manufacturing technology (me461) lecture23
TRANSCRIPT
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Manufacturing Technology
(ME461)
Instructor: Shantanu Bhattacharya
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Review of previous lecture
Type of control charts (variable control chart
and attribute control chart).
Benefits of control charts. Characteristics of data.
Alternate methods for finding the mean in
normal dataset as well as a frequency table.
Plotting of control charts.
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Checklist necessary for X and R chartsIt is helpful to visualize the decisions and calculations that must be made and the
actions that need to be taken for plotting a X and a R chart. They are the following:
1. Decisions preparatory to the control charts Some possible objectives of the charts.
Choice of variables.
Decisions on the basis of subgroups.
Decisions on size and frequency of subgroups.
Setting up the forms for recording the data.
Determining the method of measurement.2. Starting the Control Charts
Making and recording measurements and recording other relevant data.
Calculating the average X and range R of each subgroup.
Plotting the X and R charts.
3. Determining the trial control limits
Decision on required number of subgroups before control limits are calculated. Calculation of trial control limits.
Plotting the central lines and limits on the charts.
4. Drawing preliminary conclusions from the charts
Indication of control or lack of control.
Interpretation of processes in control.
Relationships between processes out of control and specification limits.
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Setting up the forms for recording the data
Layout of data should be as per convenience of calculation and analysis.
The forms should have a recording space for item of measurement, unit of measurement,
and operator remarks about tool change, operator change, machine change etc.
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Setting up the forms for recording the data
Layout of data should be as per convenience of calculation and analysis.
The forms should have a recording space for item of measurement, unit of measurement,
and operator remarks about tool change, operator change, machine change etc.
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Starting the Control charts
Making and recording the measurements and recording other relevant data:
The actual work of the control charts start with the first measurements.
Any method of measurement will have its own inherent variability; it is important that this isnot increased by mistakes in reading measurement instruments or errors in recording data.
Calculating the average and range for each subgroup:
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Plotting both the charts:Determining the trial control limits:
1. Decision is needed on the
required number of subgroups
before control limits are
calculated.
2. The fewer the subgroups used
the sooner the information thus
obtained will provide a basis foraction but the less the assurance
that the action will be sound.
Calculation of trial control limits:
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Drawing Preliminary Conclusions from the chart
Indication of control or lack of control:
Lack of control is indicated by points falling outside the control limits on either of the
charts.
When the points fall outside the control limits, we say that a process is out of control,this is equivalent to saying that assignable causes of variation are present and this is not a
constant cause system.
In contrast if none of the points fall outside the control limits then No assignable causes
are present.
Interpretation of processes in control:With the evidence from the control chart that a process is in control, we are in a
position to judge what is necessary to permit the manufacture of product that meets the
specifications for the quality characteristic charted. The control chart data gives us
estimate of :
1. The centering of the process.
2. The dispersion of the process.
Actions based on the relationship between the specifications and the centering and
dispersion of a controlled process depend somewhat on whether there are two
specification limits, a maximum or upper limit or a minimum or lower limit. This would
again depend heavily on the parameter that we choose to plot.
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Possible relationships of a process in control to upper
and lower specification limits
Case I: The spread of the process 6is
appreciably less than the difference
between the specification limits.
Case II: The spread is approximately
equal to the difference between the
specification limits.
Case III: The spread of the process is
appreciably greater than the difference
between the specification limits.
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Case I: When the process 6is appreciably less than difference between specification
limits
Frequency curves A,B,C, D and E indicate various
positions in which the process can be centered.
With any of the position A, B and C practically all
the products manufactured will meet thespecifications as long as the process stays in control.
In general when conditions A, B and C come it
represents the ideal manufacturing situation. When
the control chart shows that one of this control
chart exists, many different possible actions may be
considered depending on the relative economy.
For example it may be considered economically advisable to permit X to go out of control if it does not go
too far, i.e., the distributions may be allowed to move between positions B and C. This may avoid the cost
of frequent machine setup and of delays due to hunting of assignable causes of variation that will not be
responsible for unsatisfactory product.Or where acceptance has been based on 100% inspection, it may be economical to substitute acceptance
based on control charts.
Or where there is an economic advantage to be gained by tightening the specification limits, such action
may be considered.
With the process in the position D some points will fall above the upper specification limit. Similarly with
the process in position E some products will fall below the lower limit. In either case the obvious action is
to bring the centering of the process towards that of A.
C II Wh h 6 i l h diff b
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Case II: When the process 6is equal to the difference between
specification limitsIn this situation only the process
exactly centered between the
specification limits, as in position
A, will practically produce
everything conforming to the
specifications.
