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StatisticsBy: Mr. Danilo J. Salmorin

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COUNTING TECHNIQUES

PROBABILITY

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COUNTING TECHNIQUES

Tree diagram

Multiplication Rule

Permutation

Combination

TREE DIAGRAMa device used to list all possibilities of a sequence of events in a systemat

ic way.

Example:1. Suppose a sales rep can travel from New York to Pittsburgh by plane, train, or bus, and from Pittsburgh to Cincinnati by bus, boat, or automobile. List all possible ways he can travel from New York to Cincinnati.

Solution:A tree diagram can be drawn to

show the possible ways. • First the salesman can travel from New York to Pittsburgh by three methods.

• Then the salesman can travel from Pittsburgh to Cincinnati by bus, boat, or automobile.

• Next the second branch is paired up with the first branch in three ways

 • Finally, all outcomes can

be listed by starting at New York and following the branches to Cincinnati, as shown at the right end of the tree. There are nine ways.

2. A coin is tossed and a die is rolled. Find all possible outcomes of this sequence of events.

Solution:Since the coin can land either

heads up or tails up, and since the die can land with any one of six numbers shown face up.

Multiplication RuleIn a sequence of n events in which

the first one has k1 possibilities and the second event has k2 and the third has k3, and so forth, the total number of possibilities of the sequence will be k1 • k2 • k3 ••••• knNote: “And” in this case means to multiply

Examples:1. A paint manufacturer wishes to manufacture several different paints. The categories include:* Color Red, blue,

white, black, green, brown, yellow

* Type Latex, oil* Texture Flat, semigloss,

high gloss* Use Outdoor, indoor

How many different kinds of paint can be made if a person can select one color, one type, one texture, and one use?

Solution:A person can choose one color and one

type and one texture and one use. Since there are seven color choices, two type choices, three texture choices, and two use choices, the total number of possible different paint is

2. There are four blood types, A, B, AB, and O. Blood can also be Rh+ and Rh-. Finally, a blood donor can be classified as either male or female. How many different ways can a donor have his or her blood labeled?

Solution:Since there are four possibilities for blood type, two possibilities for Rh factor, and two possibilities for gender of the donor, there are 4 • 2 • 2 or 16, different classification categories as shown:

When determining the number of different possibilities of a sequence of events, one must know whether repetitions are permissible.

Exercises:1. Find all possible outcomes for the genders of the children in a family that has three children.2. If a baseball manager has five pitchers and two catchers, how many different possible pitcher-catcher combinations can he field?3. How many different three-digit identification tags can be made if the digits can be used more than once? If the first digit must be a 5 and repetitions are not permitted?

Two other rules that can be used to determine the total number of possibilities of a sequence of events are the permutation rule and combination rule.

These rules use factorial notation. The factorial notation uses the exclamation point. 5! Means 5•4•3•2•1 9! = 9•8•7•6•5•4•3•2•1

In order to use the formula in the permutation and combination rules, a special definition of 0! is needed. 0! = 1

Factorial FormulasFor any counting n n! = n(n-1) (n-2)…1 0! = 1

an arrangement of n object in a specific order.

PERMUTATION

Example:1. Suppose a business owner has a choice of five locations in which to establish her business. She decides to rank each location according to certain criteria, such as price of the store and parking facilities. How many different ways can she rank the five locations?

Solution:There are 5! = 54321 = 120

Different possible rankings. The reason is that she has five choices for the first location, four choices for the second location, three choices for the third location, etc.

2. Suppose the business owner in example no. 1 wishes to rank only the top three of the five locations. How many ways can she rank them?

Solution:Using the multiplication rule, she can

select any one of the five for first choice, then any one of the remaining four locations for her second choice, and finally, any one of the remaining three locations for her third choice, as shown.

Permutation Rule 1The number of permutations of n distinct objects is n!.Example:In how many ways can a photographer take pictures of five ladies in a row?Solution:

5! = 5 4 3 2 1 = 120

 

Permutation Rule 2The arrangement of n objects in a specific order using r objects at a time is called a permutation of n objects taking r objects at a time. It is written as nPr, and the formula is

nPr =

 

Example 1: five locations were taken and

then arranged in order; hence,5P5= = = = 120(Recall 0! = 1)In Example 2, three locations were selected from five locations, so n=5 and r=3; hence

5P3 = = = = 60

 

The notation nPr is used for permutations.

6P4 means or = = 360

Although example 1 and 2 were solved by the multiplication rule, they can now be solved by permutation rule.

 

2. How many ways can a basketball team schedule 3 exhibition games with 3 teams if they are all available on any of 5 possible dates?

Solution:5P3 = = (5) (4) (3) = 60

 

Example:1. Two lottery tickets are drawn from 20 for first and second prizes. Find the number of sample points in the space S.

