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Chapter 3 Basic Probability 27
CHAPTER 3
BASIC PROBABILITY
1.0 Introduction
2.0 Events
3.0 Probability
4.0 Methods to Determine Probability Values
5.0 Multiplication Theorem
6.0 Addition Theorem
7.0 Permutations and Combinations
8.0 Probability Distributions
9.0 Conditional Probability
10.0 Probability Roadmap
11.0 Venn Diagrams
“Seven Out”
Infamous last words
Chapter 3 Basic Probability 29
BASIC PROBABILITY
1.0 INTRODUCTION
Probability concepts are familiar to everyone. The weather forecaster states that the
probability of rain tomorrow is twenty percent. At the racetrack, the odds are three to one that
a certain horse will win the fifth race. Relating probability concepts to manufacturing
operations may not be as familiar as the above examples, but they work the same way.
Probability is the key to assessing the risks involved in the decision-making process. The
gambling casinos determine the probabilities for each game of chance then make the rules
so that the odds are always in their favor. The same can be done for manufactured products.
The probability of a certain number of defective parts in a large lot can be determined. Also,
the percentage of parts within a certain dimension range can be predicted. If the desired
results are not obtained, then adjustments to the process can be made. Adjustments to a
process in a manufacturing operation are analogous to changing the rules in a casino game.
The objective is to obtain the desired results.
Since a major portion of statistical quality control and statistical process control deals with
probability concepts, it is important to have a good knowledge of probability. In a
manufacturing operation, there are very few occasions when complete information is
available. Therefore, information must be generalized from samples and limited known facts.
It is sometimes surprising to discover the vast amount of information and knowledge about a
process that can be obtained from a relatively small amount of data. Probability is the
building block of statistics and statistical quality control.
2.0 EVENTS
An event is defined as any outcome that can occur. There are two main categories of events:
Deterministic and Probabilistic.
•A deterministic event always has the same outcome and is predictable 100% of
the time.
•Distance traveled = time x velocity
•The speed of light
•The sun rising in the east
•James Bond winning the fight without a scratch
•A probabilistic event is an event for which the exact outcome is not predictable
100% of the time.
•The number of heads in ten tosses of a coin
•The winner of the World Series
•The number of games played in a World Series
•The number of defects in a batch of product
In a boxing match there may be three possible events. There could be more events
depending on the question asked.
•Fighter A wins
•Fighter B wins
•Draw
QReview 30
2.1 Four Basic Types of Events
•Mutually Exclusive Events: These are events that cannot occur at the same time.
The cause of mutually exclusive events could be a force of nature or a man made
law. Being twenty-five years old and also becoming president of the United States are
mutually exclusive events because by law these two events cannot occur at the
same time.
•Complementary Events: These are events that have two possible outcomes. The
probability of event A plus the probability of A' equals one. P(A) + P(A') = 1. Any event
A and its complementary event A' are mutually exclusive. Heads or tails in one toss of
a coin are complementary events.
•Independent Events: These are two or more events for which the outcome of one
does not affect the other. They are events that are not dependent on what occurred
previously. Each toss of a fair coin is an independent event.
•Conditional Events: These are events that are dependent on what occurred
previously. If five cards are drawn from a deck of fifty-two cards, the likelihood of the
fifth card being an ace is dependent on the outcome of the first four cards.
3.0 PROBABILITY
Probability is defined as the chance that an event will happen or the likelihood that an event
will happen.
The definition of probability is
Probability
Number of Favorable Events
Number of Total Events

The favorable events are the events of interest. They are the events that the question is
addressing. The total events are all possible events that can occur relevant to the question
asked. In this definition, favorable has nothing to do with something being defective or nondefective.
What is the probability of a head occurring in one toss of a coin?
The number of favorable events is 1 (one head) and the number of total events is 2 (head or
tail). In this case, the probability formula verifies what is obvious.
Probability of a head
Number of Favorable Events
Number of Total Events
1
2
Probability numbers always range from 0 to 1 in decimals or from 0 to 100 in percentages.
Chapter 3 Basic Probability 31
3.1 Notation for Probability Questions
Instead of writing out the whole question, the following notation is used.
•What is the probability of event A occurring? = Probability (A) = P(A)
•What is the probability of events A and B occurring? = P(A and B) = P(A) and P(B)
•What is the probability of events A or B occurring? = P(A or B) = P(A) or P(B)
3.2 Probability in Terms of Areas
Probability may also be defined in terms of areas rather than the number of events.
Probability
Favorable Area
Total Area