If the distribution shifts away
this exact centering as in B or C, it
is apparent that some of the
products will fall outside the
specification limits.
Here the obvious action is to take all steps possible to maintain the centering of
the process. This usually calls for continuous use of the control charts for X and R
with subgroups at frequent intervals and immediate attention to points out ofcontrol.
If fundamental changes can be made that reduce dispersion that eases the
pressure.
Consideration should also be given to changing of the tolerances.
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Case III: When the process 6is less than the difference between
specification limits
The third type of situation
arises when thespecification limits are so
tight that even with the
process in control and
perfectly centered some
non conforming parts still
get produced as in position
A.
This primarily calls for a review of tolerances.
It also calls for a fundamental change in the process that will reducethe process dispersion.
It is still very important to maintain the centering of the process; the
curves in position B and C show how a shift in process average will
increase the non conformity.
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Possible relationships of a process in control to a single specification
I: Low value of distribution ( ) isappreciably above LSL.
II: Low value of distribution is at LSL.
III: Low value of the distribution is
appreciably below LSL.
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Possible relationships of a process in control to a single
specification
The first situation is one in which there is a margin of safety.
The second is a one in which the specification is just barely met as long as the process
stays in control.
The third is one in which some non conforming products are produced unless there is a
fundamental change in the process spread or increase in process average.
All three distributions A,B and C have the same lower value. However, distribution B with a
greater dispersion must have a greater process average than A for the low points to be at
the same level. Similarly distribution C should have a greater process average than B. It is
evident that the greater the distribution the higher the average must be for the entire
distribution to fall above the lower specification limit.
The relationship between average and cost is vital in many instances. For example in the
filling of containers a reduction in dispersion may reduce cost by reducing the average
overfill.
On the other hand, the less the dispersion, the more important it is that the processaverage does not go out of control. This is illustrated by comparing distributions A and C in
Case III. In Case III both process averages have shifted an equal amount below their
position. However, the proportion of bad product, as indicated by the area of the
distribution below the specification limit, is much greater in A than in C.
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Use on control charts by a purchaser to help suppliers improve
their processes.
Facts of the case:
A manufacturer of electronic devices had trouble with the cracking of a certain small cross shaped
ceramic insulator used in the device.
The cracking generally took place after the manufacturing operations were nearly completed and did
so in a way that made it impossible to salvage the unit.
Hence the costs resulting from each cracked insulators during manufacturing operations suggested
that others might be likely to crack under service conditions.
In an effort to improve the situation, all incoming insulators of this type were given 100% inspection.This 100% inspection failed to decrease the percentage defective units.
A simple testing was then constructed to measure the actual strength in flexure by testing insulators
to destruction. From each incoming lot of insulators 25 were tested. As the insulators came from 2
suppliers, control charts (X and R) from both suppliers were separately maintained. The tests showed
that both suppliers had approximately the same % defectives but the explanations for defectives were
different.
Supplier A had higher average strength but complete lack of anything resembling statistical control.
Supplier B on the other hand had excellent statistical control but at a level such that an appreciable
part of frequency distribution was below the required minimum strength.
This diagnosis of the situation was brought to the attention of both suppliers and they were
encouraged to exchange information about production methods. Finally product control could be
achieved.
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Milling a slot in an aircraft terminal
block.Decisions preparatory to the control chart:
High percentages of rejections for many of the parts made in the machine shop of an
aircraft company indicated the need for examination of the reasons for trouble.
As most of the rejections were for failure to meet dimensional tolerances, it was decided
to try to find the causes of trouble by the use of X and R charts.
These charts which off-course required the measurements of dimensions, were to beused only for those dimensions that were causing numerous rejections.
Among the many dimensions, the ones selected for control charts were those having
high costs of spoilage and rework and those on which rejections were responsible for
delays in assembly operations.
This example deals with one such critical dimension, the width of a slot on the duralumin
forging used as a terminal block at the end of the rear tail or an airplane.The final matching of the slot width was a milling operation. The width of the slot was
specified as 0.8750 .
The design engineers had specified this dimension with an unilateral tolerance because
of the fit requirements of the terminal block; it was essential that the slot width be at
least .8750 in and desirable that it is very close to this value.
Milli l i d
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Milling example continuedDue to the small no. of available
inspection personnel it was decided
that for each chart the sample would
be inspected would be approximately5% of the total production of the parts
in question.
It was decided that all measurements
would be made at the place of
production and it was decided that a
part be measured every 20 partsproduced. The subgroup size was thus
fixed as 5. The figure on left shows all
observations.