Solution:The total number of sample points is

20P2 = = (20) (19) = 380

 

Permutations that occur by arranging objects in a circle are called circular permutations. Two circular permutations are not considered different, unless corresponding objects in the two arrangements are preceded or followed by a different object as we proceed in a clockwise direction.

For example, if 4 people are playing bridge, we do not have a new permutation if they all move one position in a clockwise direction. By considering 1 person in a fixed position and arranging the other 3 in 3! Ways, we find that there are 6 distinct arrangements for the bridge game.

Permutation Rule 3The number of permutations of n

distinct objects arranged in a circle is (n-1)!.

Example:In how many ways can eight persons be seated in a round table?Solution:

P= (8-1)! = 7! = 5040

Permutation Rule 4The number of distinct permutations of n things of which n1 are one kind, n2 of a second kind, … , nk of a kth kind is

 

Example:How many different ways can 3 red, 4 yellow, and 2 blue bulbs be arranged in a string of Christmas tree light with 9 sockets?

Solution:The total number of distinct

arrangements is, = 1260

 

Permutation Rule 5The number of ways of partitioning a set of n objects into r cells with n1 elements in the first cell, n2 elements in the second, and so on, is

= Where n1 + n2 + … + nr = n.

 

Example:How many ways can 7 people be assigned to 1 triple and 2 double rooms?

Solution:The total number of possible

partitions would be = = 210

 

The difference between permutation and combination is that in combination, the order of arrangement of the objects is not important: by contrast, order is important in a permutation.

In several problems we are interested in the number of ways of selecting r objects from n without regard to order. These selections are called combinations. A combination creates a partition with 2 cells, one cell containing the r objects selected and the other cell containing the n-r objects that are left.

a selection of distinct objects without regard to order.

COMBINATION

Permutations CombinationsAB BA CA DA AB BCAC BC CB DB AC BDAD BD CD DC AD CD

Example:1. Given the letters A, B, C, and

D, list the permutations and combinations for selecting two letters.

Solution:The listing follow:

¤ Note that in permutations, AB is different from BA. But in combinations, AB is the same as BA, so only AB is listed. (Alternatively BA could be listed instead of AB.)

¤ The elements of a combination are usually listed alphabetically.

¤ Combinations are used when the order or arrangement is not important, as

in the selecting process.

Combination RuleThe number of combinations of r objects selected from n objects is denoted by nCr and is given by the formula

nCr =

 

The number of such combinations, denoted by , is usually shortened to , since the number of elements in the second cell must be n – r.

 

Example:2. How many combinations of four objects are there taken two at a time?

Solution:Since this is a combination

problem, the answer is 4C2 = = = = 6

 

Notice that the formula for nCr is

Which is the formula for permutation,

 

With an r! in the denominator. This r! divides out the duplicates from the number of permutations, as shown in Example 2. For each two letters, there are two permutations but only one combination. Hence, dividing the number of permutations by r! eliminates the duplicates. This result can be verified for other values of n and r. Note: nCn = 1.

Example:From 4 Republicans and 3

Democrats find the number of committees of 3 that can be formed with 2 Republicans and 1Democrat.

Solution:The number of ways of

selecting 2, Republicans from 4 is

nCr = = 6.The number of ways of

selecting 1 Democrat from 3 is nCr = = 3.

So 4C2 ● 3C1 = 6 ●3 = 18.

 

We human beings are fond of gambling. When we gamble, we bet money. Some of us do not just bet money. We bet our life. We indulge in excessive drinking of alcoholic drinks. We smoke excessively or even we tend to drive exceeding the normal speed. We don’t care about the risks when we are involved in these activities because we don’t understand the concept of probability. On the other hand, some of us may fear activities that involve little risk to health or life because these activities have been sensationalized by press and the media.

So at this point, we will learn the concept of probability. Its meaning and how it is computed.

The theory of probability grew out of the study of various games of chance using coins, dice and cards. So these devices will be used as examples.

as a general concept can be defined as the chance of an event occurring.

PROBABILITY

* Processes such as flipping a coin, rolling a die, or drawing a card from a deck are called probability experiments.

* A probability experiment is a chance process that leads to well-defined results called outcomes.

* An outcome is the result of a single trial of a probability experiment.

* A trial means flipping a coin once, rolling a die once, or the like. When a coin is tossed, there are two possible outcomes: head or tail. In the roll of a single die, there are six possible outcomes: 1,2,3,4,5, or 6. In any experiment, the set of all possible outcomes is called the sample space.

* A sample space is the set of all possible outcomes of a probability experiment.

Experiment Sample SpaceToss one coin Head, tailRoll a die 1,2,3,4,5,6Answer a true-false question True, false

Toss two coins Head-head, tail-tail, head-tail, tail-head  

¤ It is important to realize that when two coins are tossed, there are four possible outcomes, as shown in the fourth experiment above. Both coins could fall heads up. Both coins could fall tails up. Coin 1 could fall heads up and coin 2 tails up.

