Grupo 5_Probability and the Operation Characteristic Curve

Chapter 2

Probability and the Operating Characteristic Curve

Undoubtedly the most important single working tool in acceptance quality control is probabilitytheory itself. This does not mean that good quality engineers have to be accomplished probabilists orerudite mathematical statisticians. They must be aware, however, of the practical aspects ofprobability and how to apply its principles to the problem at hand. This is because most informationin quality control is generated in the form of samples from larger, sometimes essentially innite,populations. It is vital that the quality engineers have some background in probability theory. Onlythe most basic elements are presented here.

Probability

It is important to note that the term probability has come to mean different things to differentpeople. In fact, these differences are recognized in dening the probability, for there is not just onebut at least three important denitions of the term. Each of them gives insight into the nature ofprobability itself. Two of them are objectivistic in the sense that they are subject to verication,while the third is personalistic and refers to the degree of belief of an individual.

Classical Denition

If there be a number of events of which one must happen and all are equally likely, and if anyone of a (smaller) number of these events will produce a certain result which cannot otherwisehappen, the probability of this result is expressed by the ratio of this smaller number to the wholenumber of events (Whitworth 1965, rule IV). Here probability is dened as the ratio of favorable tototal possible equally likely and mutually exclusive cases.

Example. There are 52 cards in a deck of which 4 are aces. If cards are shufed so that they areequally likely to be drawn, the probability of obtaining an ace is 4=52 1=13.This is the denition of probability which is familiar from high school mathematics.

Empirical Denition

The limiting value of the relative frequency of a given attribute, assumed to be independent ofany place selection, will be called the probability of that attribute . . . . (von Mises 1957, p. 29).Thus, probability is regarded as the ratio of successes to total number of trials in the long run.

Example. In determining if a penny was in fact a true coin, it was ipped 2000 times resulting in1010 heads. An estimate of the probability of heads for this coin is .505. It would be expected thatthis probability would approach 1=2 as the sequence of tosses was lengthened if the coin were true.This is the sort of probability that is involved in saying that Casey has a .333 batting average.It implies that the probability of a hit in the next time at bat is approximately 1=3.

2008 by Taylor & Francis Group, LLC.

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15 Subjective Denition

Probability measures the condence that a particular individual has in the truth of a particularproposition, for example, the proposition that it will rain tomorrow (Savage 1972). Thus prob-ability may be thought of as a degree of belief on the part of an individual, not necessarily the samefrom one person to another.

Example. There is a high probability of intelligent life elsewhere in the universe.Here we have neither counted the occurrences and nonoccurrences of life in a number of

universes, nor sampled universes to build up a ratio of trials. This statement implies a degree ofbelief on the part of an individual who may differ considerably from one individual to another.These denitions have immediate applications in acceptance quality control. Classical probability

calculations are involved in the determination of the probability of acceptance of a lot of nite size,where all the possibilities can be enumerated and samples taken therefrom. Empirical probabilitiesare used when sampling from a process running in a state of statistical control. Here, the processcould conceivably produce an uncountable number of units so that the only way to get at theprobability of a defective unit is in the empirical sense. Subjective probabilities have been used inthe evaluation of sampling plans, particularly under cost constraints. They reect the judgment of anindividual or a group as to the probabilities involved. While sampling plans have been derivedwhich incorporate subjective probabilities, they appear to be difcult to apply in an adversaryrelationship unless the producer and the consumer can be expected to agree on the specicsubjective elements involved.There are many sources for information on probability and its denition. Some interesting

references of historic value are Whitworth (1965) on classical probability, von Mises (1957) onempirical probability, and the Savage (1972) work on subjective probability. Since the classical andempirical denitions of probability are objectivistic and can be shown to agree in the long run, andsince the empirical denition is more general, the empirical denition of probability will be usedhere unless otherwise stated or implied. When subjective probabilities are employed their nature willbe specically pointed out.

Random Samples and Random Numbers

Random samples are those in which every item in the lot or population sampled has an equalchance to be drawn. Such samples may be taken with or without replacement. That is, items may bereturned to the population once drawn, or they may be withheld. If they are withheld, the probabilityof drawing a particular item from a nite population changes from trial to trial. Whereas, if the itemsare replaced or if the population is uncountably large, the probability of drawing a particular itemwill not change from trial to trial. In any event, every item should have an equal opportunity forselection on a given trial, whether the probabilities change from trial to trial or not.This may be illustrated with a deck of cards. There are 52 cards, one of which is the ace of spades.