Example 1
A plane drops a parachutist at random on a seven by five mile field. The field contains a
two by one mile target as shown below. What is the probability that the parachutist will
land in the target area? Assume that the parachutist drops randomly and does not steer
the parachute.
Probability of hitting t arg et
Favorable Area
Total Area

2
35
.057
4.0 METHODS TO DETERMINE PROBABILITY VALUES
There are three major methods used to determine probability values.
•Subjective Probability: This is a probability value based on the best available
knowledge or maybe an educated guess. Examples are betting on horse races,
selecting stocks or making product-marketing decisions.
•Priori Probability: This is a probability value that can be determined prior to any
experimentation or trial. For example, the probability of obtaining a tail in tossing a
coin once is fifty percent. The coin is not actually tossed to determine this
probability. It is simply observed that there are two faces to the coin, one of which
is tails and that heads and tails are equally likely.
Target
5 miles
7 miles
2 miles
1 mile
QReview 32
•Empirical Probability: This is a probability value that is determined by
experimentation. An example of this is a manufacturing process where after
checking one hundred parts, five are found defective. If the sample of one
hundred parts was representative of the total population, then the probability of
finding a defective part is .05 (5/100). The question may be asked: How is it
known that this sample is representative of the total population? If repeated trials
average .05 defective, with little variation between trials, then it can be said that
the empirical probability of a defective part is .05.
5.0 MULTIPLICATION THEOREM
The multiplication theorem is used to answer the following questions:
•What is the probability of two or more events occurring either simultaneously or in
succession?
•For two events A and B: What is the probability of event A and event B occurring?
The individual probability values are simply multiplied to arrive at the answer. The word “and”
is the key word that indicates multiplication of the individual probabilities. The multiplication
theorem is applicable only if the events are independent. It is not valid when dealing with
conditional events. The product of two or more probability values yields the intersection or
common area of the probabilities. The intersection is illustrated by the Venn diagrams in
section 11.0 of this chapter. Mutually exclusive events do not have an intersection or
common area. The probability of two or more mutually exclusive events is always zero.
For mutually exclusive events:
•P(A) and P(B) = 0
For independent events:
•Probability (A and B) = P(A) and P(B) = P(A) X P(B)
For multiple independent events, the multiplication formula is extended. The probability that
five events A, B, C, D and E occur is
P(A) and P(B) and P(C) and P(D) and P(E) = P(A) x P(B) x P(C) x P(D) x P(E)
Example 2
What is the probability of getting a raise and that the sun will shine tomorrow?
Given: Probability of getting a raise = P(r) = .10
Probability of the sun shining = P(s) = .30
The events are independent.
P(raise) and P(sunshine) = P(r) x P(s) = .10 x .30 = .03 or 3%
Chapter 3 Basic Probability 33
6.0 ADDITION THEOREM
The addition theorem is used to answer the following questions:
•What is the probability of one event or another event or both events occurring?
•What is the probability of event A or event B occurring?
The word “or” indicates addition of the individual probabilities. The answers to the above
questions are different depending on whether the events are mutually exclusive or
independent.