The method to inspection to secure
data for each of the charts was to
measure two portions of the slot widthusing a micrometer screw gauge and
average these measurements.
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Milling example continuedStarting the control charts:
The actual measurements of the first 16 subgroups are shown in the table. These numbers of subgroups
correspond to a production order for 1600 of these terminal blocks.At the time of the 12thsubgroup, before the completion of this production order and before the calculation of the
central line or control limits, the quality control inspector noticed that the machine operator was occasionally
checking performance on a terminal block that had just come off the machine and was still hot.
After the 12thsubgroup the operator was asked to do measurements after some time of cooling is allowed.
Determining the trial control limits:
Calculations of the trial control limits was
made after the first 16 subgroups which
completed the production order. As shown in
the figure on the right these were done by
finding out the A2and D4factors for a
subgroup size 5.
Drawing preliminary conclusions from the
graphs:
Subgroup 1 is above the upper control limit
on the R chart.
Subgroup 10 is below the lower control limit
on the X chart. Moreover the last 10- of the 6points fall below the central line.
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Milling example continuedStarting the control charts:
The actual measurements of the first 16 subgroups are shown in the table. These numbers of subgroups
correspond to a production order for 1600 of these terminal blocks.At the time of the 12thsubgroup, before the completion of this production order and before the calculation of the
central line or control limits, the quality control inspector noticed that the machine operator was occasionally
checking performance on a terminal block that had just come off the machine and was still hot.
After the 12thsubgroup the operator was asked to do measurements after some time of cooling is allowed.
Determining the trial control limits:
Calculations of the trial control limits was
made after the first 16 subgroups which
completed the production order. As shown in
the figure on the right these were done by
finding out the A2and D4factors for a
subgroup size 5.
Drawing preliminary conclusions from the
graphs:
Subgroup 1 is above the upper control limit
on the R chart.
Subgroup 10 is below the lower control limit
on the X chart. Moreover the last 10- of the 6points fall below the central line.
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Milling Example ContinuedIt is obvious that the measurements made
are not the result of a constant system of
chance causes.
If subgroup 1 is eliminated fromconsideration, R for the remaining 15
subgroups is 536/ 15 = 36.
This gives us a revised upper control limit
D4 (R) = 2.11 (36) = 76.
Subgroup 5 falls exactly on the upper
control limit in the X chart. Hence a secondrevision of R with subgroup 5 eliminated
seems reasonable. Therefore R = 460/14 =
33. .
An estimate may be made of from the R
values. With a d2value of 2.362
corresponding to the subgroup size 5 we get
a = R/ d2 = 0.0014 in.
The value of the natural tolerance or spread
of the process is therefore 6= 6 (.0014) =
.0084 in.
This spread may be compared with
tolerance spread: U-L = .0050in
It is evident from the process that the natural tolerance of the
spread is substantially greater than the specified tolerance.
Therefore, unless the process dispersion is reduced the process
will keep producing non conforming components.
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Milling Example ContinuedIt is obvious that the measurements made
are not the result of a constant system of
chance causes.
If subgroup 1 is eliminated fromconsideration, R for the remaining 15
subgroups is 536/ 15 = 36.
This gives us a revised upper control limit
D4 (R) = 2.11 (36) = 76.
Subgroup 5 falls exactly on the upper
control limit in the X chart. Hence a secondrevision of R with subgroup 5 eliminated
seems reasonable. Therefore R = 460/14 =
33. .
An estimate may be made of from the R
values. With a d2value of 2.362
corresponding to the subgroup size 5 we get
a = R/ d2 = 0.0014 in.
The value of the natural tolerance or spread
of the process is therefore 6= 6 (.0014) =
.0084 in.
This spread may be compared with
tolerance spread: U-L = .0050in
It is evident from the process that the natural tolerance of the
spread is substantially greater than the specified tolerance.
Therefore, unless the process dispersion is reduced the process
will keep producing non conforming components.
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Revision of theory of ProbabilityDefinition:
Probability is concerned with the likelihood of an event occurring. The scale of
probability of an event varies from 0 to1.
If an event cannot occur on a trial, then the probability of its occurrence in 0.
If another event is certain to occur then its probability of occurrence is 1.
As an example suppose that a trial is drawing a piece at random from some production
line, and that the event in question is that the piece drawn is a defective or non
conforming one.Let us suppose that the probability of a defective is 0.05. This means that 5% of the
time when we draw a random piece from the line, it is defective.
The complementary event is that the piece drawn is a good one. Its probability is .95.
P(good or defective) = 1
Occurrence ratio
Suppose that we have a production process for which the probability of a piececontaining at least one minor defect is constantly 0.08.
