counting techniques and basic probability concepts · 2018-10-13 · [counting techniques and basic...

Perhaps it could be said that the reason for the study of probability in the early days is the man’s urge to gamble. Tracing back to human history, gamblers in the early days seek the help of mathematicians to help them maximize their profit in gambling. Gamblers are hoping that mathematicians would find various ways and strategies to optimize their winnings in different games of chance. Some of the great mathematicians that have helped in formulating these strategies were, Blaise Pascal, Leibniz, Pierre De Fermat, and James Bernoulli.

Because of this early development in the field of statistical theory of probability and statistical inferences, today it branches out into many other fields of study that encompassed many branches of disciplines were chances are associated. Some of these branches of studies are in politics, business, weather forecasting, scientific researches

1 Sample Space and Events In this section we review our notions of sets in algebra as applied to statistics. We also introduce some

of the terms that we will encounter in our study of probability.

A statistical experiment refers to any procedure that yields a collection of outcomes. An example of a statistical experiment could be a tossing of a coin, or a rolling of a die. In the real world scenario, statistical experiments could be, buying a television set. The buying of a television set could bring about an outcome of buying a defective TV set. Another possible outcome is that the TV set is not defective.

A sample space is the set of all possible outcomes of a statistical experiment and is represented by the symbol “S”. Each outcome in a sample space is called an element or a member of the sample space or simply a sample point.

A sample space is called a finite sample space if it had a finite number of outcomes in it, while a sample space consisting of infinite number of outcomes is called an infinite sample space.

An event is a subset of a sample space. A simple event or an elementary event is an event containing

a single outcome. The union of several simple events is what we call the compound event.

An impossible event is an event that has no outcome in it or simple an outcome that cannot occur. On the other hand we call an event a sure event if it has all the possible outcomes of the sample space,

or simply an event that will surely occur.

The null space or the empty space is a subset of the sample space that contains no elements, denoted

by the symbol .

There are several ways of writing the elements of our sample space. One of which is to list all the possible elements of that space. For instance, if we have an experiment of tossing a die once, our sample space would be, S = {1, 2, 3, 4, 5, 6} or we could just describe the elements of our sample space, that is

𝑠 = {𝑥|0 < 𝑥 < 7, 𝑤ℎ𝑒𝑟𝑒 𝑥 𝑖𝑠 𝑎𝑛 𝑖𝑛𝑡𝑒𝑔𝑒𝑟}.

Counting Techniques and Basic Probability Concepts

[COUNTING TECHNIQUES AND BASIC PROBABILITY CONCEPTS]

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The relationships between events and the corresponding sample space can be illustrated by the used of Venn-diagrams. We refer the sample space as our universal set thus is denoted by U.

1.1 Operations with Events

INTERSECTION OF TWO EVENTS

Just like ordinary set operations, we also have operations of events. We define the intersection of two

events A and B, denoted by the symbol A B, is the event containing all elements that are common to

both A and B. The elements of A B can be listed using the rule AB = {x | x A and x B}.

Example

a. Let A = {1, 2, 3, 4, 5} and let B ={2, 4, 6, 8,10}: then the intersection of A and B denoted by

A B = {2, 4} b. Let A = {the set of all positive integers greater than 5 but less than 12} and let B ={3, 6, 9} thus

A B = {6, 9}

If and whenever two events cannot occur at the same time, we define the two sets to be mutually

exclusive. Two events A and B are mutually exclusive if A B = , that is A and B have no elements in common. This is clearly illustrated using a Venn diagram.

U

A B

U

B

C

A B

Set A and B have

some elements in

common


3

It is clearly seen in the diagram the two events have no elements in common.

Example Suppose that a die is tossed and let A be the event that an even number occur and let B, the event that

an odd number occurs. We note that A = {2, 4, 6} and B = {1, 3, 5} it is obvious that the two events does

not have any element in common thus A B = . Hence, events A and B are said to be mutually exclusive.

UNION OF TWO EVENTS Another operation that we would like to look into is the union of events; we define the union of two

events similarly to how we define union of two sets. The union of two events, A and B denoted by A B is the event containing all the elements of or to B or to both. The elements of the union of A and B can be

listed or defined using the rule A B = {x | x A or x B}.

