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CHAPTER 5: ELEMENTARY PROBABILITY

CHAPTER FIVE ELEMENTARY PROBABILITY 5.1 INTRODUCTION Dear students, in the first module of the course we have discussed in detail the descriptive statistics. The concepts of probability and probability distributions are the basis for our discussions in inferential statistics that will be discussed in chapters 7 and 8. In most areas of human endeavor, there is always an element of uncertainty. If we consider weather, a sporting event, a stock transaction, an election result, or a matter relating to health, we are always faced with a certain degree of risk. Infact, according to the old adage, the only things in life that are certain are death and taxes. Therefore, we must be able to assess the degree of uncertainty in any given situation, and this is done mathematically using probability. In particular, our primary goal in this module is to understand and use inferential statistics which will allow us to make predictions and decisions based on sample data. However, any statistical prediction or decision also involves an element of uncertainty, and we need the ideas of probability to assess its accuracy.

5.2 OBJECTIVES At the end of this chapter students will be able to: 

Define basic terms which will be used to define probability;



Explain counting techniques;



Compute probability of an event.

5.3 DEFINITION OF CONCEPTS Random Experiment: An experiment is called a random experiment if when conducted repeatedly under essentially homogeneous conditions, the result is not unique but may be any one of the various possible out comes. Example 5.1: i) Tossing a well balanced coin. ii) Rolling a fair die Trial and Event: Performing of a random experiment is called a trial and out come or combination of outcomes are termed as events (cases). Example 5.2: Tossing of a coin is a random experiment or trial and getting of head or tail is an event.

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Sample Space: The collection of all the possible out comes of a random experiment is called the sample space and will be denoted by S. Example 5.3: i) In tossing a coin once the sample space is S = {H.T} ii) In rolling a die the sample space is S= {1, 2, 3, 4, 5, 6} Exhaustive cases: The total number of possible outcomes of a random experiment is called the exhaustive cases (events) for the experiment. Example 5.4: i) In throwing a coin once the exhaustive events =2 ii) In a throw of two dice the exhaustive events = 62=36 Favorable cases or events: The number of out comes of a random experiment which result in the happening of an event are termed as the cases favorable to the event. Example 5.5: i)

In a toss of two coins, the number of cases favorable to the event’ exactly one head’ is 2,viz., HT,TH and for getting ‘ two heads’ is one viz., HH.

ii)

In drawing a card from a pack of cards the cases favorable to ‘getting a diamond’ are 13.

Mutually exclusive events or cases: Events are said to be mutually exclusive if no two or more of them can happen simultaneously (i.e. A  B  , B  C  , etc ).

Example 5.6: i) In a toss of a coin the events ‘head’ and ‘tail’ are mutually exclusive because if head comes, we can’t get tail and if tail comes we can’t get head. ii) Similarly in the throw of a die, the six faces numbered 1, 2,3,4,5 and 6 are mutually exclusive. Equally Likely Cases: Two or more events that have the same probability of occurrence are said to be equally likely events. Example 5.7: In tossing of a coin (die), all the out comes viz., H, T (the faces 1, 2, 3, 4, 5, 6) are equally likely if the coin (die) is unbiased. Independent Events: Two events or more are independent if the occurrence or nonoccurrence of one doesn’t affect the occurrence or non- occurrence of the other. 2 ADU: DEPARTMENT OF STATISTICS

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5.4 COUNTING TECHNIQUES Addition Rule Suppose that a procedure, designated by 1, can be performed in n1 ways. Assume that a second procedure, designated by n2, can be performed in n2 ways. Suppose furthermore that it is not possible that both 1 and 2 are performed together. Then the number of ways in which we can perform 1 and 2 is n1+n2. Example 5.8: Suppose that we are planning a trip and are deciding between bus or train transportation. It there are three bus routes and two train routes, then there are 3+2=5 different routes available for the trip. Example 5.9: A student goes to the nearest snack to have a breakfast. He can take tea, coffee, or milk with bread, cake and sandwich. How many possibilities does he have? Solution: Bread Tea

Cake Sandwich Bread

Coffee

Cake Sandwich Bread

Milk

Cake Sandwich

 There are nine possibilities.

