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Page | 1 Module 1 Probability is a way of expressing knowledge or belief that an event will occur or has occurred. The word was derived from probity, a measure of the authority of a legal witness in a legal case in Europe, and often correlated with the witness’s nobility. In a sense, this differs much from the modern meaning of probability, which, in contrast, is used as a measure of the weight of empirical evidence, and is arrived at from inductive reasoning and statistical inference. 1.1 SAMPLE SPACE Definition 1.1 The set of all possible outcomes of a statistical experiment is called the sample space 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. Example: An experiment involves tossing a pair of dice, 1 green and 1 red, and recording the numbers that come up. If x equals the outcome on the green die and y the outcome on the red die, describe the sample space S by listing the elements (x,y). Solution: Tabulating the outcomes of tossing a pair of dice that is Red(y) 1 2 3 4 5 6 1 1, 1 1. 2 1, 3 1, 4 1, 5 1, 6 Probability and Statistics PROBABILITY
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Page 1: Probability

P a g e | 1

Module 1

Probability is a way of expressing knowledge or belief that an event will occur or has occurred.

The word was derived from probity, a measure of the authority of a legal witness in a legal case in Europe, and often correlated with the witness’s nobility. In a sense, this differs much from the modern meaning of probability, which, in contrast, is used as a measure of the weight of empirical evidence, and is arrived at from inductive reasoning and statistical inference.

1.1 SAMPLE SPACE

Definition 1.1 The set of all possible outcomes of a statistical experiment is called the sample space 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.

Example: An experiment involves tossing a pair of dice, 1 green and 1 red, and recording the numbers that come up. If x equals the outcome on the green die and y the outcome on the red die, describe the sample space S by listing the elements (x,y).Solution:Tabulating the outcomes of tossing a pair of dice that is

Red(y) 1 2 3 4 5 6

Gre

en(x

)

1 1,1 1.2 1,3 1,4 1,5 1,62 2,1 2,2 2,3 2,4 2,5 2,63 3,1 3,2 3,3 3,4 3,5 3,64 4,1 4,2 4,3 4,4 4,5 4,65 5,1 5,2 5,3 5,4 5,5 5,66 6,1 6,2 6,3 6,4 6,5 6,6

Or by listing method:S = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5),

(2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}Therefore, the Sample space consists of 36 sample points.

Probability and Statistics

PROBABILITYPROBABILITY

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1.2 EVENTS

Definition 1.2 An event is a subset of a sample space.

Definition 1.3 The complement of an event A with respect to S is the subset of all elements of S that are not in A. We denote the complement of A by the symbol A’.

Definition 1.4 The intersection of two events A and B, denoted by the symbol A ∩ B, is the event containing all elements that are common to A and B.

Definition 1.5 Two events A and B are mutually exclusive, or disjoint if A ∩ B = ø, that is, if A and B have no elements in common.

Definition 1.6 The union of the two events A and B, denoted by the symbol A ∪ B, is the event containing all the elements that belong to A or B or both.

1.3 COUNTING SAMPLE POINTS

Theorem 1.1 If an operation can be performed n1 ways, and if for each of these a second operation can be performed in n2 ways, then the two operations can be performed together in n1n2 ways.

Example: Registrants at a large convention are offered 6 sightseeing tours on each 3 days. In how many ways can a person arrange to go on a sightseeing tour planned by this convention?Solution:

Let N be the no. of ways can a person arrange to go on sightseeing tour

n1 = 6 no. of ways of sightseeing toursn2 = 3 no. of days offered for sightseeing

Therefore,N = n1n2 = (6)(3) = 18 possible ways

Theorem 1.2 If a operation can be performed in n1 ways, and if for each of these a second operation can be performed in n2 ways, and for each of the first two a third operation can be performed in n3

ways, and so forth, then the sequence of k operations can be performed in n1n2,…, nk ways.

Example: A developer of a new subdivision offers a prospective home buyer a choice of 4 designs, 3 different heating systems, a garage or carport, and a patio or screened porch. How many different plans are available to this buyer?

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Solution:Let N be the no. of different available plans

n1 = 4, choices of 4 designsn2 = 3, choices of heating systemn3 = 4, choices of car park design (garage, carport, patio or screened

porch)

N = n1 n2 n3 n4 = (4)(3)(4) = 48

Definition 1.7 A permutation is an arrangement of all or part of a set of objects.

Theorem 1.3 The number of permutation of n distinct objects is n!

Example: A contractor wishes to build 9 houses, each different in designs. In how many ways can he place these houses on a street if 6 lots are on one side of the street and 3 lots are on the opposite side?

Solution:The number of permutations of n distinct objects is n!Therefore, the number of ways = 9! = 362,880.

Theorem 1.4 The number of permutations of n distinct objects taken r at a time is

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

Solution:The total number of sample points is

Theorem 1.5 The number of permutations of n distinct objects arranged in a circle is (n-1)!.

Example: In how many ways can a caravan of 8 covered wagons from Arizona be arranged in a circle?