We then say that the probability is 0.08 for a minor defective.
Now what happens to the observed proportion of minor defectives as we continue to
sample? This observed proportion of minor defectives is what we know as the
occurrence ratio.
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Samp e Tota s Occurrencei The proportion defective only tends to
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ratio
n D n d P = d/ n
10 0 10 0 .0000
10 1 20 1 .0050
10 1 30 2 .0667
10 1 40 3 .0750
10 0 50 3 .0600
50 5 100 8 .0800
50 4 150 12 .0800
50 6 200 18 .0900
50 4 250 22 .0880
50 8 300 30 .1000
50 1 350 31 .0886
50 3 400 34 .0850
50 5 450 39 .0867
50 3 500 42 .0840
50 5 550 47 .0855
50 5 600 52 .0867
50 5 650 57 .0877
50 4 700 61 .0871
50 3 750 64 .0853
50 6 800 70 .0875
50 4 850 74 .0871
50 7 900 81 .0900
50 4 950 85 .0895
50 4 1000 89 .0890
The proportion defective only tends to
approach p =0.08.
Sometimes it gets closer, sometimes it
backs away from p.
Before the total sample size was 100,
the occurrence ratio was below .08.
Between 100 and 150 it was 0.08 and
thereafter above.
Principle:
We can say that the p= d/n is an
estimate of the constant probability of
population p. How close the estimate
will be depends on sample size n, thevalue of p and also upon chance.
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Probability laws
Consider again the production line producing pieces with a constant probability .08 of
the piece being a minor defective, and such that each piece is independent of the other
produced.
This means that the probability of the next piece being a defective is 0.08 and the piece
being good is .92 irrespective of the preceding piece.
Now let us suppose that we draw a sample of two pieces and inspect them.
The outcomes are that the sample may contain 0, 1 or 2 defectives. Let us find the
probabilities of this outcome or events.
For the probability of the samples containing no defectives, we must have good pieces
on both draws.
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Probability Laws
P b bilit L
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Probability LawsIf 2 events A and B might occur on a trial or experiment, but the occurrence of either
one prevents the occurrence of the other, then events A and B are called mutually
exclusive.
For two such events P(A) + P (B) = P( A or B, mutually exclusive events)
We have seen this in the earlier example in the 1 good case P (1 Good) = .0736 +0.0736
If one of the two events A and A is certain to occur on a trial, but both cannot
simultaneously occur, then A and A are called complementary events.
For any such pair of events
We have seen this in the example of a single draw where p was given as 0.08.
L f P b bilit
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Laws of ProbabilityTwo events A and B are independent if the occurrence or non occurrence of A does not
affect the probability of B occuring.
Whenever a process produces defectives independently, or at random, so that theprobability of a defective on the next piece does not depend upon what the preceding
pieces were like, then we have the case of independent events. Such a process is said to
be in control, that is stable, even though some non conformity is produced.
Not all processes do behave in this manner.
For example: We consider the production of 3000 piston ring castings. The sample of 100
contain 25 defectives whereas the remaining 2900 only 4. This was because the
defectives occur in bunches from a certain defect producing condition.
Particularly in this case it was found that the castings were made from stacks of molds,
and if the iron is not hot enough when poured into a stack, many castings may be
defective. Under such conditions, whether a piece is good or defective does have an
influence on the probability of the next one being defective.
If two events A and B are independent, then we have :
P(Both A and B occurring) = P(A). P(B)
Example of Dependence and Equal
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Example of Dependence and Equal
LikelihoodAs a second example of probability, let us consider drawing without replacement from a lot
of N=6 speedometers, of which 1 is defective
Let N = no. of pieces in one lot and
D= no. of defective pieces in a lot
Now consider the very simple case in which we just draw a random sample of 1 from a lot
of 6. Random means that each of the six meters is equally likely to be chosen for thesample. Probability of each is 1/6.
There are only two kind of meters good and defective with P(good) = 5/6 and
P(defective) = 1/6.
Now next consider drawing a sample of n= 2 from the lot having N =6, of which D=1 isdefective. This may contain no. of defectives either d=0 or 1.
This is a case of two consecutive drawings which are not independent. Take first the case of
the sample yielding no defectives, that is, two good meters. We need
P(2 good) = P(good, good) = P(good on first draw). P(good on 2 nddraw given good on 1st
draw) = 5/6. 4/5 = 2/3
Counting samples (Permutations and
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Counting samples (Permutations and
combinations)In combinations we consider for example we have n objects which we can distinguish
between.Now how many distinct samples, each of one, can we draw from a lot of N=10?