Example

a. Let A = {2, 3, 4, 5, 6} and let B = {3, 6, 8, 10}: then A B = {2, 3, 4, 5, 6, 8, 10}.

b. Let A = {x | x < 0, x is a real number} and B = {x | x 0, x is a real number} then the union of A and B

denoted by A B is the whole number line or simply R (all real numbers).

Lastly, we define the complement of an event with respect to the sample space. If A is an event then the complement of an event A, denoted by Ac or A’ is the event containing all the sample point of the

sample space that are not elements of event A, written in mathematical form as Ac = {x| x S and x A}.

EXERCISE Do the following. 1. Given sets: U = {x| x is a positive integer less than 12} A = {x| x is an even number less than 10} B = {x| x is a positive integer less than 6} C = {1, 5, 7, 8, 9, 10} If we consider our sample space to be the set U, list the sample points of the following

a. A’ C’

b. (AB) C

c. (AB) (CB)

d. (AC)’

e. (BC)’ A 2. Give at least 5 examples of a statistical experiment and state some possible outcomes for each. 3. Suppose a group of 5 teenagers went to a swimming trip at Batangas during a summer vacation. And further suppose that C is the event of a car malfunction occurs during the trip, V is the event that they commit a traffic violation and being given a ticket and N is the event that they could not find any vacant resort open for public swimming.


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Describe what the following regions mean. a) Region 1 b) Regions 1 and 3 together c) Region 2 d) Regions 5, 3, and 6 together e) Regions 6, 2, 7, and 8 together f) Region 8

2 Counting Techniques One of the problems that statisticians must consider and attempt to evaluate is the element of

chance associated with the occurrence of certain events when an experiment is performed. These problems belong in the field of probability. Determining the number of possible outcomes a given statistical experiment have is needed in our study of probability. Thus there is a need to find ways of counting these outcomes not by mere listing of all possibilities but some mathematical computations. Knowing the principles of counting is a prerequisite in finding the probability of certain outcomes. Consider the following examples:

1. If a coin is tossed once, how many possible outcomes are there? 2. If a die is tossed once, how many possible outcomes are there? 3. If a pair of coin is tossed once, how many possible outcomes are there? 4. If four coins are tossed once, how many possible outcomes are there?

These questions would bring about the question on the possibility of counting the number of outcomes that a certain experiment could produce. This question can be answered by the following procedures:

2.1 Tree Diagram

One way of counting the number of possible outcomes is to list down all possible outcomes systematically by means of a tree diagram. The example below shows

Example For instance, in the experiment of tossing a coin twice will produce the following outcomes:

C V

N

5 4

1

3 2

6 8

7


5

In this experiment of tossing a coin twice, the sample space contains 4 possible outcomes. Thus

the sample space of tossing a coin twice is a finite sample space. If we let E be the event such that at least one head will occur after tossing a coin twice. Then the event contains the following outcomes from our sample space.

Since event E contains more than 1 sample point or outcome, then E is said to be a compound event.

Second Toss

H

T

Sample Space

T

H

T

H

First Toss Second Toss

T H

H

T

T H

H

T

Event E

T H

H T H H


6

Suppose event F is the event where exactly one head will occur, the sample points for this event is shown below: Suppose event G is the event where exactly two tails will occur, the sample points for this event is shown below:

Example a. In an experiment of tossing a die and then tossing a coin each once, what are the possible outcomes?

Tossing a dice Tossing a coin Sample Space

T

H

H

H

H

H

H

H

T

T

T

T

T

T

Event F

T H T H

Event G

T T


7

From our illustration the total number of possible outcomes is 12. b. In an experiment of tossing a die once, if an even number occurs toss a coin once but if an odd number

occurs toss a coin twice. What are the possible outcomes? The following tree diagram shows the number of possible ways the given experiment can occur.

Tossing a dice First Toss Sample Space

T

H

H

H

H

H

T

T

T

T

Second Toss

T

H

H

T

First Toss

T

H

H H

T H T T

H T H H

T H T T

H T H H

T H T T


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2.2 Fundamental Principle of Counting Determining the total possible outcomes can be tedious when we try to complicate our experiment.