Sampling without replacement, the probability of drawing the ace of spades on the rst draw is 1 outof 52, while on the second draw it is 1 out of the 51 cards that remain, assuming it was not drawn onthe rst trial. If the cards were replaced as drawn, the probability would be 1 out of 52 on any drawsince there would always be 52 cards in the deck.Note that if the population is very large, the change in probability when samples are not replaced

will be very small and will remain essentially the same from trial to trial. In a rafe of 100,000tickets the chances of being drawn on the rst trial is 1 in 100,000 and on the second trial 1 in

99,999. Essentially, 0.00001 in each case. Few rafes are conducted in which a winning ticket isreplaced for subsequent draws.


At the core of random sampling is the concept of equal opportunity for each item in thepopulation sampled to be drawn on any trial. Sometimes special sampling structures are usedsuch as stratied sampling in which the population is segmented and samples are taken from thesegments. Formulas exist for the estimation of population characteristics from such samples. In anyevent, equal opportunity should be provided within a segment for items to be selected.To guarantee randomness of selection, tables of random numbers have been prepared. These

numbers have been set up to mimic the output of a truly random process. They are intended to occurwith equal frequency but in a random order. Appendix Table T2.1 is one such table. To use therandom number tables

1. Number the items in the population.

2. Specify a xed pattern for the selection of the random numbers (e.g., right to left, bottom totop, every third on a diagonal).

3. Choose an arbitrary starting place and select as many random numbers as needed for thesample.

4. Choose as a sample those items with numbers corresponding to the random numbers selected.

The resulting sample will be truly representative in the sense that every item in the population willhave had an essentially equal chance to be selected.Sometimes it is impractical or impossible to number all the items in a population. In such cases

the sample should be taken with the principle of random sampling in mind to obtain as good asample as possible. Avoid bias, avoid examining the samples before they are selected. Avoidsampling only from the most convenient location (the top of the container, the spigot at the bottom,etc.). In one sampling situation, an inspector was sent to the producers plant to sample the productas a boxcar was being loaded, since it was impossible to obtain a random sample thereafter. Suchstrategies as these can help provide randomness as much as the random sampling tables themselves.

Counting Possibilities

Evaluation of the probability of an event under the classical denition involves counting thenumber of possibilities favorable to the event and forming the ratio of that number to the total ofequally likely possibilities. The possibilities must be such that they cannot occur together on a singledraw; that is, they must be mutually exclusive. There are three important aids in making counts ofthis type: permutations, combinations, and tree diagrams.Suppose a lot of three items, each identied by a serial number is received, two of which are

good. The sampling plan to be employed is to sample two items and accept the lot if no defectivesare obtained. Reject if one or more are found. Thus, the sampling plan is n 2 and c 0, where n isthe sample size and c represents the acceptance number or maximum number of defectives allowedin the sample for acceptance of the lot.If the items are removed from the shipping container one at a time, we may ask in how many

different orders (permutations) the three items can be removed from the box. Suppose the serialnumbers are the same except for the last digit which is 5, 7, and 8, respectively. Enumerating theorders we have

578 875758 857785 587


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15

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15 The formula for the number of permutations of n things taken n at a time is

Pnn n! n(n 1)(n 2) . . . 1

where n!, or n factorial, is the symbol for multiplications of the number n by all the successivelysmaller integers down to one. Thus

1! 12! 2(1) 23! 3(2)(1) 64! 4(3)(2)(1) 24

and so on. It is important to note that we dene

0! 1

In the example, we want the number of permutations of 3 things taken 3 at a time, or

P33 3! 3(2)(1) 6

which agrees with the enumeration.In how many orders can we select the two items for our sample? Enumerating again:

57 8775 8578 58

The formula for the number of permutations of n things taken r at a time is

Pnr n!

(n r)!

Clearly the previous formula for Pnn is a special case of this formula. To determine the number ofpermutations of three objects taken two at a time we have

P32 3!

(3 2)! 3!1! 3(2)(1)

1 6

This makes sense and agrees with the previous result since the last item drawn is completelydetermined by the previous two items drawn and so does not contribute to the number of possibleorders (permutations).Now, let us ask how many possible orders are there if some of the items are indistinguishable

one from the other. For example, disregarding the serial numbers, we have one defective itemand two good ones. The good items are indistinguishable from each other and we may ask inhow many orders can we draw one defective and two good items. The answer may be found inthe formula for the number of permutations of n things, r of which are alike (good) and (n r) arealike (bad).Pnr,(nr) n!

r!(n r)!