The multiplication Principle If one operation can be performed in m different ways and another operation can be performed in n different ways, then the two operations when associated together can be performed in mxn ways. The result can be generalized to more than two operations. 3 ADU: DEPARTMENT OF STATISTICS

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Example 5.10: Suppose a man traveling from his home to office has a choice of 4 buses and from his office to his school he has a choice of 3 buses; then from his home to school, through his office, he can make the trip in 4x3 = 12 ways. Example 5.11: A student owns 3 pairs of sneakers, 5 pairs of jeans, and 20 shirts. If she randomly selects one of each in the morning with out regard to color or matching, how many different outfits does she have? Solution: Here there are three operations with m 3, n 5, p 20 Then the total number of outfits possible is 3 x5 x20 300 . Exercise: An automobile manufacturer offers a certain car in 12 different colors, 3 different models (two door four-door and wagon), and 4 different engines. In addition air conditioning is available. How many different car packages can be put together? Permutation Definition: A permutation of n different objects taken r at a time denoted by nPr or nPr or P (n,r) is an ordered arrangement of only r objects out of the n – objects. Theorem 5.1: The number of different permutations of n different objects taken r at a time is given by n

Pr n  n  1  n  2  ...  n  r 1  

n!  n  r !

 r n 

Example 5.12: A student has 3 books and has a book shelf that will hold only 2 of them. How many different arrangements are possible for the 2 books? Solution: Let A, B and C be the three books. Then the possible arrangements for the 2 books are: AB, BA, AC, CA, BC, and CB. Hence number of possible arrangements is equal to 3! 3! 6  3  2 !

3P2 

Example 5.13: From a committee of eight people in how many ways can we choose a chairperson and a vice- chair person, assuming one person can not hold more than one position? Solution: We are actually asking for the number of permutations of eight objects taken two at a time; that is, 8 P2 : 8

8! 8! 8.7.6!   56  8  2 ! 6! 6!

P2 

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Theorem 5.2: The number of permutations of n distinct objects, taken altogether, denoted by n Pn , is n ! . Example 5.14: If there are three letters A, B, C there are six possible permutations: ABC, BCA, CAB, ACB, BAC, CBA. This can be done by means of tree diagrams where the total number of paths as shown gives the number of distinct permutations.

Permutation B A  C A B C A C B

C A B

ABC ACB

C B A B C A

BAC BCA CAB CBA

Note : 1) If r  n, 2) n Pn  1 

n

pr 

n! n!   n!  n  r  ! 0!

n! n!  n!.Then n Pn n Pn  1  n  n 1 ! 1!

Example 5.15: In how many ways can five people be lined up to get on a bus? Solution: 5! =120 different ways Example 5.16: It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible?

Solution: The men may be seated in 5 P5 ways and the women in 4 P4 ways. Each arrangement of the men may be associated with each arrangement of the women. Hence number of arrangements 5 P5 .4 P4  5!4!   120  24   2880 5 ADU: DEPARTMENT OF STATISTICS

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Theorem 5.3: The number of permutations of n objects taken n at a time in which n 1 objects of one kind are a like, n2 of another kind are alike,…, nk of a kind are alike is given by n!  where n1  n 2  ...  nk n  n1 !n2 !...nk ! Example 5.17: In how many ways can the letters the following words be written? a)

ECONOMICS

b)

STATISTICS

Solution: a) C and O occur twice each hence the required number of ways is 9! 90720 . 2!2! b) S, T occur twice and I thrice each. Hence the required number of ways is: 10! 50, 400 3!3!2! Theorem 5.4: The number of permutations of n distinct objects arranged in a circle is  n  1 ! Circular permutations depend on the relative positions of objects. If there are n-objects we fix the position of one object and arrange the remaining n-1 objects in all possible ways. Example 5.18: In how many ways can four men be seated at a round table? Solution: Let the four men be A, B, C and D. Then the different permutations of the 4 men round a table are: A

D B C

A

A

A

A

A

C B

B C

D C

C D

B D

D

D

B

B

C

Hence the 4 men can seat round a table in  4  1 ! 3! 6 different ways.