Drawing and listing all possible outcomes would be nearly impossible for some situations. Thus we need to think of ways so that we can account for all those possibilities.

A more simple and easy way of counting the number of possible outcomes is by the use of the fundamental principle of counting. It is often referred to as the multiplication rule.

Example If a pair of dice is tossed once, how many possible outcomes are there? Since there are 6 way or 6 possible outcomes if one die is tossed and also for the second die there are (6)(6) =36 ways that the pair of dice can land. To illustrate this we have the following figure. The second dice is colored so that it can be distinguished from the first dice.

Example

In general, if there are k321 O,...,O,O,O operation to be perform such that O1 can be done in n1 ways,

O2 can be done in n2 ways, and so on until Ok can be done in nk ways, then all the operations can be

performed in k21 nnn ways

The multiplication rule states that if an operation can be performed in n ways, and if each of these a second operation can be performed in m ways then the two operations can be performed in n*m ways.


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Exercises 1. How many even three digit numbers can be formed from the digits 1, 2, 3, 4, and 5 if each digit can be used once only? 2. How many two-digit odd numbers can be formed using the numbers 2, 4, 6, 8, and 9 if repetition is allowed? 3. A college freshman must take a math course, a science course, and an English course. If she may select any of 6 Math courses, any of 5 Science courses, and any 4 English courses, how many ways can she arrange her program? 4. How many different license plates (consisting of 3 letters and three numbers) are there altogether if

i. there are no restrictions? ii. if the letters must be different?

iii. if the letters must be different and the first digit cannot be 0? iv. the letters and numbers must be different?

5. How many ways can you order a meal from a certain restaurant if you have two choices of drink (coffee or tea), three main courses to choose from (chicken, beef, or fish) and two desserts (pie or cake) where a full meal consists of a drink a main course and a dessert? 6. In the 63 UAAP season, eight sprinters from different universities are in the final match of a race. How many different ways are there to award the gold, silver, and bronze medals? 7. A committee is to elect a president, secretary, and treasurer. If the committee consists of 35 members, how many outcomes for the three positions can occur (note that no member can elected in two positions)? 8. In how many ways can a model dress up at a fashion show if she has 5 different shoes, 4 different pants, 5 different shirts, and 4 different hats?

2.3 Permutations There are times when we are interested in knowing how many ways can we arrange people things

in a row or in a line. For example, we might want to know how many arrangements are possible for six people be seated around a table or how many different orders are possible to draw 2 tickets form a total of 20 different tickets. These different arrangements are called permutations.

A Permutation is an arrangement of all or part of a set of distinct objects. For instance, the letters

of the word DOG can be arranged in six different ways namely: DOG, DGO, OGD, ODG, GOD, and GDO. Using the fundamental principle of counting we could actually compute for the number of ways we can arrange the letters without listing the possible outcomes. Our task is to arrange the letters but we can view it this way. We want to fill out three positions for the three letters. Filling out the first one, we have three choices, all the letters, D,O,G. For the second position we only have two choices since we already placed one of the letters in the first position. Finally we are to fill up the last position; however we now only have one letter to choose from. Thus in effect we have (3)(2)(1) = 6 ways of arranging the letters.

What if we have 4 letters, for instance, the word BOLT. Following the above procedure, we can get the following computations (4) (3)(2)(1) = 24 ways.


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In general, if we have n distinct letters or objects to

arrange, we can arrange those letters or objects in n (n-1) (n-2) (n-3)… (3)(2)(1) ways or simply n! (Read as “n factorial”) ways.

Consider the following situation. We have four different objects; say A, B, C, and D but we only

want to arrange 2 objects a time. This would give us the following outcomes: AB, AC, AD, BA, BC, BD, CA, CB, CD, DA, DB, and DC. Cleary we have 12 ways of arranging 2 objects out of the four. Again using the fundamental principle of counting we can 2 positions to fill up. For the first one, we have four choices and for the second position we only have 3 choices since we already used one of the objects to fill in the first position. Thus we have the following computation: (4) (3) = 12 ways.

