7 fundamentals of probability

Besterfield: Quality Control, 8th ed.. © 2009 Pearson Education, Upper Saddle River, NJ 07458.All rights reserved

Quality ControlQuality Control

Chapter 7- Chapter 7- Fundamentals of Fundamentals of

ProbabilityProbability

Chapter 7- Chapter 7- Fundamentals of Fundamentals of

ProbabilityProbabilityPowerPoint presentation to accompanyPowerPoint presentation to accompany

BesterfieldBesterfieldQuality Control, 8eQuality Control, 8e

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PowerPoint presentation to accompanyPowerPoint presentation to accompany BesterfieldBesterfield

Quality Control, 8eQuality Control, 8e

PowerPoints created by Rosida PowerPoints created by Rosida CoowarCoowar


Lesson OutcomesLesson Outcomes Upon completion of this chapter, the reader is

expected to be able to:1. Define probability using the frequency definition.2. Know the seven basic theorems of probability.3. Identify the various discrete and continuous probability distributions.4. Calculate the probability of non-conforming units occurring using the

Hypergeometric, Binomial and Poisson distributions.5. Know when to use the Hypergeometric, Binomial and Poisson

distributions. Upon completing this course, the students should be able to:

describe important concepts pertaining to quality, quality system and total quality management apply the principles of probability and statistics in quality planning, control and improvement and

utilize appropriate statistical tools and techniques undertake complete problem-solving cycle using appropriate quality improvement tools and

techniques design a quality system and incorporate important quality elements necessary for its proper

functioning implement the concept of total quality in engineering and non-engineering processes


Likelihood, chance, tendency, and trend The chance that something will happen

Examples:1. If a Nickel is tossed, the probability of a head is ½ and the

probability of the tail is ½

2. When a die is tossed on the table, the probability of one spot is 1/6, the probability of two spots is 1/6,.....

3. We are drawing a card from a deck of cards. The probability of a spade is 13/52

Definition of ProbabilityDefinition of Probability


The area of each distribution is equal to 1 The are under the normal distribution curve, which

is a probability distribution, is equal to 1 The total probability of any situation will be equal to

1 The probability is expressed as a decimal (the

probability of a head is 0.5) An event is a collection of outcomes (six-sided die

has six possible outcomes)



When the number of outcomes is known or when the number of outcomes is found by experimentation:

P(A) = NA/N

where:P(A) = probability of an event A occurring to 3 decimal places

NA=number of successful outcomes of event A

N= total number of possible outcomes

DefinitionDefinition of Probability of Probability


A part is selected at random from a container of 50 parts that are known to have 10 nonconforming units. The part is returned to the container and a record of the number of trials and the number of nonconforming is maintained. After 90 trials, 16 nonconforming units were recorded. What is the probability based on known outcomes and on experimental outcomes?

Known outcomes = P(A) = NA/N = 10/50 = 0.200

Experimental outcomes = P(A) = NA/N = 16/90 = 0.175

DefinitionDefinition of Probability of Probability


The probability calculated using known outcomes is the true probability, and the one calculated using experimental outcomes is different due to the chance factor

For an infinite situation (N), the definition would always lead to a probability of zero

In the infinite situation the probability of an event occurring is proportional to the population distribution



Theorem 1

Probability is expressed as a number between 1 and 0, where a value of 1 is a certainty that an event will occur and a value of 0 is a certainty that an event will not occur

Theorem 2

If P(A) is the probability that event A will occur, then the probability that A will not occur is:

P(not A) = 1- P(A)

Theorems of ProbabilityTheorems of Probability


If the probability of finding an error on an income tax return is 0.04, what is the probability of finding an error-free or conforming return?