Obviously the answer is 10. So, we call this a combination of N objects taken 1 at a time,
or in symbols :
C(N,1) = N.
Next consider samples of two, from say four good pieces (g1, g2, g3 and g4).
Then the number of distinct unordered samples may be found from the number of
distinct ordered samples. For example: for ordered samples
g1g2, g2g1, g1g3, g3g1, g1g4, g4g1, g2g3, g3g2, g2g4, g4g2, g3g4, g4g3
The no. of unordered samples are only half as much , that is, six, because, for example,
the one unordered pair g2g4 corresponds to two ordered pairs g2g4 and g4g2.
Now let us consider lots of 10 distinct pieces. The number of possible ordered samples,
each of two is 10.9, because there are 10 choices for the first piece and having made a
choice their remain 9 choices for the second piece. So we have 90 ordered samples and
exactly of this unordered. (45)
C ti l (P t ti d
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Counting samples (Permutations and
combinations)Now let us go to a sample of 3 from a lot of 10. The no. of distinct ordered samples is
10.9.8; 10 choices for the first, 9 choices for the second and 8 choices for the third. But
now six of this ordered samples correspond to just one unordered sample. For example:
g1g4g6, g1g6g4, g4g1g6, g4g6g1, g6g1g4, g6g4g1 all correspond to g1g4g6 unordered
sample.
Hence the number of distinct or unordered samples or combinations is 10.9.8/6 = 120
Ordered samples are also called permutations and they are calculated by the general
formulae
P(n,r) = n! / (n-r)! In the earlier case P(10,3) = 10!/ 7! = 10.9.8 = 720
We call the number of distinct unordered samples a combination and is given by thegenera formulae C(n,r) = n!/ r! (n-r)! In earlier case this would be = 10!/ 3!. 7! = 10.9.8/
3.2.1 = 120.
C t d d t D f t d d f ti
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Counted data: Defects and defectivesInspecting or testing n pieces , we may search for defects or non conformances in the
n pieces and record the total number of such defects. This is measuring quality by a
count of defects.
In inspecting or testing n pieces, we may consider whether each of n pieces does
contain any defects. Each piece having one or more defects is called a defective. This
measure is known as the count of defectives.
We shall consistently use the following symbols here and in later discussions:
n = number of units in a sample.
d = number of defective units in the sample of n units.
p = d/n= sample of fraction defective = proportion of defective units in the sample.
q= 1-p = sample fraction good
d= np = no. of defective units in the sample.
Binomial distribution for defectives:
Suppose we had a process with population fraction defective of p = .10. For a sample size
of 4 there could be
Either 0, 1,2, 3 or 4 defectives can exist.
Binomial distribution for defectives
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Binomial distribution for defectivesFirst find the probability of drawing a sample with all pieces good.
So for P(d=0) = P(4 good) = [P(good)]4 =(0.9)4 = .6561
Thus about 2/3 of the time, we draw a sample of n=4 pieces from the process, all four will
be good ones.
Now next we seek P (3 good, 1 defective) .
P (3 good, 1 defective) = P(g,g,g,d) + P(g,g,d,g) + P (g,d,g,g) + P (d,g,g,g) = (.9)(.9)(.9)(.1) + (.9)(.9)(.1)(.9) + (.9) (.1)(.9)(.9) + (.1) (.9)(.9)(.9) = 4(.9)^3 (.1) = .2916.
Next consider samples with d=2; they have two defectives and two good ones. How many
distinct orders are there for such samples? Six: ggdd. gdgd, gddg, dgdg, ddgg, dggd, the
probability for each one of these sample outcomes is (0.9)2(0.1)2.
P(d=2) = 6 (0.9)2
(0.1)2
= .0486
Similarly for d=3 there are only 4 orders of sampling results
P(d=3) = 4 (0.9)2(0.1)2 =.0036
Finally for d =4, all four must be defective
P(d=4) = .0001
Binomial distrib tion for Defecti es
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Binomial distribution for DefectivesThe sum of all the 4 outcomes are 1.
We can also represent the various coefficients viz, 4,6,4 of the products of the powers of .9
and .1 as
C(4,1) =4, C(4,2) = 6, C(4,3)= 4 respectively.
This reasoning enables us to write all the four probabilities in 1 formulae
P(d) = C(4,d) (0.9)4-d(0.1)d
This is also a representative of the dthterm of a Binomial distribution of n =4 and the p =0.9
and q= 0.1.
In general
P(d) = C(n,d) (p)n-d
(q)d
d P(d) P= d/n
0 .6561 .00
1 .2916 .25
2 .0486 .50
3 .0036 .75
4 .0001 1.00
Total 1 0000