Mutually exclusive events do not have an intersection or common area. The individual
probabilities are simply added to arrive at the answer. For mutually exclusive events:
•P(A or B) = P(A) or P(B) = P(A) + P(B)
•P(A or B or C or D) = P(A) + P(B) + P(C) + P(D)
For two independent events, the intersecting or common area must be subtracted or it will be
included twice. (Refer to the Venn diagram in section 11.0).
Probability (A or B) = P(A) or P(B) = P(A) + P(B) – P(A X B)
For three independent events:
P(A or B or C) = P(A) + P(B) + P(C) – P(A X B) – P(A X C) – P(B X C) + P(A X B X C)
Example 3
What is the probability of getting a raise or that the sun will shine tomorrow?
Given: Probability of getting a raise = P(r) = .10
Probability of the sun shining = P(s) = .30
P(raise) or P(sunshine) = P(r) or P(s) = P(r or s)
P(r or s) = P(r) + P(s) - = .10 + .30 - = .40 - .03 = .37 or 37%
The word “and” is associated with the multiplication theorem and the word “or” is
associated with the addition theorem.
7.0 COUNTING TECHNIQUES - PERMUTATIONS AND COMBINATIONS
Permutations and combinations are simply mathematical tools used for counting. In many
cases, it may be cumbersome to count the number of favorable events or the number of total
events when solving probability problems. Permutations and combinations help simplify the
task.
7.1 Permutations
A permutation is an arrangement of things, objects or events where the order is
important. Telephone numbers are special permutations of the numerals 0 to 9 where
each numeral may be used more than once. The order defines each unique telephone
number.
QReview 34
In the following example, it is assumed that each object is unique and cannot be used
more than once. The letters A, B, and C may be arranged in the following ways:
ABC BAC CAB
ACB BCA CBA
This is an ordered arrangement, because ABC is different than BCA. Since the order of
the letters makes a difference, each arrangement is a permutation. From the above
example, It is concluded that there are six permutations that can be made from three
objects. The general formula for permutations is
n Pr = n
n r
!
( )!
n = The total objects to arrange
r = The number of objects taken from the total to be used in the arrangements
By definition: 0! = 1 and 1! = 1
Example 4
Using the permutation formula and the three letters A, B and C, how many permutations
can be made using all three letters?
3P3 =
3!
3 - 3 !
=
3 x 2
0!
=
6
1
= 6
( )
Example 5
How many permutations can be made by using two out of the three letters?
3P2 =
3!
3 - 2 !
=
3 x 2
1!
=
6
1
= 6
( )
The permutations are
AB BA BC
AC CA CB
Example 6
There are three different assembly operations to be performed in making a certain part.
There are nine people working on the floor. How many different assembly crews can be
formed?
Chapter 3 Basic Probability 35
This may be stated as the number of permutations that can be made from nine objects
used three at a time.
9P3 =
9!
(9 - 3)!
=
9!
6!
=
9 x 8 x 7 x 6!
6!
= 504
7.2 Combinations
A combination is a grouping or arrangement of objects where the order does not make a
difference.
The arrangement of the letters ABC is the same as BCA. The number of combinations
that can be made by using three letters, three at a time, is one. This can be expanded to
state that the number of combinations that can be made by using n letters, n at a time, is
one. A hand of five cards consisting of a Jack, a Queen, a King, and two Aces is the
same as a Queen, two Aces, a Jack and a King. The order in which the cards were
received makes no difference. There is only one combination that can be made by using
five cards, five at a time.
The formula for combinations is
n r
C n r
P
r
n
r r