Example:1. Find the sample space for rolling two dice.Solution: Since each die can land in six different ways, and two dice are rolled, the sample space can be represented by a rectangular array. The sample space is the list of pairs of numbers in the chart

Die 1 Die 21 2 3 4 5 61 (1,1) (1,2) (1,3) (1,4) (1,5) (1,6)2 (2,1) (2,2) (2,3) (2,4) (2,5) (2,6)3 (3,1) (3,2) (3,3) (3,4) (3,5) (3,6)4 (4,1) (4,2) (4,3) (4,4) (4,5) (4,6)5 (5,1) (5,2) (5,3) (5,4) (5,5) (5,6)6 (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Outcomes when two dice are rolled.

2. Find the sample space for drawing one card from an ordinary deck of cards.

Solution: Since there are four suits (hearts, clubs, diamonds, and spades) and 13 cards for each suit (ace through king), there are 52 outcomes in the sample space.

An event consists of the outcomes of a probability experiment.

An event can be one outcome or more than one outcome. For example, if a die is rolled and a 6 shows, this result is called an outcome, since it is a result of a single trial. An event with one outcome is called a simple event. The event of getting an odd number when a die is rolled is called a compound event, since it consists of three outcomes or three simple events. In general, a compound event consists of two or more outcomes or simple events.

There are THREE BASIC TYPES OF

PROBABILITY:1. Classical probability2. Empirical or relative frequency probability3. Subjective probability

CLASSICAL PROBABILITY - uses sample spaces to determine the numerical probability that an event will happen. One does not actually have to perform the experiment to determine that probability. Classical probability is so named because it was the first type of probability studied formally by mathematicians in the 17th and 18th centuries.

* Classical probability assumes that all outcomes in the sample space are equally likely to occur.

* Equally likely events are events that have the same probability of occurring

Formula for Classical ProbabilityThe probability of any event E is

 

This probability is denoted byP(E) =

This probability is called classical probability, and it uses the sample space S.

 

Rounding Rule for ProbabilitiesProbabilities should be expressed as

reduced fractions or rounded to two or three decimal places. When the probability of an event is an extremely small decimal, it is permissible to round the decimal to the first nonzero digit after the point.

Example:1. For a card drawn from an ordinary deck, find the probability of getting a king.

Solution:Since there are 52 cards in a

deck and there are 4 kings, P(king) = = .

 

2. If a family has three children, find the probability that all children are girls.Solution:

The sample space for the gender of children for a family that has three children is BBB, GBB, BBG, BGB, GGG, GGB, GBG, and BGG. Since there is one way in eight possibilities for all three

children to be girls,P(GGG) =

 

3. A card is drawn from an ordinary deck. Find these probabilities.a. Of getting a jack.b. Of getting the 6 of clubs.c. Of getting a 3 or a diamond.Solution:a. There are 4 jacks and 52 possible outcomes. Hence,

P(jack) = = b. Since there is only one 6 of clubs, the

probability of getting a 6 of clubs isP(6 of clubs) =

 

c. There are four 3s and 13 diamonds, but the 3 of diamonds is counted twice in this listing. Hence, there are 16 possibilities of drawing a 3 or a diamond, so

P(3 of diamond) = = There are four basic probability rules.

These rules are helpful in solving probability problems, in understanding the

nature of probability, and in deciding if your answers to the problems

are correct.

 

Probability Rule 1The probability of any event E is a number (either a fraction or decimal) between and including 0 and 1. This is denoted by 0 < P(E) < 1Rule 1 states that probabilities cannot be negative or greater than one.Probability Rule 2If an event E cannot occur (i.e., the event contains no members in the sample space), the probability is zero.Example:When a single die is rolled, find the probability of getting a 9

Solution: Since the sample space is 1,2,3,4,5, and 6, it is impossible to get a 9, Hence, the probability is P(9)= = 0.

 

Probability Rule 3If an event E is certain, then the probability of E = 1In other words, if P(E) = 1, then the event E is certain to occur.Example:When a single die is rolled, what is the probability of getting a number less than 7?Solution:Since all outcomes 1,2,3,4,5, and 6, are less than 7, the probability isP(number less than 7) = = 1The event of getting a number less than 7 is certain.

 

Outcome 1 2 3 4 5 6

Probability

Sum + + + + + = = 1

Outcome 1 2 3 4 5 6

Probability

Sum

Probability Rule 4The sum of the probabilities of the outcomes in the sample space is 1.Example:In the roll of a fair die, each outcome in the sample space has a probability of 1/6. Hence, the sum of the probabilities of the outcomes is as shown.

Another important concept in probability theory is that of complementary events. When a die is rolled, for instance, the sample space consists of the outcomes 1,2,3,4,5, and 6. The event E of getting odd numbers consists of the outcomes 1,3, and 5. The event of not getting an odd number is called the complement of event E, and it

consists of the outcomes 2,4, and 6.