P(not A) = 1.000 – P(A) = 1.000 – 0.040 = 0.960


Table 7-1 Inspection Results by Supplier

Supplier Number Conforming

Number Nonconforming

Total

X 50 3 53

Y 125 6 131

Z 75 2 77

Total 250 11 261


One EventOut or Two

or More Events

MutuallyExclusive

Theorem 3

Not MutuallyExclusive

Theorem 4

Two or More EventOut or Two

or More Events

Independent

Theorem 6

Dependent

Theorem 7


Figure 7-2 When to use Theorems 3,4,6 and 7


Theorem 3

If A and B are two mutually exclusive events (the occurrence of one event makes the other event impossible), then the probability that either event A or event B will occur is the sum of their respective probabilities:

P(A or B) = P(A) +P(B)

This is the “additive law of probability”.

Theorem 4

If event A and event B are not mutually exclusive, then the probability of either event A or event B or both is given by:

P(A or B or both) = P(A) +P(B) – P(both)

Events that are not mutually exclusive have some outcomes in common



If the 261 parts described in Table 7-1 are contained in a box, what is the probability of selecting a random part produced by supplier X or by suppler Z?

P(X or Z) = P(X) + P(Z) = 53/261 + 77/261 = 0.498

What is the probability of selecting a nonconforming part from supplier X or a conforming part from supplier Z?

P(nc. X or nc. Z) = P(nc. X) + P(nc. Z) = 3/261 + 75/261 = 0.299



If the 261 parts described in Table 7-1 are contained in a box, what is the probability of a randomly selected part will be from supplier X or a nonconforming unit?

P(X or nc. or both) = P(X) + P(nc.) + P(X and nc.)

= 53/261 + 11/261 – 3/261= 0.234

A health inspector examines 3 products in a subgroup to determine if they are acceptable. From past experience it is known that the probability of finding no nonconforming units in the sample is 0.990, the probability of 1 nonconforming unit in the sample of 3 is 0.006, and the probability of finding 2 nonconforming units in the sample of 3 is 0.003. What is the probability of finding 3 nonconforming units in the sample of 3?

P(0) + P(1) + P(2) + P(3) = 1.000

0.990 + 0.006 + 0.003 + P(3) = 1.000 thus, P(3) = 0.001



Theorem 5

The sum of the probabilities of the events of a situation is equal to 1.000

P(A) + P(B) + …..+ P(N) = 1.000

Theorem 6

If A and B are independent events (one where its occurrence has no influence on the probability of the other event or events), then the probability of both A and B occurring is the product of their respective probabilities:

P(A and B) = P(A) X P(B)



If the 261 parts described in Table 7-1 are contained in a box, what is the probability that 2 randomly selected parts will be from supplier X and supplier Y? Assume that the first part is returned to the box before the second part is selected (called with replacement)

P(X and Y) = P(X) and P(Y) = 53/261 x 131/261 = 0.102

Assume that in the above problem, the first part was not returned to the box before the second part was selected. What is the probability?

P(X and Y) = P(X) x P(Y/X) = 53/261 x 131/260 = 0.102

Since the first part was not returned to the box, there was a total of only 260 parts in the box



What is the probability of choosing both parts from supplier Z?

P(Z and Z) = P(Z) and P(Z) = 77/261 x 76/260 = 0.086

Since the first part was from supplier Z, there are only 76 from supplier Z of the new total of 260 in the box

If the 261 parts described in Table 7-1 are contained in a box, what is the probability that two randomly selected parts (with replacement) will have one conforming part from supplier X and one conforming part from supplier Y or supplier Z?

P[co. X and (co. Y or co. Z)] = P(co. X)[P(co. Y) + P(co. Z)]

= (50/261) x (125/261 + 75/261) = 0.147



Theorem 7

If A and B are dependent events, the probability of both A and B occurring is the probability of A and the probability that if A occurred, then B will occur also:

P(A and B) = P(A) X P(B\A)

P(B\A) is defined as the probability of event B, provided that event B has occurred



1.Simple multiplication

If an event A can happen in any of a ways or outcomes and, after it has occurred, another event B can happened in b ways or outcomes, the number of ways that both events can happen is ab

2.Permutations

A permutation is an ordered arrangement of a set of objects

Example: The word “cup”…… cup, cpu, upc, ucp, puc, and pcu.