Combinations Definition: A combination of n different objects taken r at a time denoted by n  n Cr or nC r or   or C  n, r  is a selection of only r objects out of the n -objects with out any r  regard to the order of the arrangement. Here the order of arrangement is immaterial. Theorem 5.5: The number of combinations of n distinct objects selected r at a time is given by nC r 

n! n pr  r ! n  r  ! r ! 6

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Example 5.19: There are 3 combinations and 6 permutations of the letters A, B, C taken 2 at a time. i .e. 3C 2  3,

3P2  6 AB BA AC CA BC CB

AB AC BC

Example 5.20: The number of ways in which a committee of 5 people can be selected form 12 people is 12

C5 

12! 12 X11 X10 x9 x8 x7!  5! 12  5 ! 5!7!  792

Example 5.21: Out of 5 men and 3 women a committee of three is to be formed a)

Consisting of two men and one woman. b)

With out any restriction.

In how many ways can this be done? 5 Solution: a) 2 men out of 5 can be selected in C2  10 ways, 1 woman out of 3 can be 3 selected in C1 3 ways. By the multiplication principle, the committee can 5! 3! x 10 X 3 30 ways . be formed in 2!  5  2  ! 1!  3  1  !

8! 8.7.6 8  56 ways . b) 3 people can be selected out of a total of 8 in C3  3!5! 6 Remark: The numbers of combinations of n. objects taken r at a time are also known as n binomial coefficients. That is the expansion of  x  y  is

 x  y

n

n C o x n n C 1x n 1 y n C 2x n 2 y 2 ... n C r x n r y r  ... n C n y n n

 nC r x n  r y r r 0

5.5 DEFINITION OF PROBABILITY There are three approaches to the definition of probability. 1. Classical (Mathematical) definition of probability If a trial (experiment) results in N exhaustive, mutually exclusive and equally likely events and out of these N , only m of them are favorable to the happening of an event E, then the probability of occurrence of E usually denoted by P(E), is given by 7 ADU: DEPARTMENT OF STATISTICS

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Number of favourable events toE Exhaustive Events m  N

P (E ) 

Remarks: 1. 0 m N , since m and N are non- negative integers. 0 m N   N M N  0  P  E  1 

Hence probability of any event is a number lying between 0 and 1. P(E)=0 then E is called an impossible or null event. If P(E) =1, then E is called a certain event (sure event). 2. If A and A’ are complementary events, then P  A   P  A '  1 Example 5.22: A fair coin is tossed twice. Find the probabilities of the events: a) A= two heads b) E=at least one tail Solution: Here the sample space is S  HH , HT , TH , TT In this case each of the N=4 elements of S are exhaustive, mutually exclusive and equally likely.

a) A= two heads = {HH} Number of favourable events to A Exhaustive Events 1  4

P  A  

b) E=at least one tail = {HT,TH,TT} Hence, n E 3 and P  E  

n E 3  n S  4

Example 5. 23: A bag contains 3 red, 6white and 7 blue balls. What is the probability that the two balls drawn are white and blue? Solution: Total number of balls = 3+6+7=16. Since 2 balls can be drawn out of 11 balls in

16

C2 ways. Exhaustive number of cases = 16 C2 8

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One white out of 6 can be drawn in

6

C1 ways and one blue out of 7 can be drawn in

7

C1

ways. Total number of favorable case = 6 C 1.7 C 1 . 6

 Probability that the two balls drawn are white and blue =

C1 .7 C1 42 7   16 120 20 C2

Example 5.24: A uniform die is thrown at random. Find the probability that the number on it is: a) 5

b) greater than 4

c) even

Solution: Here S= {1, 2, 3, 4, 5, 6} a) the number on it is 5= A= {5} Then P  A  

n A

n S 



1 6

b) The number on it is greater than 4 Thus, P  E  

n E 

2 1   n S  6 3

c) The number on it is even = B= {2, 4, 6}. Hence P  B  

nB

3 1   n S  6 2

Example 5.25: A set of 10 items consists of 4 defective and 6 non- defective items. If 3 of these are selected at random, what is the probability that: a) all the selected items will be non-defective? b) one of selected items will be non-defective? c) all of the selected items will be defective ? Solutions: Total number of items = 10. 3 items out of 10 can be selected in 10 C3 ways. a) 3 non- defective items out of 6 can be selected in

6

C3 ways.