! 
!
(n )! !
n = Total objects to arrange
r = Number of objects taken from the total to be used in the arrangements
The symbol for number of combinations is often shown as
( ) ( ) r
n
r
n
n r where = C
When the symbol appears in a formula, the number of combinations is to be computed
using the combination formula.
r
n =
n!
(n - r)! r!
Example 7
From the three letters A, B and C, how many combinations can be made by using two
out of the three letters?
2
3 =
3!
(3 - 2)!2!
=
3 x 2 x 1
2 x 1 x 1
= 3
QReview 36
The combinations are
AB AC BC
BA is the same as AB
CA is the same as AC
CB is the same as BC
Example 8
Ten parts have been manufactured. Two parts are to be inspected for a critical
dimension. How many different sample arrangements can be made?
If the parts are labeled 1 to 10, then parts 1 and 5 make one arrangement, parts 3 and 7
make another, 6 and 8 another, etc. The listing of the various arrangements can be
completed and total arrangements counted. The combination formula can perform this
task and save a considerable amount of time.
The total arrangements or combinations that can be made:
2
10 10
10 2 2
10
8 2
10 9 8
8 2 1
45


!
( )! !
!
! !
!
x !
x x
x x
The permutation and combination formulas are very useful tools in evaluating and solving
probability problems. It is often necessary to count the number of favorable and total
events that can occur. Without these counting techniques, this would be a very
cumbersome and sometimes impossible task.
8.0 PROBABILITY DISTRIBUTIONS
Probability distributions and their associated formulas and tables allow us to solve a wide
variety of problems in a logical manner. Probability distributions are classified as discrete or
continuous. Three discrete distributions will be reviewed in this chapter. Continuous
distributions are covered in the next chapter. Probability distributions are used to generate
sampling plans, predict yields, arrive at process capabilities, determine the odds in games of
chance and many other applications.
The three discrete distributions that will be reviewed:
•The Hypergeometric Probability Distribution
•The Binomial Probability Distribution
•The Poisson Probability Distribution
One of the most difficult tasks for a beginning student in probability is to know which
distribution or formula to use for a specific problem. A roadmap is given in section 10.0 of
this chapter to assist in the task.
The quality engineer may be asked to calculate the probability of the number of defects or the
number of defective units in a sample. There is a difference between the two phrases. A
defect is an individual failure to meet a requirement. A defective unit is a unit of product that
contains one or more defects. Many defects can occur on one defective unit.
Chapter 3 Basic Probability 37
8.1 The Hypergeometric Probability Distribution
The hypergeometric distribution is the basic distribution of probability. The
hypergeometric probability formula is simply the number of favorable events divided by
the number of total events. It can be described as the true basic probability distribution of
attributes. To use the hypergeometric formula, the following values must be known.
N = The total number of items in the population (lot size)
n = The number of items to be selected from the population (sample size)
A = The number in the population having a given characteristic
B = The number in the population having another characteristic
a = The number of A that is desired to occur
b = The number of B that is desired to occur
The hypergeometric probability formula is

P(a and b) = a
A
b
B
n N
Example 9
An urn contains fifteen balls, five red and ten green. What is the probability of obtaining
exactly two red and three green balls in drawing five balls without replacement?
This question may also be stated as:
•What is the probability of obtaining two red balls?
•What is the probability of obtaining three green balls?
All three questions are the same. When setting up the problem, all events must be
considered regardless of how the question is asked.
In this case, the probability of a single event is not constant from trial to trial. This is the
same as sampling without replacement. The outcome of the second draw will be
affected by what was obtained on the first draw. The number of favorable events and the
number of total events must be computed.
The number of ways that red balls may be selected:
2
5 =
5!
3! 2!
=
5 x 4 x 3 x 2 x 1
3 x 2 x 1 x 2 x 1
= 10
The number of ways that green balls may be selected:
3
10 =
10!
7! 3!
=
10 x 9 x 8 x 7!
7! x 3 x 2 x 1
= 120
QReview 38
The total number of ways to select a sample of five balls from a population of fifteen balls:
5
15 =
15!
10! 5!
=
15 x 14 x 13 x 12 x 11 x 10!
10! x 5 x 4 x 3 x 2 x 1
= 3003

Then: P(2r and 3g) = =
10 x 120
3003
= .3996 2
5
3
10
5
15
This is a specific application of the hypergeometric probability formula. Many similar
problems may be solved using this method. To use the hypergeometric formula, the
population must be small enough so that the number of items with the characteristics in
question can be determined.
Example 10
A box contains ten assemblies of which two are defective. A sample of three assemblies
is selected at random. What is the probability that the two defective parts will be
selected? (For this to occur there must be two defective parts and one good part in the
sample.)