In general, we can arrange n distinct objects taken r at a time in n(n-1)(n-2)…(n-r+1) ways. We can simply represent the product by the symbol nPr = n!/(n-r)!

Example

Suppose we have 3 different math books. In how many ways can these books be arranged in a bookshelf that has a slot for 3 more books?

The number of

permutations of n distinct

objects is n!

M2 M1 M2 M3

M1 M3

M2 M3

M2

M1 M3 M2 M1

M3 M1 M2

M1 M3

Math 1

Math

2

Math 3

If we are to arrange n distinct object taken r at a time

!!

rn

nPrn


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What if we do not want to arrange things or objects in a straight line or in a linear manner? What if we want to arrange objects around a circle? Can we still use the procedures above?

The answer is no. Let us consider arranging 3 distinct objects around a circle. One way of doing so is to first fix one object at a certain position and then arrange the other objects.

Note that the following arrangements are equivalent.

There are only 2 ways in arranging the 3 objects around a circle. Try arranging 4 distinct objects

around a circle and you will arrive at the following computations (3)(2)(1) = 6 ways.

Example a. If we are to arrange 7 people around a circular table, how many ways can we do so?

Answer: (7-1)! = 6! = 720 ways b. In how many ways can we arrange 4 people around a committee table if two of them need to be

seated next to each other?

In general, the number of permutations of n distinct objects in a circle is (n-1)!.

A

C B

A

C B

A

C B

B

A C

Equivalent


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One way of solving this problem is a group together object that needs to be grouped and treating them, the group, as one object to be arranged. Suppose person A and B needs to be seated next to each other all the time, grouping the two, I will now have to arrange 3 objects in a circular manner.

In this case, we will now have to perform two operations, arranging the three objects (group of A and B, person C, and person D) and then I would still have to arrange person A and B within their group. Applying the fundamental principle of counting, we just need to multiply the number of ways we can perform these operations. Since there are 3 objects that need to be arranged in a circular manner, we have (3-1)! = 2! = 2. There are two ways of arranging 3 objects in a circle. As for arranging person A and B, we also have two ways of arranging them. Thus, there are 4 ways of arranging 4 people in a circular manner such that two of those people need to be seated next to each other.

From our previous discussions we limited ourselves from distinct objects. What if we have common objects or certain objects cannot be distinguished from other objects? For instance if we are to arrange the letters in the word “ALL”, in how many ways can we do it?

Listing all possible outcomes, we have the following: ALL, LAL, and LLA . Clearly we can see that there are only three ways in doing so. What if we are to arrange the letters of the word BALL. The following are the possible outcomes: BALL, BLAL, BLLA, ABLL, ALBL, ALLB, LABL, LALB, LBAL. LBLA, LLAB, and LLBA, thus we have 12 ways.

In general, the number of distinct permutations of n objects, n1 of which are of one kind, n2 of which are of another kind, until nk of the kth kind is

!!..nn!n k21

!n

Let us check our examples above using the formula. In arranging the letters of the word ALL, there are 3 objects to be arranged so our n is 3, n1 = 2. Thus we have the following 3!/2! = 3. In the example of arranging the letters of the word BALL, we have the following: 4!/2! = 12.

Exercise 1. Evaluate the following:

a) 13 P b) 45 P c) 06 P

A B

C D


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d) 99 P e) 87 P f) 35 P

2. A group of 5 teens went to a movie. In how many ways can they be arranged in a row of 5 seats? 3. In how many ways can you arrange 2 students in two chairs if you can choose from 4 students? 4. How many ways can the 5 starting position on a basketball team with 9 men who can play any

position? 5. How many distinct permutations can be made from the letters of the word “ENGINEER”? 6. In how many ways can 5 different trees be planted around a circular lot? 7. In how many ways can 4 boys and 5 girls sit in a row if the boys and girls must alternate? 8. Four married couples have bought 8 seats in a row for a concert. In how many different ways can

the be seated a. With no restrictions? b. If each couple is to sit together? c. If all the men sit together to the right of all the women?