Let A be the event that all will be non-defective. 6 Total way in which A occurs  C3

 P(all the selected items will be non-defective) 6

 10

C3 20 4   C3 120 6

6 4 b) 1 item out of 6 can be selected in C1 ways and 2 items out of 4 can be selected in C2 ways.

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6 C1 x4 C 2 10C3 6 x6 36   120 120 3  10

 Re quired probability 

c) 3 defective items out of 4 can be selected in 4 C3 ways. Then P(all of the selected items will be defective 4

 10

C3 C3

4  120 1  30

2. Relative Frequency (statistical) definition of probability Definition: If an event A occurs n times in a series of N independent random trials, then the probability of the event A is P A   

Number of times event A occurred Number of times experiment was run n N

Frequency of A Total number of trials Re lative frequency of A 

Example 5.26: Ten of the 500 randomly selected cars manufactured at a certain auto factory are found to be lemons. Assuming that the lemons are manufactured randomly. What is the probability that the next car manufactured at this auto factory is a lemon? 10 ADU: DEPARTMENT OF STATISTICS

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Solution: Here total number cars in the sample are 500 and number of lemons is 10.  P  next car is a lemon  

n 10  0.02 N 500

Example 5.27: The following data show the length of life of wholesale grocers in a particular city: Length of life (years)

Percentage of wholesales

0.5

65

5.10

16

10.15

9

15.25

5

25 and over

5

Total

100

a) During the period studied, what is the probability that an entrant to this profession will fail with in five years? b) That he will survive at least 25 years? c) How many years would he have to survive to be among the 10 percent longest survivors? Solution: Here the exhaustive number of cases =100 a) An entrant to the profession of whole sale grocery will fail with in five ears if his life (in this profession) is less than 5 years and the favorable numbers of cases for this event (from the above table) are 65. Hence required probability 

65 0.65 100

b) A wholesale grocers will survive at least 25 years if his life in this profession is over 25 years and the favorable number of cases for this event (from the above table ) are 5. 5 0.05 Hence required probability  100 c) From the above table we see that the number of wholesale grocers with age over 15 years is

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5+5 =10, and therefore, the probability that a grocer survives at least 15 years is 10  0.10 Thus in order to be among the 10 percent longest survivors, the grocer should 100 survive at least 15 years in the business life. 3. Axiomatic Approach to the definition of probability Let S be a sample space associated to random experiment. A set function denoted by P(.) defined in a sample space S, is called a probability measure ( or simply probability) if it satisfies the following axioms. 1. P  A  0 A  S 2.P  S  1 3. If A1, A2, A3,…, An are mutually exclusive subsets of S, then n P  Ai  i1

 n  P     Ai   i1   n

 P  Ai  , We call P  A  is the probability of A. i1

Example 5.27: The distribution of blood types in the United States is roughly 41% type A,9% type B, 4% type AB, and 46% type O. An individual is brought in to an emergency room and is to be blood-typed. What is the probability that the type will be A, B or AB?

Solution: The sample space for this experiment is S= {A, B, AB, O} The Sample points are not equally likely so the classical approach to probability is not applicable. Let A 1, A2, and A3 denote the events that the patient has type A, B and AB blood, respectively. The events A1, A2, and A 3 are mutually exclusive and we are looking for P  A1  A2  A3  . By axiom 3, P (A 1 A 2 A 3)  P (A 1)  P (A 2)  P (A 3) 0.14  0.09  0.04 0.54

5.6 SOME PROBABILITY RULES 12 ADU: DEPARTMENT OF STATISTICS

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5.6.1 ADDITION RULE OF PROBABILITY Theorem 5.6: The probability of occurrence of at least one of two events A and B is given by P  A  B  P  A   P  B   P  A  B  .

Example 5.28: In a group of 120 students, 60 are taking math, 40 are taking psychology, and 15 are taking both. A randomly chosen student from this group is selected. a) What is the probability that he or she is taking either math or psychology? b) What is the probability that he or she is taking neither one? Solution: Here P(taking math) 

60 120

P( taking...