The complement of an event E is the set of outcomes in the sample space that are not included in the outcomes of event E. The complement of E is denoted by Ē.Example:Find the complement of each event.a. Rolling a die and getting a 4.b. Selecting a letter of the alphabet and getting a vowel.c. Selecting a month and getting a month that begins with a J.

d. Selecting a day of the week and getting a weekday.

Solution:a. Getting a 1,2,3,5, and 6b. Getting a consonantc. Getting February, March, April, May, August, September, October, November, and Decemberd. Getting a Saturday and Sunday

The outcomes of an event and the outcomes of the complement make up the entire sample space. For two coins are tossed, the sample space is HH, HT, TH and TT. The complement of “getting all heads” is not “getting all tails,” the complement of the event “all heads” is the event “getting at least one tail.”

Since the event and its complement make up the entire sample space, it follows that the sum of the probability of the event and the probability of its complement will equal to 1. That is, P(E) + P(Ē) = 1. In the previous example, let E = all heads, or HH, and let Ē = at least one tail, or HT, TH, TT. Then P(E) = and P(Ē) = ; hence, P(E) + P(Ē) = + = 1.

The rule for complementary events can be stated algebraically in three ways.

 

Rule for Complementary Events P(Ē) = 1 – P(E) or P(E) = 1 – P(Ē) or P(E) + P(Ē) = 1

Stated in words, the rule is: If the probability of an event or the probability of its complement is known, then the other can be found by subtracting the probability from 1. This rule is important in probability theory because at times the best solution to a problem is to find the probability of the complement of an event and then subtract

from 1 to get the probability of the event itself.

Example:If the probability that a person lives in an industrialized country of the world is , find the probability that a person does not live in an industrialized country.

Solution:P (not living in an industrialized country) = 1 – P (living in an industrialized country) = 1 - =

 

The difference between classical and empirical probability is that classical probability assumes that certain outcomes are equally likely while empirical probability relies on actual experience to determine the likelihood of outcomes. In empirical probability, one might actually roll a given die 6000 times and observe the various frequencies and use these frequencies to determine the probability of an outcome. Suppose, for example, that a researcher asked 25 people if they liked the taste of a new soft drink. The responses were classified as “yes” “no” or “undecided.” The results were categorized in a frequency distribution, as shown.

Response FrequencyYes 15No 8Undecided 2

Total 25

Probabilities now can be compared for various categories. For example, the probability of selecting a person who liked the taste is , or , since 15 out of 25 people in the survey answered “yes.”

 

Formula for Empirical ProbabilityGiven a frequency distribution, the probability of an event being in a given class isP(E) = =

This probability is called EMPIRICAL PROBABILITY

and is based on observation

 

Example:1. In the soft-drink survey just described, find the probability that a person responded “no.”Solution:

P(E) = = 2. In a sample of 50 people, 21 had type O blood, 22 had type A blood, 5 had type B blood, and 2 had type AB blood. Set up a frequency distribution and find the following probabilities:a. A person has type O blood.b. A person has type A or type B blood.c. A person has neither type A nor type O blood.d. A person does not have type AB blood.

 

Solution:

a. P (O) = = b. P (A or B) = + = c. P (neither A nor O) = + = d. P(not AB) = 1 - = =

 

Type FrequencyA 22B 5AB 2O 21

Total 50

SUBJECTIVE PROBABILITY - uses a probability value based on an educated guess or estimate, employing opinions and inexact information.

In subjective probability, a person or group makes an educated guess at the chance that an event will occur. This guess is based on the person’s experience and evaluation of a solution. For example, a sportswriter may say that there is a 70% probability that the ADMU Blue Eagle will win the UAAP next year. A physician might say that on the basis of her diagnosis, there is 30% chance that the patient will need an operation. A seismologist might say there is an 80% probability that an earthquake will occur in a certain area.

Two events are mutually exclusive if they cannot occur at the same time.

In another situation, the events of getting a 4 and getting a 6 when a single card is drawn from a deck are mutually exclusive events, since a single card cannot be both a 4 and a 6. On the other hand, the events of getting a 4 and getting a heart on a single draw are not mutually exclusive, since one can select the 4 of hearts when drawing a single card from an ordinary deck.

Example:1. Determine which events are mutually exclusive and which are not when a single die is rolled.a) Getting an odd number and getting an even

number.b) Getting a 3 and getting an odd number.c) Getting an odd number and getting a number

less than 4.d) Getting a number greater than 4 and getting a

number less than 4.Solution:e) The events are mutually exclusive, since the

first event can be 1,3, or 5, and the second event can be 2,4, or 6.

f) The events are not mutually exclusive, since the first event is a 3 and the second can be 1,3, or 5. Hence, 3 is contained in both events.