9. How many ways can 6 people be lined up to get on a bus if: a. 3 persons insist on following each other? b. If 2 certain persons refuse to follow each other?

10. In a family Noche Buena, how many ways 8 family members be seated around a table if: a) Four persons insist of sitting next to each other? b) Two persons refuse to sit next to each other? 11. A travel agency offers 8 different tour packages and a choice of one among three tour guides: Brenda,

Sheryl, and Ginny. Find the number of ways in which a tourist could choose a tour package and a tour guide for a much needed and well-deserved vacation.

12. A certain cellular phone brand comes in 5 different models with each model available in 6 different colors. If the manufacturer of this cell phone wishes to display his product showing all possible combinations of the model and color, how many cell phone units should he put on display?

13. A witness to a bank robbery reported to the police that the car plate number of the getaway car contained the letters UTR followed by three digits, the last one of which was a 0. The witness could not remember the first two digits but he is sure that they are nonzero and the three digits are different from one another. How many car plate numbers at the most will have to be checked by the police?

14. A one peso coin, 10-centavo coin and a 25-centavo coin are tossed together with a die, how many possible outcomes are there?

15. If all questions are answered (i.e., blank answers not allowed) in a true-false quiz consisting of 5 questions, how many ways are there of answering the entire quiz?

16. If a new car buyer is faced with a choice of 3 body styles, 3 engines types and 8 different colors, in how many different ways can he choose a body style, an engine, and a color for his car?

17. At a certain museum there are four entrance gates and 6 exits. In how many ways may two men enter together, but leave by different exits?

18. How many three-digit numbers can be formed from the integers 1, 2, 3, 4, 5, and 6 if (a) Repetition of digits is allowed? (b) Repetition of digits is not allowed? (c) The number must be even and repetition of digits is allowed? (d) the number must be even and repetition of digits is not allowed? (e) The number must be less than 500?

19 If a multiple-choice test consists of 10 questions each with 5 possible answers of which only one is correct,

(a) in how many different ways can a student check off one answer to each question?


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Selection 2 Selection 3

(b) in how many different ways can a student check off one answer to each question and get all the answers wrong?

2.4 Combinations

In permutations, we are concerned in arranging object. What if we disregard the arrangements and we wish to find the number of ways of selecting objects regardless of their arrangement? These selections are called combinations.

The number of ways in selecting “r” distinct objects at a time out of “n” distinct objects is given by

!r!rn

!nC rn

Example a). For instance we have the 3 colored poker chips in a jar, a red, a yellow and a black chip. In how many ways can we select three chips from the jar?

Clearly we only have 3 choices in choosing one letter at a time we could only have 3 different ways to select a letter, namely A, B and lastly C.

Thus, we say that there are only 3 possible combinations of selecting on chip from the given jar. In how many ways can we select 2 chips from the jar? The following are the only possible selection:

Selection 1

R B Y

Selection 1

B

Selection 2 Selection 3

R Y

B Y R

R

Y

B


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Note that we are not particular with the arrangement and we are just after the objects to be selected so the following selections are equivalent.

What if we select 3 chips? The only possible selection is {Red, Yellow, Black}

b). In a committee of 4 members, in how many ways can you select 3 members to form a subcommittee?

Using our formula we have,

41*2*3*1

1*2*3*4

!3!34

!4C34

Exercise 1. Evaluate the following: a) 5C3 b) 7C3 c) 10C9 d) 4C4 e) 20C1 f) 20C0 2. From 4 red, 5 green and 6 yellow apples, how many selections of 9 apples are possible if 3 red apples,

3 green, and 3 yellow apples are to be selected? 3. Anika is a college freshman and a freshman must take a science course, a humanities course, an

elective course, and a mathematics course from the first term of the school year. If she may select any of the 6 science courses, any of the 4 humanities courses, 5 elective courses, and any of the 4 mathematics courses that are still open for enrollment, how many ways can she arrange her program for the term?