Solution: The total number of sample points is

(n - 1)! = (8 – 1)! = 7! = 5040 ways

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Theorem 1.6 The number of distinct permutations of n things of which n1 are of 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 lights with 9 sockets?

Solution:The total number of distinct arrangements is

Theorem 1.7 The 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 forth, 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

Theorem 1.8 The number of combinations of n distinct objects taken r at a time is

Example: From 4 Republicans and 3 Democrats find the number of committees of 3 that can be formed with 2 republicans and 1 Democrat.

Solution:The number of ways of selecting 2 Republicans from 4 is

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The number of ways of selecting 1 Democrat from 3 is

Using Theorem 1.1, we find the number of committees that can be formed with 2 Republicans and 1 Democrat to be

N = (6)(3) = 18.

1.4 PROBABILITY OF AN EVENT

Definition 1.8 The probability of an event A is the sum of the weights of all sample points in A. Therefore,

0 ≤ P(A) ≤ 1, P(ø) = 0, and P(S) = 1.Theorem 1.9 If an experiment can result in any one of N different equally

likely outcomes, and if exactly n of these outcomes correspond to event A, then the probability of event A is

Example: A box contains 500 envelopes of which 75 contain $100 in cash, 150 contain $25, and 275 contain $10. An envelope may be purchased for $25. What is the sample space for the different amounts of money? Assign probabilities to the sample points and then find the probability that the first envelope purchased contains less than $100.Solution: The sample space for the different amount of money is S = {$10, $25, $100}.If n1 = 75 contains $100n2 = 150 contains $25n3 = 275 contains $10N = 500 envelopesThe probabilities to the sample points are:

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The probability that the first envelope purchased contains less than $100 is:

1.5 ADDITIVE RULES

Theorem 1.10 If A and B are any two events, then

P (A∪ B) = P (A) + P (B) – P (A∩ B).

Example: The probability that Paula passes mathematics is 2/3, and the probability that she passes English is 4/9. If the probability of passing both courses is ¼, what is the probability that Paula will pass at least one of these courses?

Solution:If M is the event “passing mathematics”

E is the event “passing English”,Then, P (M) = 2/3, P (E) = 4/9, and P (M∩E) = 1/4By additive rule, the probability that Paula will pass at least one of the

courses is

Corollary 1 If A and B e mutually exclusive, thenP (A∪ B) = P (A) + P (B).

Example: 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

A = {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)}B be the event that 11 comes up.

B = {(5,6),(6,5)}

Since all sample points are equally likely, we have

The events A and B are mutually exclusive since a total of 7 and 11 cannot both occur on the same toss. Therefore,

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Corollary 2 If A1,A2,A3,…,An are mutually exclusive, thenP (A1∪ A2∪…∪ An) = P (A1) + P (A2) +…+P (An).

Example: If the probabilities are, respectively, 0.09, 0.15, 0.21, and 0.23 that a person purchasing a new automobile will choose the color green, white, red, or blue, what is the probability that a given buyer will purchase a new automobile that comes in one of those colors?

Solution:Let G, W, R, and B be the events that a buyer selects, respectively, a green,

white, red, or blue automobile. Since these four events are mutually exclusive, the probability is

Corollary 3 If A1,A2,A3,…,An is a partition of a sample space S, thenP (A1∪ A2∪…∪ An) = P (A) + P (A2 +…+P (An)

= P(S) = 1.

Theorem 1.11 For three events A, B, then C, P (A ∪ B ∪ C) = P (A) + P (B) + P(C) – P (A ∩ B) – P (A ∩ C) – P (B∩C) + P (A ∩B ∩C).

Theorem 1.12 If A and A’ are complementary events, then P (A) + P (A’) = 1

Example: If the probabilities that an automobile mechanic will service 3, 4, 5, 6, 7, or 8 or more cars on any given workday are, respectively, 0.12, 0.19, 0.28, 0.24, 0.10, and 0.07, what is the probability that he will service at least 5 cars on his next day at work?

Solution:Let E be the event that at least 5 cars are serviced

E’ be the event that fewer than 5 cars are servicedSince P (E’) = 0.12 + 0.19 = 0.31, therefore,

P (E) = 1 - P (E’) = 1 – 0.31 = 0.69

1.6 CONDITIONAL PROBABILITY

Definition 1.9 The conditional probability of B, given A, denoted by P(B⃒A), is defined by

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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) = 0.82; and the probability that it departs and arrives on time is P(D∩A)=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 has arrived on time.

Solution:(a) The probability that a plane arrives on time given that it departed on

time is

(b) The probability that a plane departed on time given that it has arrived on time is

Definition 1.10 Two events A and B are independent if and only if

P(B⃒A) = P(B) and P(A ⃒B) = P(A).

Otherwise, A and B are dependent.

Example: Consider an experiment in which 2 cards are drawn in succession from an ordinary deck, with replacement. The event are defined asA: the first card is an ace,B: the second card is a spade.