Counting of EventsCounting of Events

!

( )!nr

nP

n r


A witness to a hit-and-run accident remembered the first 3 digits of the license plate out of 5 and noted that the last 2 were numerals. How many owners of automobiles would the police have to investigate?

ab = (10)(10) = 100ab = (10)(10) = 100

If the last 2 were letters, how many would need to be If the last 2 were letters, how many would need to be investigated?investigated?

ab = (26)(26) = 676ab = (26)(26) = 676



3. Combinations If the way the objects are ordered is

unimportant, then we have a combination: !

!( )!nr

nC

r n r


Example: The word “cup” has 6 permutations when the 3 objects are taken 3 at a time. There is only one combination, since the same three letters are in different order.


How many permutations are there of 5 objects taken 3 at a time?

In the license plate example, suppose the witness further remembers that the numerals were not the same.


601.2

1.2.3.4.5

)!35(

!5

)!(

!

5

3

P

P rn

nn

r

901....7.8

1....8.9.10

)!210(

!10

)!(

!

5

3

P

P rn

nn

r


An interior designer has 5 different colored chairs and will use 3 in a living room arrangement. How many different combinations are possible?


101.2.1.2.3

1.2.3.4.5

)!35(!3

!5

)!(!

!

5

3

P

C rnr

nn

r


Hypergeometric Probability Distribution

1. Occurs when the population is finite and the random sample is taken without replacement

2. The formula is constructed of 3 combinations (total, nonconforming, and conforming):

3. Mean and standard deviation of the distribution:

( )D N Dd n d

Nn

C CP d

C

Discrete Probability Discrete Probability DistributionsDistributions

N

nD

1

))(1(

N

nNND

NnD


A lot of 9 thermosets located in a container has 3 nonconforming units. What is the probability of drawing 1 nonconforming unit in a random sample of 4?

N = 9, D = 3, n = 4, and d = 1

Similarly, P(0) = 0.119, P(2) = 0.357, and P(3) = 0.048. Since there are only 3 nonconforming units in the lot, P(4) is impossible. The sum of the probabilities must equal to 1.000.


476.0

)!49(!4!9

)!36(!3!6

)!13(!1!3

)(

)( 9

4

39

14

3

1

dP

dPCCC

CCCN

n

DN

dn

D

d


Binomial Probability Distribution

1. It is applicable to discrete probability problems that have an infinite number of items or that have a steady stream of items coming from a work center

2. It is applied to problems that have attributes

3.3. Applicable provided outcomes are constant and Applicable provided outcomes are constant and trials are independenttrials are independent

1 2 2( 1)( ) .........

2n n n n nn n

p q p np q p q q




Figure 7-6 Distribution of the number of tails for an infinite number of tosses of 11 coins


Binomial Probability Distribution cont’d

3. See Figure 7-6. Since p=q, the distribution is symmetrical regardless of the value of n, however, when p is not equal to q, the distribution is asymmetrical

4. In quality work p is the portion or fraction nonconforming and is usually less than 0.15

5. Mean and standard deviation of this distribution:

0 0!

( )!( )!

d n dnP d p q

d n d

Discrete Probability Discrete Probability DistributionDistributionss

pn0

)1(00ppn


A random sample of 5 hinges is selected from a steady stream of product from a punch press, and the proportion nonconforming is 0.10. What is the probability of 1 nonconforming unit in the sample? What is the probability of 1 or less? What is the probability of 2 or more?

q0 = 1 – p0 = 1.000 – 0.100 = 0.900

Discrete Probability Discrete Probability DistributionDistributionss

328.0)!15(!1

!5

)!(!

!)1( 90.010.0

15

0

1

000

qpdnd

dnd

nP

918.0328.0590.0)1()0()__1( PPlessorP

590.0)!05(!0

!5)0( 90.010.0

05

0

0

0

P

082.0918.0000.1)__1()()__2( lessorPTPmoreorP


Binomial Probability Distribution cont’d.