Y R

Y R Equivalent

Y R B

Selection 1

A B C

Selection 2

A B D

Selection 3

A D C

Selection 4

B

A

C D


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4. Out of 4 mathematicians and 6 scientists, a committee consisting of 2 mathematicians and 3 scientists is to be formed. How many possible committees can be formed if :

a) Any mathematician and any scientist can be included? b) One particular scientist must be in the committee? c) Two particular mathematicians must be in the committee? d) Two particular scientists must not be in the committee?

5. In selecting an ace, king, queen, and jack from an ordinary deck of 52 cards, how many ways may we choose if the cards must be of

(a) Different suits? (b) The same suit?

6. A box of jelly bears contains 52 jelly bears of which 19 are white, 10 are brown, 7 are pink, 3 are purple, 5 are yellow, 2 are orange, and 6 are green. In how many ways can you select 9 pieces of candy at random from the box, without replacement, if:

(a) 3 white jelly bears are selected? (b) 3 white, 2 brown, 1 pink, 1 yellow and 2 green jelly bears are selected?

7. A man wishes to give his 12 different books to his 3 friends, so that the first shall receive 5, the second 4 and the third 3 books. In how many ways can he give the books?

3 Probabilty of an Event

The likelihood of the occurrence of an event resulting from a statistical experiment is

evaluated by means of a set of real numbers called weights or probabilities ranging from 0 to 1.

To every point in the sample space we assign a probability such that the sum of all probabilities

for all the sample point in the sample space is 1.

We define the probability of an event A to be the sum of the probabilities of all sample

points in A. Properties of probability: 1) 1)(0 AP that is the probability of an event A only takes values on the interval [0, 1]

2) P() = 0 that is the probability of a null event is always 0. 3) P(S) = 1 that is the probability of the sample space is always 1. Since all possible outcomes in the experiment is included in its sample space then it is certain that the event which is in this case the sample space is bound to happen. The following are some approaches that have been taken in answering the question on defining what probability is and how do we assign probabilities to outcomes. In our formal study of probabilities, we will be using the classical approach since this approach is based on theoretical or mathematical foundations.

NOTE: If an experiment can result in any one of N different equally likely outcomes, and if exactly n of these

outcomes corresponds to event A, then the probability of the event A is given by N

nAP )(

CLASSICAL Approach - In this approach, the fundamental assumption is that all possible outcomes in the given experiment are equally likely to occur. RELATIVE FREQUENCY Approach – When experiments are repeated then an events probability is derived as the proportion of times the event occurs in the long run. SUBJECTIVE Approach – In this approach, probabilities are interpreted as personal in such a way that probabilities are geared to the person’s beliefs regarding to the uncertainties involved.


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Example 1. If a card is drawn from an ordinary deck, find the probability that a heart card is drawn. Solution: Clearly, there are 13 heart cards, thus there are 13 ways of selecting a heart among the 52 cards. So if we

let A be the event of selecting a heart card then 13n and N = 52, then the probability of selecting a

heart card is 4

1

52

13

N

nAP

2. If a card was randomly drawn from an ordinary deck of playing cards, what is the probability that the card is a letter card?(A letter card consists of the aces, jacks, kings and queens) Solution: If we let B be the event of selecting a letter card, as shown in the illustration above, there are 16 letter different cards. Since we are selecting just one card out of the 16 possible letter cards, using our formula

for combination we have,

16!1!15

!16C

116 . The same is true when we look at our process of selecting

a card, since there are 52 cards and we are only selecting one of them, we have,

52

!1!51

!52

!1!152

!52C

152

. Thus n = 16 and since we are selecting a card from an ordinary deck

of playing cards, there are N = 52 cards that can be chosen. Hence, the probability of selecting a letter card from an ordinary deck of playing cards is given by

13

4

52

16

N

nBP


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3. A mixture of candies contains 6 mints, 4 toffees, and 3 chocolates. If a person makes a random selection of one of these candies, find the probability of getting a mint. Solution: Let M be the event of selecting a mint. Since there are only 6 mints and we are to select one, there are 6

possible ways of selecting a mint, 6!5

!6C16 . Since we are selecting a candy out of the 13 candies the

number of possible ways of selecting a candy out of the 13 candies is given by

13

!12

!13

!1!113

!13C113

. Thus our n = 6 and N = 13. Hence, the probability of selecting a mint in the

mixture of candies is 13

6MP .

