELATIONSHIP BETWEEN THE NUMBER OF COMBINATIONS AND THE NUMBER OF PERMUTATIONS

!r

PC

nrn

r

93

EXAMPLEM+

A printed circuit board may be purchased from five suppliers. In how many ways can three suppliers be chosen from the five?

Solution: Number of ways is

101.2.1.2.3

1.2.3.4.5

)!35(!3

!535

C

94

EXAMPLEM

Two city council members are to be selected from a total of five to form a subcommitte to study the city’s traffic problems.

(a) How many different subcommittee are possible?

(b) If all possible council members have an equal chance of being selected, what is the probability that members Santos and Cruz are both selected?

95

SOLUTION Since the order in forming the

subcommitte is not important, the no. of possible subcommitte is

(a)

(b) Since Santos and Cruz comprise only one subcommitte, the probability that they are both selected is 1/10 = 0.10

10!3!2

!5

)!25(!2

!525

C

96

EXAMPLE Fifteen people entered a talent contest.

The top 3 contestants will each win Php2,000, and everyone else will get an honorable mention.

a. In how many different ways can 3 winners be chosen

b. In how many different ways can 12 people be chosen for honorable mention?

97

SOLUTION

455!12!3

!15C

3r 15,n )!(!

!C .

315

rn

rnr

na

proven! was84 pageon formula:note

12r 15,n 455!3!12

!15C . 1215 b

98

EXAMPLE Five manufacturers produce a certain

electronic device, whose quality varies from manufacturer to manufacturer. If you were to select three manufacturers at random, what is the chance that the selection would contain exactly two of the best three?

99

SOLUTION The total no. of simple events can be

counted as the no. of ways to choose three of the five manufacturers, or

Since the manufacturers are selected at random, any of these 10 simple events will be equally likely, with probability 1/10.

10!2!3

!535 C

100

SOLUTION CONT’D…

Event A: Exactly two of the ‘best’ three The number of events in A, nA, can be

counted in two steps because event A will occur when you select two of the ‘best’ three and one of the two ‘not best.’

Number of ways to accomplish the 1st stage: 3

!1!2

!323 C

101

SOLUTION CONT’D…

Number of ways to accomplish the 2nd stage:

Applying mn rule, nA = 3(2) = 6 of the simple events in

event A

2!1!1

!212 C

10

6)(

n

nAP A

102

MORE EXAMPLES Problem:

A die is rolled. Specify the following:a. the sample space, Sb. event A, which is the event of rolling an

even numberc. event B, which is the event of rolling a

number greater than 4

Solution:a. S = {1, 2, 3, 4, 5, 6}b. A = {2, 4, 6}c. B = {5, 6}

103

MORE EXAMPLES CONT’D…

Problem:A die is rolled. Find the probability of each

event. (a) getting a 4 (b) getting an odd number © getting 1, 2, 3, 4, 5, or 6 (d) getting a 7

Solution:a. P(getting 4) = 1/6b. P(getting an odd number) = 3/6 = ½c. P(getting 1,2,3,4,5, or 6) = 6/6 = 1d. P(getting 7) = 0/6 = 0

104

A SuccessGetting an outcome in a specified event

A FailureGetting an outcome that is not in a

specified event

The odds than an event will occur, or the odds in favor of an event, are expressed as the ratio of the number of successes to the number of failuresOR Odds in favor of an event = (number of successes):(number

of failures)

105

EXAMPLE Two coins are tossed. Specify the

following:a. the sample space, Sb. the probability of getting exactly 2

headsc. the odds in favor of getting exactly 2

headsd. the probability of getting exactly 1

heade. the odds in favor of getting exactly 1

head

106

SOLUTION a. S = {HH, HT, TH, TT} b. P(getting exactly 2 heads) = ¼

Refer to the tree diagram on page 40 c. The odds in favor of getting exactly 2

heads are 1:3 Just 1 outcome is a success, while 3 are

failures d. P(getting exactly 1 head) = 2/4 = ½ e. The odds in favor of getting exactly 1

head are 2:2 or 1:1Two outcomes are successes and 2 are

failures107

PROBLEM A bridge deck consists of 52 cards, with

13 in each of four suits: spades, hearts, diamonds, and clubs. Each suit consists of an ace, nine cards numbered 2 through 10, and three face cards (jack, queen, and king). If 3 cards are picked from a bridge deck, what is the probability that they will all be diamonds?

108

SOLUTION The sample space, S, is the set of all

possible 3-card hands in a 52-card deck. Number of outcomes in S:

100,22!49!3

!52

C Sin outcomes of . 352

No

109

SOLUTION CONT’D…

Let A be the event of picking 3 diamonds from the 13 in the deck

Number of outcomes in A:

2863!10!

13!

C A in outcomes of . 313

No

110

SOLUTION CONT’D…

111

013.0850

11

22,100

286

diamonds) 3 ()(

pickingPAP