5. As the sample size gets larger, the shape of the curve will become symmetrical even though p is not equal to q

6. It requires that there be two and only two possible outcomes (C, NC) and that the probability of each outcome does not change

7. The use of the binomial requires that the trials be independent

8. It can be approximated by the Poisson when Po≤0.10 and nPo≤5

9. The normal curve is an excellent approximation when Po is close to 0.5 and n/N>̳� 0.10



Poisson Probability Distribution

1. It is applicable to many situations that involve observations per unit of time

2. It is also applicable to situations involving observations per unit amount

3. In each of the preceding situations, there are many equal opportunities for the occurrence of an event

4. The Poisson is applicable when n is quite large and Po is small

5. When Poisson is used as an approximation to the binomial, the symbol c has the same meaning as d has in the binomial and hypergeometric formulas



Poisson Probability Distribution

6. When nPo gets larger, the distribution approaches symmetry

7. Table C in the Appendix

8.8. Mean and standard deviation:Mean and standard deviation:


00( )( )

!

cnpnp

P c ec

p

p

n

n

0

0


Note: x is c while t is np0


Table C


The average count of billing errors at a local bank per 8-h shift is 1.0. What is the probability of 2 billing errors? The probability of 2 or more?

From Table C, for np0 value of 1.00,

P(2) = 0.184

P(1 or less) = 0.736

P(2 or more) = 1.000 – P(1 or less)

= 1.000 – 0.736

= 0.264



Poisson Probability Distribution cont’d.

9. The Poisson probability is the basis for attribute control charts and for acceptance sampling

10. It is used in other industrial situations, such as accident frequencies, computer simulation, operations research, and work sampling

11. Uniform (generate a random number table), Geometric, and Negative binomial (reliability studies for discrete data)

11. The Poisson can be easily calculated using Table C

12. Similarity among the hypergeometric, binomial, and Poisson distributions can exist



Normal Probability Distribution (as covered in Chapter 4 and 5)

1.When we have measurable data

2.The normal curve is a continuous probability distribution

3.Under certain condition the normal probability distribution will approximate the binomial probability distribution

4.The Exponential probability distribution is used in reliability studies when there is a constant failure rate

5.The Weibull distribution is used when the time to failure is not constant

Continuous Probability Continuous Probability DistributionsDistributions


If the operating life of an electric mixer, which is normally distributed, has a mean of 2200 h and a standard deviation of 120 h, what is the probability that a single mixer will fail to operate at 1900 h or less?

From the table of areas under normal curve, for a Z value of -2.5, area = 0.0062. Therefore, the probability of an electric mixer failing is

P(failure at 1900 h or less) = 0.0062

Continuous Probability Continuous Probability DistributionsDistributions

5.2120

22001900

X iZ


1. Poisson distribution can be easily calculated using Table C and should be used whenever appropriate

2. Hypergeometric is used for finite lots of stated size N

3. Binomial approximates hypergeometric when n/N </= 0.10

4. Poisson approximates hypergeometric when n/N </= 0.10, p0

</= 0.10, and np0 </= 5

5. Normal approximates hypergeometric when n/N </= 0.10

6. Normal approximates binomial

7. Binomial is used for infinite situations or when there is a steady stream and infinite situation is assumed

8. Poisson approximates binomial when p0 </= 0.10 and np0 </= 5

9. Normal curve approximates binomial excellently when p0 is

close to 0.5 and np0 >/= 10

Distribution InterrlationshipDistribution Interrlationship


10. Approximation is still good even though p0 deviates from 0.5

as long as np0 >/= 5 and n increases to 50 or more for

values of p0 as low as 0.10 and as high as 0.90

Distribution InterrlationshipDistribution Interrlationship


THE END

IQ + EQ + SQ = TQ

• “We have indeed created man in the best of molds”. (At-Tin: 4)• Taqwim: mold, symmetry, form, nature, constitution

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