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Chapter 7

Probability

I Three Views of Probability

A. Subjective-Personalistic View

1. Probability of event A, p(A), is a measure of the

strength of one’s expectation that event A will or

will not occur.

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B. Classical or Logical View

1. Probability of event A is the number of events

favoring A, nA, divided by the total number of

equally likely events, nS,

p(A) = nA/nS

2. Probability is based on a logical analysis.

3. p(A) is always a number between 0 and 1 because nA ≤ nS.

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C. Empirical Relative-Frequency View

1. Probability of event A is a number approached by

the ratio nA/n as the total number of observations,

n, approaches infinity.

2. If a head is obtained 12 times in 20 tosses, the

best estimate of the probability of heads is nA/n =

12/20 = .6. If a head is obtained 120 times in 200

tosses, our confidence in the estimate 120/200 =

.6 is even greater.

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II Basic Concepts

A. Experiment

B. Compound and Simple Events

1. Compound events in tossing a die

Event A—observe an odd number

Event B—observe an even number

Event C—observe a number less than 4

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2. Simple events in tossing a die

Event E1—observe a 1

Event E2—observe a 2

Event E3—observe a 3

Event E4—observe a 4

Event E5—observe a 5

Event E6—observe a 6

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III Graphing Simple and Compound Events

A. Euler Diagram

1. Sample space, S, with nS sample points, Ei

• • •

• • •

S

E1

E2

E3

E4 E6

E5

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B. Formal Properties of Probability

1. 0 ≤ p(Ei) ≤ 1 for all i

2. p Ei i1

ns 1

3. p(S) = 1

IV Probability of Combined Events

A. Union of Events A and B: Set of Elements that Belongs to A or B or to Both A and B

B. Intersection of Events A and B: Set of Elements that Belongs to Both A and B

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V Addition Rule of Probability

A. p(A or C) = p(A) + p(C) – p(A and C)

• • •

• • •

S

E1

E2

E3

E4 E6

E5

C

AA and C

1. p(A or C) = 3/6 + 3/6 –2/6 = 2/3

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B. Addition Rule of Probability for Mutually Exclusive Events where p(A and B) = 0

1. p(A or B) = p(A) + p(B)

• • •

• • •

S

E1

E2

E3

E4 E6

E5

B

A

2. p(A or B) = 3/6 + 3/6 = 1

3. Collectively exhaustive events: probability of union equals 1

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VI Multiplication Rule of Probability

A. Conditional Probability

1. p(A | C) = p(A and C)/p(C)

2. p(C | A) = p(A and C)/p(A)

p( A | C) p( A and C)

p(C)

nA and C

nS

/

nC

nS

nA and C

nC

p(C | A) p( A and C)

p( A)

nA and C

nS

/

nA

nS

nA and C

nA

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3. Example of conditional probability for events A and C.

p( A | C)

p( A and C)

p(C)

nA and C

nC

2

3

• • •

• • •

S

E1

E2

E3

E4 E6

E5

C

AA and C

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B. Multiplication Rule of Probability

1. p(A and C) = p(A) p(C | A) = p(C) p(A | C)

2. p(A and C) = (3/6)(2/3) = (3/6) (2/3) = 1/3

• • •

• • •

S

E1

E2

E3

E4 E6

E5

C

AA and C

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C. Statistical Independence

1. Events A and C are independent if p(A | C) = p(A)

2. Events A and C are not independent because

p(A | C) ≠ p(A)

2/3 ≠ 3/6

• • •

• • •

S

E1

E2

E3

E4 E6

E5

C

AA and C

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3. Events A (obtain a H on the toss of a coin) and B

(obtain a 5 on the roll a die) are statistically

independent because p(B | A) = p(B).

• H 1 • H 2 • H 3 H• 4 • H 5 • H 6

• T 1 • T 2

• T 3 • T 4•

T 5 • T 6

nS = 12

nA = 6

nB = 2

n = 1A and B

p(B | A) nA and B / nA 1 / 6

p(B) nB / nS 2 / 12 1 / 6

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D. Multiplication Rule for Statistically Independent Events where p(B|A) = p(B)

1. p(A and B) = p(A) p(B)

• H 1 • H 2 • H 3 H• 4 • H 5 • H 6

• T 1 • T 2

• T 3 • T 4•

T 5 • T 6

nS = 12

nA = 6

nB = 2

n = 1A and B

2. p(A and B) = (6/12)(2/12) = 1/12

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VII Counting Simple Events

A. Fundamental Counting Rule

1. If an event can occur in n1 ways and a second

event in n2 ways and each of the first event’s n1

ways can be followed by any of the second’s n2

ways, then the number of ways that event 1

followed by event 2 can occur is n1 × n2.

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2. Rolling a die (event 1) and tossing a coin (event 2)

n1 n2 = (6)(2) = 12

3. k events, say rolling k = 3 dice

n1 n2, . . . , nk = (6)(6)(6) = 216

B. Permutation of n Objects Taken n at a Time( nPn )

1. n factorial, n! = n(n – 1)(n – 2) . . . (1)

2. nPn = n! = n(n – 1)(n – 2) . . . (1)

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3. Consider n objects and a box with n compartments

1 2 3 n. . .

n ways n Š 1 ways n Š 2 ways 1 way

4. The first compartment can be filled with any of the

n objects, the second with any of n – 1 remaining

objects, and the nth compartment with the one

remaining object.

5. According to the fundamental counting rule

nPn = n(n – 1)(n – 2) . . . (1)

− −

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6. For n = 5 objects

5P5 = (5)(4)(3)(2)(1) = 120

C. Permutation of n Objects Taken r at a Time( nPr )

1. Consider n = 5 objects and a box with r = 3

compartments

1 2 3

5 ways 5 Š 1 ways 5 Š (3 Š 1) ways − − −

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2. The number of ways that n = 5 objects can be

placed in r = 3 ordered compartments is given

by

5P3 5(5 1)(5 31) 5(4)(3) 60

3. Alternative formula

n Pr

n!

(n r)!

5 P3

5!

(5 3)!

(5)(4)(3)(2)(1)

(2)(1)

120

260

)1()2)(1( rnnnnPrn

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D. Combination of n Objects Taken r at a Time( nCr )

nCr n Pr

r!

n!

(n r)!

r!

n!

r!(n r)!

1. nCr ignores the order of the objects by dividing

nPr by the number of ways that r elements can be

arranged which is r!.

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2. The number of ways that n = 5 objects can be

placed in r = 3 compartments ignoring the order of

the objects is given by

nCr

n!

r!(n r)!

5C3

5!

3!(5 3)!

(5)(4)(3)(2)(1)

(3)(2)(1) (2)(1) 10