Since the first card is replaced, our sample space for both the first and second draws consists of 52 cards, containing 4 aces and 13 spades. Hence

P (A⃒B) = P (B ⃒A) =

That is P(A ⃒B) = P(A) or P(B⃒A) = P(B). When this is true the events A and B are said to be independent.

1.7 MULTIPLICATIVE RULES

Theorem 1.13 If in an experiment the events A and B can both occur, then

P (A∩ B) = P (A)P(B⃒ A).Probability and Statistics

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Example: One bag contains 4 white balls and 3 black balls, and a second bag contains 3 white balls and 5 black balls. One ball is drawn from the first bag and placed unseen in the second bag. What is the probability that a ball mow drawn from the second bag is black?

Solution:Let B1, B2, and W1 represent, respectively, the drawing of a black ball from bag 1, a black ball from bag 2, and a white ball from bag 1. We are interested in the union of the mutually exclusive events B1 ∩ B2 and W1 ∩ B2.

The probability of drawing black ball from bag 2 after knowing that it was a black ball drawn from bag 1

The probability of drawing black ball from bag 2 after knowing that it was a white ball drawn from bag 1

The probability of drawing a black ball from the second bag is

Theorem 1.14 Two events A and B are dependent if and only ifP (A ∩ B) = P (A) P (B).

Example: A small town has one fire engine and one ambulance available for emergencies. The probability that the fire engine is available when needed is 0.98, and the probability that the ambulance is available when called is 0.92. In the event of an injury resulting from a burning building, find the probability that both the ambulance and the fire engine will be available.Solution:Let A and B represent the respective events that the fire engine and the ambulance are available. Then

P(A∩ B) = P(A)P(B) = (0.98)(0.92) = 0.9016Theorem 1.15 If, in an experiment, the events A1, A2, A3, …, Ak can occur,

then

P(A1∩ A2∩ A3∩…∩ Ak)= P(A1)P(A2 ⃒A1)P(A3 ⃒A1∩A2)…P(Ak ⃒A1 ∩A2∩…∩Ak-1).

If the events A1, A2, A3, …, Ak are independent, then

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P(A1∩ A2∩ A3∩…∩ Ak)= P(A1)P(A2 )P(A3 )…P(Ak ).

Example: Three cards are drawn in succession, without replacement, from an ordinary deck of playing cards. Find the probability that the event A1∩ A2∩ A3 occurs, where A1 is the event that the first card is a red ace, A2 is the event that the second card is a 10 or a jack, and A3 is the event that the third card is greater than 3 but less than 7.

Solution:First we define the eventsA1: the first card is a red ace,A2: the second card is a 10 or jack,A3: the third card is greater than 3 but less than 7.

Now

And hence,

1.8 BAYES’ RULE

Theorem 1.16 If the events B1, B2, …, Bk constitute a partition of the sample space S such that P(Bi) ≠ 0 for i = 1, 2,…,k, then for any event A of S,

Example: Three members of a private country club have been nominated for the office of president. The probability that Mr. Adams will be elected is 0.3, the probability that Mr. Brown will be elected is 0.5, and the probability that Ms. Cooper will be elected is 0.2. Should Mr. Adams be elected, the probability for an increase in membership fees is 0.8. Should Mr. Brown or Ms. Cooper be elected, the corresponding probabilities for an increase in fees are 0.1 and 0.4. What is the probability that there will be an increase in membership fees?

Solution: Consider the following events:

A: membership fees are increased,B1: Mr. Adams is elected,B2: Mr. Brown is elected,

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B3: Ms. Cooper is elected.

Applying the rule of elimination, we can writeP (A) = P(B1)P(A⃒B1) + P(B2)P(A⃒B2) + P(B3)P(A⃒B3).

Referring to the tree diagram that follows, we find that the three branches give probabilities

P(B1)P(A⃒B1) = (0.3)(0.8) = 0.24P(B2)P(A⃒B2) = (0.5)(0.1) = 0.05P(B3)P(A⃒B3) = (0.2)(0.4) = 0.08

and hence P(A) = 0.24 + 0.05 + 0.08 = 0.37.

Theorem 1.17 (Bayes’ Rule) If the events B1, B2,…,Bk constitute a partition of the sample S, where P(Bi) ≠0 for i = 1,2,…,k, then for any event A in S such that P(A)≠0,

for r = 1, 2, …, k.

Example: With reference to example of Theorem 1.18, if someone is considering joining the club but delays his decision for several weeks only to find out the fees have been increased, what is the probability that Ms. Cooper was elected president of the club?

Solution:Using Bayes’ rule to writeP(B3 ⃒A) =

Probability and Statistics

B1

B2

B3

Tree Diagram

P(B1) = 0.3

P(A ⃒B1) = 0.8

P(B2) = 0.5 P(A ⃒B2) = 0.1

P(A ⃒B3) = 0.4P(B3) = 0.2

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And then substituting the probabilities calculated in example of Theorem 1.18, we have

ADDITIONAL PROBLEMS:

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