THE LIKELIHOOD THAT SOMETHING WILL HAPPEN Probability THE LIKELIHOOD THAT SOMETHING WILL HAPPEN

Sweet statistics where 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.

M&M Colors by count Color Count Brown 91 Yellow 112 Red 102 Blue 151 Orange 137 Green 99 Total 692

M&M Colors by percentages Brown 13.2 Yellow 16.2 Red 14.7 Blue 21.8 Orange 19.8 Green 14.3 Total 100.0

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.

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.

Percentages and probability numbers are “the same, but different 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.

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

experiment The process by which an observation (or measurement) is obtained

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

Sample space THE SET OF ALL POSSIBLE OUTCOMES OF AN EXPERIMENT or the set of all simple events

event A collection of one or more simple events or any subset of the sample space

Simple event An event that cannot be decomposed

Mutually exclusive events Two events are mutually exclusive if, when one event occurs, the other cannot, and vice versa

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.

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?

Tree diagram A drawing that schematically represents the various possible outcomes of an experiment

Example 1 Experiment: Events: Toss a coin and observe the face that appears Events: Event A: observe a head Event B: observe a tail

Example 2 Experiment: Events: 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

note When event E1 occurs, event E2 cannot occur These 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

Example 3 Experiment Events: Sample space, S = {E1, E2} or S = {H, T} 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} Toss H “Root” T

Example 4 Experiment: Events: 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 toss Event E2: Observe a head in the 1st toss and a tail in the 2nd toss Event E3: Observe a tail in the 1st toss and a head in the 2nd toss Event E4: Observe a tail in the 1st toss and a tail in the 2nd toss

Continuation… Experiment: Tree diagram: Toss a single coin two times and observe the result Tree diagram: Toss 2 Toss 1 H H T “Root” H T T

Sample space, s Number of possible outcomes: 4

EXAMPLE 5 Experiment: Tree diagram: Toss a single coin three times and observe the result Tree diagram: Toss 1 Toss 2 Toss 3 H H T H H T “Root” T H H T T H T T

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

EXAMPLE 6 Experiment: Number of possible outcomes: 36 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

Example 7 Experiment Events: Event A: Observe an odd number 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

Example 8 Experiment Events: Sample space, S = {E1, E2, E3 ,E4} Record a person’s blood type. Events: Event E1 : Blood type A Event E2 : Blood type B Event E3 : Blood type AB Event E4 : Blood type O Sample space, S = {E1, E2, E3 ,E4}

Example 9 Experiment Events: 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-}

solution Tree diagram A B AB O Blood Rh factor type Outcome A+ + _ A-

Probability of an event THE RELATIVE FREQUENCY IN WHICH AN EVENT CAN BE EXPECTED TO OCCUR

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

If an experiment is repeated a large number of times (i. e 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

Probability of an event a

The closer P(A) is to 1, the more likely it is that A will occur Since P(A) behaves like a relative frequency, P(A) must be a proportion lying between 0 and 1 P(A) = 0 The event A never occurs P(A) = 1 The event A always occurs The closer P(A) is to 1, the more likely it is that A will occur

Mutually exclusive simple events Requirements for simple-event probabilities Each probability must lie between 0 and 1 The sum of the probabilities for all simple events in S equals 1

example Tree diagram: Toss two fair coins and record the outcome. Find the probability of observing exactly one head in the two tosses. Second coin First coin H H T “Root” H T T

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

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?

solution Tree diagram P(A) = P(both candies are red) Event A: both candies are red A = {R1R2 , R2R1) P(A)= 1/6 + 1/6 = 1/3 2nd choice Simple event probability 1st choice R1R2 1/6 R2 R1Y R1 1/6 Y R2R1 1/6 R2 R1 Root or start Y R2Y 1/6 Y R1 1/6 YR1 R2 1/6 YR2

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)

Tree diagram Toss 1 Toss 2 Toss 3 H H T H H T “Root” T H H T T H T T

example What 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)

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)

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

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

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

Two conditions that must be satisfied 1. Include all simple events in the sample space 2. Assign realistic probabilities to the simple events

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?

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

Solution, cont’d… The event of interest consists of two simple events P(person is either type A or type AB) =P(A) + P(AB) = 0.41 + 0.04 = 0.45

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

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.

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)?

Exercises cont’d… 4. Draw a tree diagram representing the tossing of three coins.

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?

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.

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?

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

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.

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

Style-color combinations 1 1 2 3 4 1 2 2 3 4 1 3 2 3 4

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)

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

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

example How many simple events are in the sample space when three coins are tossed? Solution: Each coin can land in one of two ways Therefore, The number of simple events = 2(2)(2) = 8

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?

solution Let Total number of ways to construct a complete route n1 = number of routes from A to B = 3 n2 = number of routes from B to C = 4 n3 = number of routes from C to D = 3 Total number of ways to construct a complete route The 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

Counting rule for Permutation a 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

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?

solution Total number of events

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?

solution Total number of tests

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:

example How many different ways can 8 people be seated in a row of 5 chairs? Solution:

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:

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:

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

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

There are 11 letters: 1 M, 4 I’s, 4 S’s and 2 P’s 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

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

Circular permutations n distinct objects are arranged in a circle

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

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.

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 = 24 Permutation of the two persons treated as one 2! = 2.1 = 2 Therefore, 24(2) = 48 ways

Counting rule for combinations A COMBINATION OF n ELEMENTS OF A SET TAKEN r AT A TIME IS ANY r-ELEMENT SUBSET OF THE GIVEN SET The number of distinct combinations of n distinct elements taken r at a time denoted , is

Relationship between the number of combinations and the number of permutations

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

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?

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

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?

solution

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?

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.

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:

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

More examples Problem: Solution: A die is rolled. Specify the following: a. the sample space, S b. event A, which is the event of rolling an even number c. event B, which is the event of rolling a number greater than 4 Solution: S = {1, 2, 3, 4, 5, 6} A = {2, 4, 6} B = {5, 6}

More examples cont’d… Problem: Solution: 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) = 1/6 b. P(getting an odd number) = 3/6 = ½ c. P(getting 1,2,3,4,5, or 6) = 6/6 = 1 d. P(getting 7) = 0/6 = 0

A Success Getting an outcome in a specified event A Failure Getting 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 failures OR Odds in favor of an event = (number of successes):(number of failures)

example Two coins are tossed. Specify the following: a. the sample space, S b. the probability of getting exactly 2 heads c. the odds in favor of getting exactly 2 heads d. the probability of getting exactly 1 head e. the odds in favor of getting exactly 1 head

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 1:1 Two outcomes are successes and 2 are failures

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?

solution The sample space, S, is the set of all possible 3-card hands in a 52-card deck. Number of outcomes in S:

Solution cont’d… Let A be the event of picking 3 diamonds from the 13 in the deck Number of outcomes in A:

Solution cont’d…