2. Determine which events are mutually exclusive and which are not when a single card is drawn from a deck.a) Getting a 7 and getting a jack.b) Getting a club and getting a king.c) Getting a face card and getting an ace.d) Getting a face card and getting a spade.

Solution:Only the events in parts a and c are mutually

exclusive.

c) The events are not mutually exclusive since the first event can be 1,3, or 5, and the second can be 1,2, or 3. Hence, 1 and 3 are contained in both events.d) The events are mutually exclusive, since the first event can be 5 or 6, and the second event can be 1,2, or 3.

Addition Rule is used when the events are mutually exclusive.

Addition Rule 1When two events A and B are mutually exclusive, the probability that A or B will occur isP(A or B) = P(A) + P(B)Example:1. A restaurant has 3 pieces of apple pie, 5 pieces of cherry pie, and 4 pieces of pumpkin pie in its dessert case. If a customer selects a piece of pie for dessert, find the probability that it will be either cherry or pumpkin.

Solution:Since there is a total 12 pieces of pie,P(cherry or pumpkin) = P(cherry) + P(pumpkin)

The events are mutually exclusive.2. At a political rally, there are 20 Republicans, 13 Democrats, and 6 Independents. If a person is selected at random, find the probability that he or she is either a Democrat or Independent.Solution:

P(Democrat or Independent) = P(Democrat) + P(Independent) =

 

Addition Rule 2If A and B are not mutually exclusive, then P(A or B) = P(A) + P(B) – P(A and B)Note: This rule can be also be used when the events are mutually exclusive, since P(A and B) will always equal 0. However, it is important to make a distinction between the two situations.Example:1. In a hospital unit there are eight nurses and five physicians. Seven nurses and three physicians are females. If a staff person is selected, find the probability that the subject is a nurse or a male.

Solution:

The probability isP(nurse or male) = P(nurse) + P(male) – P(male

nurse) = + - =

 

Staf Females Males TotalNurses 7 1 8

Physicians 3 2 5

Total 10 3 13

Exercises:1.Determine whether the following events are

mutually exclusive.a.Roll a die: Get an even number, and get a

number less than 3.b.Roll a die: Get a prime number, and get an

odd number.c.Roll a die: Get a number greater than 3,

and get a number less than 3.d.Select a student in your class: The student

has blond hair, and the student has blue eyes.

e.Select a student in your college: The student is a sophomore, and the student is a business major.

f. Select any course: It is a calculus course, and it is an English course.

g.Select a registered voter: The voter is a Republican, and the voter is a Democrat.

2. On New Year’s Eve, the probability of a person driving while intoxicated is 0.32, the probability of a person having a driving accident is 0.09, and the probability of a person having a driving accident while intoxicated is 0.06. What is the probability of a person driving while intoxicated or having a driving accident?Solution:P(intoxicated or accident) = P(intoxicated) + P(accident) – P(intoxicated and accident) = 0.32 + 0.09 – 0.06 = 0.35

3. A furniture store decides to select a month for its annual sale. Find the probability that it will be April or May. Assume that all months have an equal probability of being selected.4. At a convention there are seven mathematics instructors, five computer science instructors, three statistics instructors, and four science instructors. If an instructor is selected, find the probability of getting a science instructor or a math instructor.5. An automobile dealer has 10 Fords, 7 Ferrari, and 5 Benz on his used-car lot. If a person purchases a used car, find the probability that it is a Ford or Ferrari.

6. On a small college campus, there are five English professors, four mathematics professors, two science professors, three psychology professors, and three history professors. If a professor is selected at random, find the probability that the professor is the following:

a.An English or psychology professorb.A mathematics or science professorc. A history, science, or mathematics professord.An English, mathematics, or history professor

7. The probability of a California teenager owning a surfboard is 0.43, of owning a skateboard is 0.38, and of owning both is 0.28. If a California teenager is selected at random, find the probability that he or she owns a surfboard or skateboard.8. On any given day, the probability of a tourist visiting Indian Caverns is 0.80 and of visiting the Safari Zoo is 0.55. The probability of visiting both places on the same day is 0.42. Find the probability that a tourist visits Indian Caverns or visits Safari Zoo on any given day.

9. A single card is drawn from a deck. Find the probability of selecting the following:

a.A 4 or a diamondb.A club or a diamondc.A jack or a black card

10. In a statistics class there are 18 juniors and 10 seniors: 6 of the senior are females, and 12 of the juniors are males. If a student is selected at random, find the probability of selecting the following:

d.A junior or a femalee.A senior or a femalef. A junior or a senior

Product Company ACompany

BCompany

CDresses 24 18 12

Blouses 13 36 15

11. A woman’s clothing store owner buys from three companies: A, B, and C. The most recent purchases are shown here

If one item is selected at random, find the following probabilities.

a.It was purchased from company A or is a dress.

b.It was purchased from company B or company C

c.It is a blouse or was purchased from company A.