P(2 defective parts) P = =
1 x 8
120
(2) = .0667 2
2
1
8
3
10
8.2 The Binomial Probability Distribution
The binomial probability formula is used when events are classified in two ways such as
good/defective, red/green, go/no-go, etc. The prefix Bi means two. The events or trials
must be independent. When the binomial formula is used, it is assumed that the lot size
is infinite and the probability of a single success is constant from trial to trial.
The binomial probability formula is be used to answer the following question: What is the
probability of x successes in n trials where the probability of a single success is p? .
The binomial formula is
P x p p x
n ( ) ( ) x (1 )nx
Example 11
A coin is tossed five times. (This is the same as a sample size of five). What is the
probability of obtaining exactly two heads in the five tosses?
It is known, by prior knowledge, that the probability of a single success (probability of a
head in one toss of a coin) is fifty percent. The question is looking for two successes or
two heads in five tosses of a coin. A success is the outcome that is desired to occur.
Chapter 3 Basic Probability 39
For this example:
•The number of trials = n = 5
•The probability of a single event = p = 1/2
•The number of successes that the question is seeking (x = 2).
To arrive at the answer to the question the values are entered in the binomial formula.
P(2Heads) = ( )2 ( )3 = 10 x x = .3125 or 31.25%
2 5
½ ½ 1/ 4 1/ 8
Example 12
In manufacturing screwdrivers, it was empirically determined that the process yields, on
average, 5% defective product. What is the probability that in a sample of ten
screwdrivers there are exactly three defective units?
n = 10, p = .05, x = 3
P(3 defective units) = (.05)3 (.95 )7 = 120 x .000125 x .6983 = .0105 or 1.05%
3
10
Example 13
A company produces electronic chips by a process that normally averages 2% defective
products. A sample of four chips is selected at random and the parts are tested for
certain characteristics.
a) What is the probability that exactly one chip is defective?
P(1 defective chip) (1) 4 (.02)1(.98)3 4(.02)(.9412) .0753
1 P 
b) What is the probability that more than one chip is defective?
More than one defective chip in a sample of four means two, three or four defective
chips. The probability of each may be calculated using the binomial formula.
P(more than 1 defective chip) = P(2) or P(3) or P(4) = P(2) + P(3) + P(4)
In any trial or sample, the sum of the probabilities of the individual events always
equal one. In this problem: P(0) + P(1) + P(2) + P(3) + P(4) = 1
P(more than 1 defective) = 1 - = 1 - = .0023
QReview 40
8.3 The Poisson Probability Distribution
The Poisson distribution is the mathematical limit to the binomial distribution and may be
used to approximate binomial probabilities. The Poisson is also a distribution in its own
right when solving problems involving defects per unit rather than fraction defectives.
Tables showing subsets of Poisson probabilities appear in many textbooks. The tables
greatly simplify the solution of many problems. The most extensive Poisson table is
Poisson's Exponential Binomial Limit by E. C. Molina. The tables were developed in the
1920s and published in 1949.
If n is large and p is small so that n times p (np) is a positive number less than five, then
the Poisson is a good approximation to the binomial. The value p and the ratio n/N should
be less than 0.10.
When solving binomial problems with the Poisson formula, the terms n, x and p are the
same as in the binomial formula. The task is to calculate the probability of x successes in
n trials, where the probability of a single success is p. Remember that p is a fraction
defective when used to approximate the binomial, and p is defects per unit when counting
the number of defects instead of the number of defective units.
In some cases neither n nor p is given, but the product np may be given. If p is a fraction
defective then np is the average number of defective units in the sample. If p is in terms
of defects per unit then np is the average number of defects in the sample.
The Poisson formula is
Example 14
In making switches, it has been determined by empirical studies that there is, on
average, one defect per switch. What is the probability of selecting a sample of five
switches that contains zero defects? There are two methods to solve this problem. The
first method is to use the above formula where x = 0, n = 5, and p = 1, therefore
np = 5 x 1 = 5.
P
e
( ) or
( )
!
.
0 . .
5
0
00674
1
00674 674%
5 0


The second and most widely used method is to use the Poisson tables that are
published in most statistics books. To use the tables, find the value of x in the leftmost
column, then find the value of np on the top row and read P(x) at the intersection of the
two values.
The Poisson table value for P(0) = .006738 or .674%
Example 15
In a paper making operation it was found that each 1000 foot roll contained, on average,
one defect. One roll is selected at random from the process.
P x
e
x
np x
( )
(np)
!