Exercise 1. Out of 4 mathematicians and 6 scientists, a committee of 5 members is to be formed that would be the judge to a national science fair. Find the probability that:

a) all members of the committee are mathematicians. b) All members of the committee are scientists. c) Exactly 1 mathematician is part of the committee d) Exactly one scientist is in the committee.

2. From a group of 5 men and 6 women, if we are to select 6 people from the group, find the probability that

a) All the men are selected. b) exactly 3 men are selected c) Exactly 1 woman is in the committee

3. A basket contains 4 red, 5 green and 6 yellow apples. If April selects 3 apples from the basket without replacement what is the probability that no red apples are selected?

4. Sunny Corporation had a shipment of 12 television sets which contains 3 defective sets. If Chiz Company bought 4 television sets from this shipment,

a) what is the probability that Chiz Company bought no defective television set. b) what is the probability that exactly 2 defective sets was bought by the company c) what is the probability that all defective set was bought by the company

5. If a coin is tossed three times, what is the probability that exactly 2 head occur. 6. A professor in philosophy prepared 10 easy, 15 average, and 20 difficult questions for a class exam. If a student chooses one question at random, what is the probability that the student selected a) an easy question b) an average question c) not a difficult question d) a difficult question 7. If two fair dice are tossed, what is the probability that the sum of the outcome of the dice is a) 12 b) 1 c) 5 d) 7

3.1 Laws on Probability


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Sometimes it is easier to calculate the probabilities of certain events based from known probabilities of other events. This is to simplify the calculations. The first law on probability is what we call the additive law of probability or simply additive rule.

If A and B are mutually exclusive events then the additive rule can be modified. Since we know that A

and B are mutually exclusive then A B = . We know from a fact that the probability of a null event, P

() = 0, thus P (AB) = P (A) + P (B). Example 1. The probability that a student passes statistics is 2/3, and the probability that he passes English is 4/9. If the probability of passing at least 1 course is 4/5, what is the probability that he will pass both courses? 2. What is the probability of getting a total of 7 or 11 when a pair of dice is tossed? Solution: Let A be the event that 7 occurs and B be the event that 11 comes up. Now a total of 6 occur for 6 of the 36 sample points and a total of 11 occur for only 2 of the sample points. Since all sample points are equally likely, we have P (A) = 6/36 or 1/6 and P (B) = 2/36 or 1/18. The events A and B are mutually exclusive,

since the total of 7 and 11 cannot occur at the same toss. Therefore, P (AB) = P (A) + P(B) = 1/6 + 1/18 = 2/9.

Example A coin is tossed 6 times in succession. What is the probability that at least 1 head occurs? Solution: Let E be the event that at least 1 head occurs. The sample space s consists of 26 = 64 sample points, since each toss can results in 2 outcomes. Now, P (E) = 1- P (Ec), where Ec is the event that no head occurs. This can happen only one way- when all of the tosses result in a tail. Therefore, P (Ec) = 1/64 and, P (E) = 1- 1/64 = 63/64.

Exercises 1. If a letter is chosen at random from the English alphabet, find the probability that the letter

a) Is a vowel b) Precedes the letter j c) Is after the letter g

2. If A and B are mutually exclusive events and P(A) = 0.3 and P(B) = 0.5, find

a) P (AB) b) P (Ac)

c) P (AcB)

If A and B are any two events, then )()()()( BAPBPAPBAP

Complementary events

If A and Ac are any two events, then 1)()( cAPAP


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3. A pair of dice is tossed. Find the probability of getting a total of 8? At most a total of 5? 4. Two cards are drawn in succession from a deck without replacement. What is the probability that both cards are greater than 2 and less than 8? 5. If 3 books are picked at random from a shelf congaing 7 Harry Potter books, 4 Lord of the Ring Book, and a Novel, what is the probability that

a) The Novel is selected? b) 2 Harry Potter books and 1 Lord of the Rings Book are selected?