Two events A and B are independent if the fact that A occurs does not affect the probability of B occurring.

Multiplication Rule 1When two events are independent, the

probability of both occurring isP(A and B) = P(A) * P(B)

Example:1. A coin is flipped and a die is rolled. Find the probability of getting a head on the coin and a 4 on the die. 

Solution:P(head and 4) = P(head) * P(4) = * =

 

2. A card is drawn from a deck and replaced; then a second card is drawn. Find the probability of getting a queen and then ace. 

Solution:P(queen and ace) = P(queen) * P(ace) = * = =  

3. An urn contains three red balls, two blue balls, and five white balls. A ball is selected and its color noted. Then it is replaced. A second ball is selected and its color noted. Find the probability of each of the following.a. Selecting two blue ballsb. Selecting a blue ball and then white ballc. Selecting a red ball and then a blue ballSolution:

a. P(blue and blue) = P(blue) * P(blue) = * = = b. P(blue and white) = P(blue) * P(white) = * = = c. P(red and blue) = P(red) * P(blue) = * = =

 

 

4. Approximately 9% of men have a type of color blindness that prevents them from distinguishing between red and green. If 3 men are selected at random, find the probability that all of them will have this type of red-green color blindness. 

Solution: 

P(C and C and C) = P(C) * P(C) * P(C) = (0.09)(0.09)(0.09) = 0.000729 

When the outcome or occurrence of the first event affects the outcome or occurrence of the second event in such a way that the probability is changed, the events are said to be dependent.

Multiplication Rule 2  When two events are dependent, the probability of both occurring is,

P(A and B) = P(A) * P(B/A) Example: 1.In a shipment of 25 microwave ovens, 2 are defective. If two ovens are randomly selected and tested, find the probability that both are defective if the first one is not replaced after it has been tested. 

Solution:P(D1 and D2) = P(D1) * P(D2/D1) = * = =

 

2. The World Wide Insurance Company found that 53% of the residents of a city had homeowner’s insurance with the company. Of these clients, 27% also had automobile insurance with the company. If a resident is selected at random, find the probability that the resident has both homeowner’s and automobile insurance with the World Wide Insurance Company. Solution:P(H and A) = P(H) * P(A/H) = (0.53)(0.27) = 0.1431

3. Box 1 contains two red balls and one blue ball. Box 2 contains three blue balls and one red ball. A coin is tossed. If it falls heads up, box 1 is selected and a ball is drawn. If it falls tails up, box 2 is selected and a ball is drawn. Find the probability of selecting a red ball.

Solution:With the use of tree diagram, the sample

space can be determined. First, assign probabilities to each branch. Next, using the multiplication rule, multiply the possibilities for each branch.

P(B2)

 

Finally, use the addition rule, since a red ball can be obtained from box 1 or box 2.P(red) = + = + =

Note: The sum of all probabilities will always be equal to 1.

 

Conditional ProbabilityThe conditional probability of an event B in

relationship to an event A was defined as the probability that event B occurs after event A has already occurred.

The conditional probability of an event can be found by dividing both sides of the equation for multiplication rule 2 by P(A), as shown:

P(A and B) = P(A) • P(B/A) = = P(B/A)

 

Formula for Conditional Probability

The probability that the second event B occurs given that the first event A has occurred can be found by dividing the probability that both events occurred by the probability that the first event has occurred. The formula is

P(B/A) =

 

Example:1. A box contains black chips and white chips. A person selects two chips without replacement. If the probability of selecting a black chip and white chip is , and the probability of selecting a black chip on the first draw is , find the probability of selecting the white chip on the second draw, given that the first chip selected was a black chip.

Solution:LetB = selecting a black chip W = selecting a white chipThen

P(W/B) = = = = • =

 

Other Examples

Examp[es:1. A box contains 24 transistors, four of which are defective. If four are sold at random, find the following probabilities.a. Exactly two are defective.b. None is defective.c. All are defective.d. At least one is defective.Solution:There are 24C4 ways to sell four transistors, so the denominator in each case will be 10,626.a. Two defective transistors can be selected as 4C2 and two nondefective ones as 20C2

P(exactly 2 defectives) = = =  

 

b. The number of ways to choose no defective is 20C4. Hence,P(no defective) = = =  c. The number of ways to choose four defectives from four is 4C4, or 1. Hence,P(all defectives) = =  d. To find the probability of at least one defective transistor, find the probability that there are no defective transistors, and then subtract that probability from 1P(at least 1 defective) = 1 – P(no defectives) = 1 - = 1 - =

 

A binomial experiment is a probability experiment that satisfies the following four requirements:1. Each trial can have only two outcomes or

outcomes that can be reduced to two outcomes. These outcomes can be considered as either success or failure.

2.There must be a fixed number of trials.3.The outcomes of each trial must be

independent of each other.4. The probability of success must remain

the same for each trial.