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15 and for example, the answer is

P32,(21) P32,1 3!2!1!

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

3

Enumerating them we have

B G G

G B G

G G B

The reader may notice the similarity of the formula

Pnr,(nr) n!

r!(n r)!

and the classic formula for the number of combinations (groups) which can be made from n thingstaken r at a time. The formula is

Cnr n!

r!(n r)!

and shows how many distinct groups of size r can be formed from n distinguishable objects. If wephrase the question, in how many ways can we select two objects (to be the good ones) out ofthree, we have

Good Bad

Group 1 57 875 8

Group 2 78 587 5

Group 3 85 758 7

or

C32 3!

2!(3 2)! 3!2!1!

3

Thus we see

Pnr,(nr) Cnr

In general, the combinatorial formula may be used to determine the number of groupings of variouskinds. For example, the number of ways (groups) to select 4 cards from a deck of 52 to form handsof 4 cards (where order is not important) is

13 17 25 152! 52 51 50 49 48!

////

C524 4! 48! 4 3 2 1 48! 270,725


Using the classical denition of probability, then, the probability of getting a hand containing allfour aces is

P(four aces) number of four ace handsnumber of four card hands

1270,725

Here we have counted groups where order in the group is not important.

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15 In the same way, probabilities can be calculated for use in evaluating acceptance sampling plans.The plan given in the earlier example was sample size 2; accept when there are no defectives in thesample. That is n 2 and c 0. To evaluate the probability of acceptance when there is onedefective in the lot of N 3, we would proceed as follows:

1. To obtain probability of acceptance, we must count the number of samples in which we wouldobtain 0 defectives in a sample of 2.

2. The probability is the quantity obtained in step 1 divided by the total number of samples of 2that could possibly be obtained.

Then

1. To obtain samples of 2 having no defectives, we would have to select both items from the twoitems which are good. The number of such samples is C22 1.

2. There are C32 3 different unordered samples. So the probability of accepting Pa with thissampling plan is

Pa C22

C32 1

3

The third tool in counting possibilities in simple cases such as this is the tree diagram. Figure 2.1shows such a diagram for this example, for the acceptance (A) and rejection (R) of samples ofgood (G) and bad (B) pieces. Each branch of the tree going downward shows a given samplepermutation. We see that 1=3 of these permutations lead to lot acceptance. Counting the permuta-tions we have

P32 3!1! 6

possible samples, of which

P22 2!0! 2

Start

BFirst draw

Second draw

Acceptance decision

G

G G G BB G

GR R R A R A

FIGURE 2.1: Tree diagram.


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15 lead to acceptance. Then, the probability of acceptance is

Pa P22

P32 2

6 1

3

which shows that the probability of acceptance can be obtained by using either permutations orcombinations.

Probability Calculus

There are certain rules for manipulating probabilities which sufce for many of the elementarycalculations needed in acceptance control theory. These are based on recognition of two kinds of events.

Mutually exclusive events. Two events are mutually exclusive if, on a single trial, the occurrence ofone of the events precludes the occurrence of the other.

Independent events. Two events are stochastically independent if the occurrence of them on a trialdoes not change the probability of occurrence of the other on that trial.

Thus the events head and tail are mutually exclusive in a single trial of ipping a coin. They arealso not independent events since the occurrence of either on a trial drives the probability ofoccurrence of the other on that trial to zero.In contrast the events ace and heart are not mutually exclusive in drawing cards since they can

occur together in the ace of hearts. Further, they are also independent since the probability ofdrawing an ace is 4=52 1=13. If you know that a heart was drawn, the probability of the card beingalso an ace is still 1=13. Note that the events face card and queen are not independent. Theprobability of drawing a queen is 4=52 1=13; however, if you know a face card was drawn, theprobability of that card being a queen is now 4=12 1=3.Trials are sometimes spoken of as being independent. This means the sampling situation is such that

the probabilities of the events being investigated do not change from trial to trial. Flips of a coin aresuch as this in that the odds remain 50:50 from trial to trial. If cards are drawn from a deck and notreplaced the trials are dependent, however. Thus, the probability of a queen of hearts is 1=52 on therstdraw from a deck, but it increases to 1=51 on the second draw assuming it was not drawn on the rst.Kolmogorov (1956) has developed the entire calculus of probabilities from a few simple axioms.