6. If a permutation of the word “Math” is selected at random, find the probability that the permutation selected

a) Begins with consonant; b) Ends with a vowel;

3.2 Conditional Probability and Independent Events

The probability of an event B occurring when it is known that some event A has already occurred is called a conditional probability, and is denoted by, P (B|A). The symbol P (B|A) is read as “the probability that B occurs given that A occurs ” or simply as “the probability of B given A”. The probability of B given A is defined by the equation,

0)(,)(

)()|( AifP

AP

BAPABP

Example 1. Suppose that our sample space S is the population of adults in the small town Cavite who have completed the requirements for a college degree. We shall categorize them according to sex and employment status.

One of these individual is to be selected at random for a free admission at the National Conference on Education.

Employed Unemployed Total

Male 360 140 500

Female 240 260 500

Total 600 300 1000

Note: The probability of A given B is defined by the equation

0)(,)(

)()|( BifP

BP

BAPBAP


21

If we let M be the event of selecting a man and let E be the event of choosing a person that is employed.

We can see that, 5

3

600

360)E|M(P . Using the equation to find the conditional probability we have

the following,

5

3

600

360

600

1000

1000

360

1000

6001000

360

)E(P

)EM(P)E|M(P

2. Suppose our sample space S is the set of 4th year high school student in a small town who took the UPCAT college entrance examination. If we categorize them according to sex and the result of their examination (whether the students passed UPCAT or not)

Passed Failed Total

Male 30 110 140

Female 25 65 90

Total 55 175 230

If we select one student at random for an interview, what is the probability that the selected student is a male if we already knew that he passed the UPCAT test? What is the probability that the student selected passed the UPCAT test if it is known that the student selected is a male?

We let M be the event that a male student is chosen and let A be the event that the chosen student passed the exam.

11

6

55

30A|MP

14

3

140

30M|AP

Notice that M|APA|MP .

Example The probability that a regularly scheduled flight departs on time is P(D) = 0.83, the probability that it

arrives on time is P(A) = 0.92, and the probability that it departs and arrives on time is P(DA) = 0.78. Find the probability that a plane a) arrives on time given that it departed on time; b) departed on time given that it arrived on time

Solution

a) 94.083.0

78.0

)(

)()|(

DP

DAPDAP

Two events are said to be independent of each other of the occurrence of one has no influence on the probability of the other. Thus we have the following conditional probabilities: P (B|A) = P (B) and P (A|B) =P (A).


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b) 85.092.0

78.0

)(

)()|(

AP

DAPADP

Exercises

1. A class in advanced physics is comprised of 12 juniors, 32 seniors, and 12 graduate students. The final grade showed that 4 of the juniors, 11 of the seniors and 6 graduate students receive a grade of 90 from the course. If a student is chosen at random from this class and is found to have earned a grade of 90, what is the probability that the student selected is a senior? 2. A random sample of 200 adults are classified below according to sex and level of education attained.

Male Female Total

Elementary 38 45 83

Secondary 28 50 78

College 22 17 39

Total 88 112 200

If a person is picked at random from this group, find the probability that

a) The person is male given that the person has a secondary education b) The person does not have a college degree given that the person is a female c) The person is male, given that the person reached elementary level only

3. A pair of dice is thrown. If it is known that one die shows a 3, what is the probability that a) the other die shows a 2? b) The total of both dice is greater than 7?

4. A card is drawn from an ordinary deck and we are told that it is black. What is the probability that the card is higher than 3 but less than 9? 5. Sunny Corporation had a shipment of 12 television sets which contains 3 defective sets. If Chiz Company bought 4 television sets from this shipment, a) what is the probability that at least 1 defective set was bought by Chiz Company? b) what is the probability that at least 2 defective sets was purchased? 6. The following table gives a two-way classification of all basketball players at a state university who began their college careers between 1988 and 1998, based on gender and whether or not they graduated.

Graduated Did not Graduate Total

Male 126 55 181

Female 133 32 165

Total 259 87 346

If one of these players is selected at random, find the following probabilities.

a) the player is female and graduated b) the player is female or did not graduate c) the player is male and graduated d) the player is male or graduated

counting techniques and basic probability concepts · 2018-10-13 · [counting techniques and basic...

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