* A binomial experiment and its results give rise to a special probability distribution called the binomial distribution.

* The outcomes of a binomial experiment and the corresponding probabilities of these outcomes are called a binomial distribution.

* In binomial experiments, the outcomes are usually classified as successes or failures. For example, the correct answer to a multiple-choice item can be classified as a success, but

any of the other choices would be incorrect and hence classified as a failure.

Notation for the Binomial Distribution

P(S) The symbol for the probability of successP(F) The symbol for the probability of failurep the numerical probability of successq The numerical probability of failureP(S) = p and P(F) = 1 – p = qn The number of trials

X The number of successesNote that 0 < X < n.

The probability of a success in a binomial experiment can be computed with the ff. formula.Binomial Probability Formula

In a binomial experiment, the probability of exactly X successes in n trials is.

P(X) = px qn-X

Example:1. A coin is tossed three times. Find the probability of getting exactly two heads.

Solution:HHH, HHT, HTH, THH, TTH, THT, HTT,

TTTThis answer is or 0.375.

 

Looking at the problem in Example 1 from the standpoint of a binomial experiment, one can show that it meets the four experiments.1. There are only two outcomes for each trial, heads or tails.2. There is a fixed number of trials (three).3. The outcomes are independent of each other (the outcome of one toss in no way affects the outcome of another toss).

4. The probability of a success (heads) is in each case.

 

In this case, n = 3, X = 2,p = . Hence, substituting in the formula gives

P(2 heads) = ( ( = = 0.375Which is the same answer obtained by using the

sample space.The same example can be used to explain the

formula. First, note that there are three ways to get exactly two heads and one tail from a possible eight ways. They are HHT, HTH, and THH. In this case, then, the number of ways of obtaining two heads from three coin tosses is 3C2, or 3. In general, the number of ways to get X successes from n trials without regard to order is

nCX =

 

2. If a student randomly guesses at five multiple-choice questions, find the probability that the student gets exactly three correct. Each question has five possible choices.

Solution:In this case n = 5, X = 3, and p = , since there is one chance in five of guessing a correct answer. Then,

P(3) = ( = 0.05

 

Recall that in order for an experiment to be binomial, two outcomes are required for each trial. But if each trial in an experiment has more than two outcomes, a distribution called the multinomial distribution must be used. For example, a survey might require the responses of “approve,” “disapprove,” or “no opinion.” In another situation a person may have a choice of one of five activities for Friday night, such as a movie, dinner, baseball game, play, or party. Since these situations have more than two possible outcomes for each trial, the binomial distribution cannot be used to

compute probabilities.

The multinomial distribution can be used for such situations if the probabilities for each trial remain constant and the outcomes are independent for a fixed number of trials. The events must also be mutually exclusive.

Formula for the Multinomial Distribution

If X consists of events E1, E2, E3, ….., Ek, which have corresponding probabilities p1, p2, p3, …… pk of occurring, and X1 is the number of times E1 will occur, X2 is the number of times E2 will occur X3 is the number of times E3 will occur, etc., then the probability that X will occur isP(X) = P1X1 P2X2 …….. PkXk

Where X1 + X2 + X3 + … + Xk = n, and P1 + P2 + P3 … + Pk = 1

 

Example:1. In a large city, 50% of the people choose a movie, 30% choose dinner and a play, and 20% choose shopping as a leisure activity. If a sample of five people is randomly selected, find the probability that three are planning to go to a movie, one to a play, one to a shopping mall.Solution:n = 5, X1 = 3, X2 = 1, X3 = 1, P1 = 0.50, P2 = 0.30, and P3 = 0.20. Substituting in the formula gives

P(X) = (0.50)3 (0.30)1 (0.20)1 = 0.15

 

Again, note that the multinomial distribution can be used even though replacement is not done, provided that the sample is small in comparison with the population.

2. In a music store, a manager found that the probabilities that a person buys zero, one, or two or more CDs are 0.3, 0.6, and 0.1, respectively. If six customers enter the store, find the probability that one won’t buy any CDs, three will buy one CD, and two will buy two or more CDs.Solution:N = 6, X1 = 1, X2 = 3, X3 = 2, P1 = 0.3, P2 = 0.6, and P3 = 0.1. Then,

P(X) = (0.3)1(0.6)3(0.1)2= 60 (0.3) (0.216) (0.01) = 0.03888

 

Thus the multinomial distribution is similar to the binomial distribution but has the advantage of allowing one to compute probabilities when there are more than two outcomes for each trial in the experiment. That is, multinomial distribution is a general distribution, and the binomial distribution is a special case of the multinomial distribution.

When sampling is done without replacement, the binomial distribution does not give exact probabilities, since the trials are not independent. The smaller the size of the population, the less accurate the binomial probabilities will be.

For example, suppose a committee of four people is to be selected from seven women and five men. What is the probability that the committee will consist of three women and one man?