Crudely stated and somewhat condensed, they are as follows:

1. The probability of an event, E, is always positive or zero, never negative: P(E) 0.2. The sumof the probabilities of events in the universeU, or population towhichEbelongs, is one:

P(U) 1:

3. If events A and B are mutually exclusive, the probability of A or B occurring is

P(A or B) P(A) P(B)

From the axioms, the following consequences can be obtained:4. The probability of an event must be less than or equal to one, never greater than one: P(E) 1.5. The probability of the null set (no event occurring) is zero: P(no event) 0.


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15 6. The probability of an event not occurring is the complement of the probability of the event:

P(not E) 1 P(E)The most useful rules in dealing with probabilities are the so-called

General rule of addition. Shows the probability of A or B occurring on a single trial.

P(A or B) P(A) P(B) P(A and B)

Clearly, if A and B are mutually exclusive, the term P(A and B) 0 and we have the so-calledspecial rule of addition.

P(A or B) P(A) P(B)

for A and B mutually exclusive

General rule of multiplication. Shows the probability of A and B both occurring on a single trialwhere P(BjA) is the conditional probability of B given A is known to have occurred

P(A and B) P(A)P(BjA) P(B)P(AjB)

Clearly, if A and B are independent, the factor P(BjA)P(B) since the probability of B isunchanged even if we know A has occurred (similarly for P(AjB). We then have the so-calledspecial rule of multiplication

P(A and B) P(A)P(B), A and B independentThis is sometimes used as a test for the independence of A and B since if the relationship holds, A

and B are independent.These rules can be generalized to any number of events. The special rules become

P(A or B or C or D) P(A) P(B) P(C) P(D), A, B, C, D mutually exclusiveP(A and B and C and D) P(A)P(B)P(C)P(D), A, B, C, D independent

and so on. These are especially useful since they can be employed to calculate probabilities overseveral independent trials. The general rule for addition is

P(A or B or C or D) P(A) P(B) P(C) P(D) P(AB) P(AC) P(AD) P(BC) P(BD) P(CD) P(ABC) P(ABD) P(ACD) P(BCD) P(ABCD)

alternating additions and subtractions of subtractions of each higher level of joint probability, whilethat for multiplication becomes

P(A and B and C and D) P(A)P(BjA)P(CjAB)P(DjABC)when there are four events. Each probability multiplied is conditional on those which went before.These rules may be illustrated using the example given earlier involving the computation of

the probability of acceptance Pa of a lot consisting of 3 units when one of them is defective and thesampling plan is n 2, c 0. Acceptance will occur only when both the items in the sample are

good. If we assume random samples are drawn without replacement, the events will be dependentfrom trial to trial. We need the probability of a good item on the rst draw and a good item on thesecond draw.


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15 Let A {event good on rst draw} and B {event good on second draw}then

P(A) 23

P(BjA) 12

since there are only two pieces left on the second draw. Applying the general rule of multiplication:

Pa P(A)P(BjA) 2312

1

3

which agrees with the result of the previous section.

Pa C22

C32 1

3

Now, what if the items were put back into the lot after inspection and the next sample drawn?This is a highly unusual procedure in practice, but serves as a model for some of the probabilitydistributions developed later. It simulates an innite lot 1=3 defective since, using this method ofinspection the lot would never be depleted. Under these conditions the special rule of multiplicationcould be employed since the events would be independent of each other from trial to trial. We obtain

Pa P(A)P(B) 2323

4

9

This makes sense since the previous method depleted the lot and made it more likely to obtain thedefective unit on the second draw.Further, suppose two such lots are inspected using the procedure of sampling without replace-

ment. What is the probability that at least one will be accepted? That is, what is the probability thatone or the other will be passed? Here, let C {event rst lot is passed} and D {event second lot ispassed}, then the probability both lots are passed is

P(C and D) 13

13

1

9

using the special rule of multiplication since they are inspected independently. Then the probabilityof at least one passing is

P(C or D) P(C) P(D) P(C and D) 1

3 13 19 5

9

The probability of not having at least one lot pass is

P(both fail) 1 P(C or D) 1 59 4

9

which could have been calculated using the special rule of multiplication as

P(both fail) [1 P(C)] [1 P(D)]

23

23

49


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15 Finally, suppose there are ve inspectors: V, W, X, Y, Z, each with the same probability of selection.The lot is to be inspected. What is the probability that the inspector chosen is X, Y, or Z? Since inthis case the use of the inspectors is mutually exclusive, the special rule of addition may be used

P(X or Y or Z) P(X) P(Y) P(Z) 1

5 15 15

35

These are a few of the tools of probability theory. Fortunately, they have been put to use by theoristsin the design of the methods of acceptance quality control to develop procedures which do notrequire extensive knowledge of the subject for application. These methods are presented here insubsequent chapters. Nevertheless, to gain a true appreciation for the subtleties of acceptancesampling, a sound background in probability theory is invaluable.