Chapter 3 Basic Probability 41
a) What is the probability that this roll contains zero defects?
Use the Poisson table where x = 0 and np = 1. The Poisson table value
for P(0) = .368.
b) What is the probability that the roll contains exactly three defects?
The Poisson table value for P(3) = .061
c) What is the probability that this roll contains more than one defect?
P(more than one defect) = P(2) + P(3) + P(4) + … + P()
= 1 -
= 1 - = .264
Example 16
In manufacturing the Que model car, a study determined that on average there are three
defects per car. What is the probability of buying a Que with less than three defects?
P(less than 3 defects) = P(0) + P(1) + P(2)
Use the Poisson tables and find P(0), P(1) and P(2) where np = 3
P(less than 3 defects) = .049 + .149 + .224 = .422
9.0 CONDITIONAL PROBABILITY
Conditional probability is defined as the probability of an event occurring if another has
occurred or has been specified to occur simultaneously, and the outcome of the first event
affects the probability of the second event. Conditional events are not independent.
The probability of B occurring given that A has already occurred is stated as P(B/A), where
the symbol / means “given that.”
The formulas for conditional probability are shown below. These are known as Bayes
Formulas.
P B A
P A B
P A
( / )
( & )
( )

P A B
P A B
P B
( / )
( & )
( )

Since the two formulas have a common term P(A & B), they may be used together to solve
many problems involving conditional probability.
QReview 42
Conditional events are not independent so P(A & B) is not equal to P(A) X P(B). From Bayes
formulas:
P(A & B) = P(B/A) P(A)
and P(A & B) = P(A/B) P(B)
Example 17
A lot of fifteen items contains five defective items. Two items are drawn at random. What
is the probability that the second item drawn will be defective?
Let A = event that first item is defective
Let A' = event that first item is good
Let B = event that second item is defective
The question stated in probability terms is P(B) = ?
P(A) = 5/15, P(A') = 10/15
P(B) = P(A & B) or P(A' & B) P(first item defective & second item defective) or
P(first item good & second item defective)
P(B) = P(B/A) P(A) or P(B/A') P(A')
P(B) = P(B/A) P(A) + P(B/A') P(A')
P(B) = (4/14)(5/15) + (5/14)(10/15)
P(B) = (20/210) + (50/210) = 70/210 = .333
Example 18
It has been found that 10% of certain relays have bent covers and will not work. If 40%
have bent covers, what is the probability that a relay with a bent cover will not work?
Let A = event that relays have bent covers
Let B = event that relays will not work
Given: P(A & B) = .10, P(A) = .40
The first formula of the conditional probability formulas, Bayes formulas, gives the
following solution:
P(B / A) =
P(A and B)
P(A)
=
.10
.40
= .25
Chapter 3 Basic Probability 43
10.0 PROBABILITY ROADMAP
Are the events
Mutually Exclusive?
P(A) & P(B) = 0
P(A) or P(B) = P(A) + P(B)
Are the events
Independent?
i.e. does P(A/B) = P(A)?
P(A) & P(B) = P(A) X P(B)
P(A) or P(B) = P(A) + P(B) -
Sampling with
Replacement
Sampling without
Replacement
Hypergeometric

1. Binomial
2. When n is

large & p is
small, use
Poisson
Formula or
Tables.
Are the events
Conditional?
(Is one event
dependent on
the other?)
P(A) & P(B) = P(B/A) X P(A)
P(A) & P(B) = P(A/B) X P(B)
Yes
Yes
Yes
No
No
QReview 44
11.0 VENN DIAGRAMS
Venn diagrams show the events and corresponding probabilities in graphical form. The
events are shown as circles and the shaded area within the circles represent the
probabilities.
2 Events, P(A) & P(B)
= P(AB)
2 Events, P(A) or P(B) =
P(A) + P(B) - P(AB)
We must subtract one P(AB)
or it is counted twice.
3 Events
P(A) & P(B) & P(C)
= P(ABC)
2 Mutually Exclusive
Events. P(A) or P(B) =
P(A) + P(B)
(No Intersection)