To solve this problem, one must find the number of ways a committee of three women and one man can be selected from seven women and five men. This answer can be found by using combinations; it is

7C3 5C1 = 35 5 = 175

Next, find the total number of ways a committee of four people can be selected from 12 people. Again, by the use of combinations, the answer is

12C4 = 495Finally, the probability of getting a

committee of three women and one man from seven women and five men is

P(X) = =

 

The results of the problem can be generalized by using a special probability distribution called the hypergeometric distribution. The hypergeometric distribution is a distribution of a variable that has two outcomes when a sampling is done without replacement.

Formula for the Hypergeometric Distribution

Given a population with only two types of objects (females and males, defective and nondefective, successes or failures, etc.), such that there are a items of one kind and b items of another kind and a + b equals the total population, the probability P(X) of selecting without replacement a sample of size n with X items of type a and n –X items of type b is

P(X) =

 

The basis of the formula is that there are aCX ways of selecting the first type of items, bCn-X ways of selecting the second type of items, and (a+b)Cn ways of selecting n items from the entire population.

Example:1. Ten people apply for a job as assistant

manager of a restaurant. Five have completed college and five have not. If the manager selects three applicants at random, find the probability that all three are college graduates.

Solution:Assigning the values to the variable givesa = 5 college graduates n = 3b = 5 nongraduates X = 3and n – X = 0. Substituting in the formula gives

P(X) = = =

 

2. A recent study found that four out of nine houses were underinsured. If five houses are selected from the nine houses, find the probability that exactly two are underinsured.Solution:In this problema = 4 b = 5 n = 5 X = 2 n-X=3Then,

P(X) = = =  

 

Exercises: 

1. When two dice are rolled, find the probability of gettinga. A sum of 6 or 7b. A sum greater than 8c. A sum less than 3 or greater than 8d. A sum that is divisible by 3e. A sum of 16f. A sum less than 112. The probability that Sam will be accepted by the college of his choice and obtain a scholarship is 0.35. If the probability that he is accepted by the college is 0.65, find the probability that he will obtain a scholarship given that he is accepted by the college.3. The probability that John has to work overtime and it rains is 0.028. John hears the weather forecast, and there is a 50% chance of rain. Find the probability that he will have to work overtime given that it rains.

Probability Problems

A geology professor has 5 silicates, 7 pyrites and 8 carbonates in a rock collection. He picks 6 rocks at random for a student to analyze. Find the probability that the professor picked 2 silicates, 1 pyrite and 3 carbonates. Express your answer in lowest fraction.Answer:

Solution:If there are 20 rocks in all, the number of ways picking 6 of these is = = 38,760The number of ways of picking 2 silicates, 1 pyrite and 3 carbonates is = = 3920Hence, the probability of picking these rocks is or approximately 0.101.

 

Russian roulette is played by two players as follows. One bullet is placed into empty cylinder of a six-gun and the cylinder is spun. The first player holds the gun to his head and pulls the trigger. If he does not die, he wins the game. If he dies, then the second player takes his turn by spinning the cylinder, putting the gun to his head, and pulling the trigger. If he dies the first player wins. If he does not die, then the game is played again. What is the probability that the second player will die?Answer:Probability of the second player will die is (5/6) (1/6) = 5/36 Hint: Compute the probability that the first player will die. Then assuming that the first player did not die, compute the probability that the second player will die.

A student elects to solve a problem about compound interest using either arithmetic methods, algebraic methods or calculus methods with corresponding probabilities of 2/9, 4/9 and 1/3. With each of these methods, the respective probabilities of obtaining the correct answer are 9/10, 7/8 and 3/4 . Determine the probability that the student obtains the correct answer. Solution:P(correct) = (choosing arithmetic and being correct OR choosing algebra and being correct OR choosing calculus and being correct)

= 2/9 x 9/10 + 4/9 x 7/8 + 1/3 x ¾= 2/10 + 7/18 + ¼=

 

Debra is taking two college entrance exams, in English and in Math. The probability that she will pass the English exam is 0.75. The probability that she will fail the Math is 0.20. The probability that she will pass both exams is 0.65. What is the probability that Debra will pass either the Math or the English exam? Solution:P(A or B) = P(A) + P(B) – P(A and B)P(English or Math) = P(English) + P(Math) – P(Math and English)P(English) = 0.75P(not Math) = 0.20P(Math) = 1 – 0.20 = 0.80P(English or Math) = 0.75 + 0.80 – 0.65 = 1.55 – 0.65 = 0.90

In how many ways can a committee of 5 be formed with 3 seniors and 2 juniors as its members if there are 6 seniors and 5 juniors to choose from? Solution: 6C3 * 5C2 = 20 * 10 = 200

“We cannot hope that many children will learn MATHEMATICS unless we find a way to

share our enjoyment and show them its beauty as well as its utility.”

Thank you for listening! ^_^

END Of

Discussion

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