Operating Characteristic Curve

A fundamental use of probability with regard to acceptance sampling comes in describing thechances of a lot passing sampling inspection if it is composed of a given proportion defective. Thevery simplest sampling plan is, of course, as follows:

1. Sample one piece from the lot.

2. If the sampled piece is good, accept the lot.

3. If the sampled piece is defective, reject the lot.

This plan is said to have a sample size n of one and an acceptance number of zero since the samplemust contain zero defectives for lot acceptance to occur; otherwise, the lot will be rejected. That is,n 1, c 0. Now, if the lot were perfect, if would have no chance of rejection since the samplewould never contain a defective piece. Similarly, if the lot were completely bad there would be noacceptances since the sample piece would always be defective. But what if the lot were mixeddefective and good? This is where probability enters in. Suppose one-half of the lot was defective,then the chance of drawing out a defective piece from the lot would be 50:50 and we would have50% probability of acceptance. But it might be one-quarter defective leading to a 75% chance foracceptance, since there are three-quarters good pieces in the lot. Or again, the lot might be three-quarters defective leading to a 25% chance of nding a good piece. Since the lot might be any of amultitude of possible proportions defective from 0 to 1, how can we describe the behavior of thissimple sampling plan? The answer lies in the operating characteristic (OC) curve which plots theprobability of acceptance against possible values of proportion defective. The curve for thisparticular plan is shown in Figure 2.2.We see that for any proportion defective p, the probability of acceptance Pa is just the comple-

ment of p; that is

Pa 1 pThis is only true of the plan n 1, c 0. Thus the OC curve stands as a unique representation of the

performance of the plan against possible alternative proportions defective. A given lot can have onlyone proportion defective associated with it. But we see from the curve that lots which have aproportion defective greater than 0.75 have less than a 25% chance to be accepted and those lots


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15 with less than 0.25 defective pieces will have greater than a 75% chance of pass. The OC curve

00

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

P a

0.2 0.4 0.6 0.8 1.0 p

FIGURE 2.2: OC curve, n 1, c 0.gives at a glance a characterization of the potential performance of the plan, telling how the plan willperform for any submitted fraction defective.Now consider the plan n 5, c 0. The OC curve can be easily constructed using the rules for

manipulation of probabilities given above. First, however, let us assume we are sampling from avery large lot or better yet from the producers process so the probabilities will remain essentiallyindependent from trial to trial. Note that the probability of acceptance Pa for any proportiondefective p can be computed as

Pa (1 p)(1 p)(1 p)(1 p)(1 p) (1 p)5

since all the pieces must be good in the sample of 5 for lot acceptance. To plot the OC curve wecompute Pa for various values of p

p (1 p) Pa.005 .995 .975.01 .99 .951.05 .95 .774.10 .90 .590.20 .80 .328.30 .70 .168.40 .60 .078.50 .50 .031

and graph the result as in Figure 2.3.


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15 00

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

P a

0.2 0.30.1 0.4 0.5 p

FIGURE 2.3: OC curve, n 5, c 0.We see from Figure 2.3 that if the producer can maintain a fraction defective less than .01 theproduct will be accepted 95% of the time or more by the plan. If product is submitted which is 13%defective, it will have 50:50 chance of acceptance, while product which is 37% defective has only a10% chance of acceptance by this plan. It is conventional to designate proportions defective havinga given probability of acceptance as probability points. Thus, a fraction defective having probabilityof g is shown as pg. Particular probability points may be designated as follows:

Pa Term Abbreviation Probability Point

.95 Acceptable quality level AQL p.95

.50 Indifference quality IQ p.50

.10 Lot tolerance percent defective(10% limiting quality)

LTPD [LQ(.10)] p.10

Designation of these points gives a quick summary of plan performance. The term acceptablequality level (AQL) is commonly used as the 95% point of probability of acceptance, although mostdenitions do not tie the term to a specic point on the OC curve and simply associate it with ahigh probability of acceptance. The term is used here as it was used by the Columbia StatisticalResearch Group in preparing the Navy (1946) input to the JAN-STD-105 standard. LTPD refers tothe 10% probability point of the OC curve and is generally associated with percent defective. Theadvent of plans controlling other parameters of the distribution led to the term limiting quality (LQ),usually preceded by the percentage point controlled. Thus, 10% limiting quality is the LTPD.The OC curve is often viewed in the sense of an adversary relationship between the producer and

the consumer. The producer is primarily interested in insuring that good lots are accepted while theconsumer wants to be reasonably sure that bad lots will be rejected. In this sense, we may think of a


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15 0

0.1

PQL CQL

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0a

b

P aproducers quality level (PQL) and associated producers risk a and a consumers quality level(CQL) with associated consumers risk b. Viewed against the OC curve the PQL and CQL appear asin Figure 2.4.Plans are often designated and constructed in terms of these two points and the associated risks.

As indicated above, the risks are often taken as a .05 for the producers risk and b .10 for theconsumers risk.The OC curve sketches the performance of a plan for various possible proportions defective. It is

plotted using appropriate probability functions for the sampling situation involved. The probabilityfunctions are simply formulas for the direct calculation of probabilities which have been developedusing the appropriate probability theory.

References

Kolmogorov, A. N., 1956, Foundations of the Theory of Probability, 2nd ed., Chelsea, New York.Savage, L. J., 1972, Foundations of Statistics, 2nd ed., John Wiley & Sons, New York.United States Department of the Navy, 1946, General Specications for Inspection of Material, Superintendent

of Documents, Washington, DC, 1946. Appendix X, April 1, 1946; see also U.S. Navy MaterialInspection Service, Standard Sampling Inspection Procedures, Administration Manual, Part D, Chapter 4.

von Mises, R., 1957, Probability, Statistics and Truth, 2nd ed., Macmillan, New York.Whitworth, W. A., 1965, Choice and Chance, Hafner, New York.

p

FIGURE 2.4: PQL and CQL.


Problems

1. A lot of 50 items contains 1 defective unit. If one unit is drawn at random from the lot, what isthe probability that the lot will be accepted if c 0?

2. A bottle of 500 aspirin tablets is to be randomly sampled. The tablets are allowed to drop out

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15 one at a time to form a string, those coming out rst at one end, those last at the other.A random number from 1 to 1000 is selected and divided by 2, rounding up. The tablet in thecorresponding numerical position is selected. Is this procedure truly random?

3. Two out of six machines producing bottles are bad. The bottles feed in successive order intogroups of six which are scrambled during further processing and packed in six-packs. In howmany different orders can the two defective bottles appear among the six?

4. Six castings await inspection. Two of them have not been properly nished. The inspector willpick two and look at them. How many groups of two can be formed from the six castings?How many groups of two can be formed from the two defective castings? What is theprobability that the inspector will nd both castings looked at are bad?

5. Form a probability tree to obtain the probability that the inspector will nd both castings badin Problem 4.

6. Use the probability calculus to nd the probability that the inspector will nd two bad castingsin selecting two. Why is it not 2=6 2=6 4=36 1=9? What is the probability that they areboth good? What is the probability that they are both the same? What type events allow theseprobabilities to be added?

7. At a given quality level the probability of acceptance under a certain sampling plan is .95.If the lot is rejected the sampling plan is applied again, just to be sure, and a nal decision ismade. What is the probability of acceptance under this procedure?

8. Draw the OC curve for the plan n 3, c 0. What are the approximate AQL, IQ, and LTPDvalues for this plan?

9. In a mixed acceptance sampling procedure two types of plans are used. The rst plan is usedonly to accept. If the lot is not accepted, the second plan is used. If both type plans havePQL .03, CQL .09 with a .05 and b .10. What is the probability of acceptance of themixed procedure when the fraction defective is .09?

10. At the IQ level the probability of acceptance is .5. In ve successive independent lots, what isthe probability that all fail when quality is at the IQ level? What is the probability that all pass?What is the probability of at least one failure? 2008 by Taylor & Francis Group, LLC.

Chapter 2: Probability and the Operating Characteristic CurveProbabilityClassical DefinitionEmpirical DefinitionSubjective Definition

Random Samples and Random NumbersCounting PossibilitiesProbability CalculusOperating Characteristic CurveReferencesProblems