INTRODUCTION TO
ENGINEERING RELIABILITY
Lecture Notes for MEM 361
Albert S. D. Wang
Albert & Harriet Soffa Professor
Mechanical Engineering and Mechanics Department
Philadelphia, PA 19104
Drexel University
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Introduction to Engineering Reliability - Lecture Notes | MEM 361, Study notes of Mechanical Engineering
Material Type: Notes; Class: Engineering Reliability; Subject: Mechanical Engr & Mechanics; University: Drexel University; Term: Unknown 1989;
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INTRODUCTION TO
ENGINEERING RELIABILITY
Lecture Notes for MEM 361
Albert S. D. Wang
Albert & Harriet Soffa Professor
Mechanical Engineering and Mechanics Department
Philadelphia, PA 19104
Drexel University
Chapter - I Introduction I - 1
CHAPTER I. INTRODUCTION
Why Engineering Reliability?
In a 1985 Gallop poll, 1000 customers were asked: what attributes are most important to you in
selecting and buying a product?
On a scale from 1 to 10, the following were the results from the poll:
* Performance quality 9.
* Lasts a long time (dependability) 9.
* Good service 8.
* Easily repaired 8.
* Warranty 8.
* Operational safety 8.
* Good look 7.
* Brand reputation 6.
* Latest model 5.
* Inexpensive price 4.
It is interesting to note that the top five attributes are related to “Engineering Reliability”; the
study of which is the essence of this course.
I-1 Basic Notions in Engineering Reliability.
Engineering Product. An “engineering” product is designed, manufactured, tested and deployed
in service. The product may be an individual part in a large operating system; or it may be the
system itself. In either case, the product is expected to perform the designed functions and last
beyond it’s designed service life.
Product Quality. It is a “quantitative" measure of the engineered product to meet it’s designed
functional requirements, including the designed service life.
Example 1-1: A bearing ball is supposed to have the designed diameter of 10mm. But an inspection of a large sample off the production line finds that the diameters of the balls vary from 9.91mm to 10.13mm, although the average diameter from all the balls in the sample is very close to being 10mm.
Now, the bearing ball is an engineering product; it’s diameter is one “quality measure”. The diameter of any given ball off the production line is uncertain, although it is “likely” to be between 9.91mm and 10.13mm.
Example 1-2: A color TV tube is designed to have an operational life of 10,000 hours. After-sale data shows that 3% of the tubes were burnt out within the first 1000 hours; 6.5% were burnt out within 2000 hours.
Here, the TV tube is an engineering product and the operating life in service is one “quality measure”. For any one given TV tube off the production line, it’s operating life (the time of failure during
Chapter - I Introduction I - 3
Note the difference in notation between f(x) and F(x) and their meanings; f(x) is termed the “ probability density function ” while F(x) is termed the “ cumulative distribution function ”. We shall discuss these functions and their mathematical relationships in Chapter II.
Probability of Failure. If the quality of a product X is measured by it’s time of failure during
service, then the value of X is in terms of time, t. Let the range of t be 0< t<∝; and the associated
probability for X=t is f(t).
Discussion: In Example 1-2, the operating life of the TV tube may be described by the probability function f(t), such as shown below:
Here, the TV tube is designed for a life of 10000 hours of operation; the chance for a given tube to last 10000 hours is better than any other t-values. Based on the sample data, there is a 3% of failure up to t=t*= 1000 hours; thus, we have
The above is indicated graphically by the shaded area under the f(x) curve in the interval 0 ≤ t ≤ t*.
Here, note the relation between f(t) and F(t).
Product Reliability. Let the quality of a product be measured by the time-to-failure probability
density function, f(t); the probability of failure up to t = t* is given by the cumulative distribution
function, F(t). Then, the probability for “ non failure ” before t = t is:
F(X > t) = 1- F(t) = ∫ f(t)dt
t*
The term F(X>t*), which is associated with the probability density function f(t), represents the
probability of survival , or known as the reliability function. A precise definition of the latter
will be fully discussed in Chapter IV.
Discussion: In Example 1-2, the data shows that 3% failed before t ≤ 1000 hrs and 6.5% before
t ≤ 2000 hrs. Hence, in terms of the reliability function, we write:
R(1000) = ∫^ f(t)dt = 0.
R(2000) = ∫ f(t)dt = 0.
t, hrs.
(design target)
f(t)
t*
f(t*)
F ( X ≤ t * ) = F (t* ) = ∫ f(t )d t
0
t *
Chapter - I Introduction I - 4
If we want to know the service life t* for which no more than 5% failure (or 95% or better reliability), we can determine t* from the following relation:
R(t*) = ∫^ f(t)dt = 0.
t*
Clearly, one attempts to answer all the questions regarding product reliability; and this can be
done when the mathematical form of f(t) is known. In fact, the knowledge of f(t) or the pursuit of
it, is one of the central elements in the study of engineering reliability.
I-2. Probability, Statistics and Random Variables.
In the preceding section, we used terms such as “random variable”, “probability” of occurrence
(say, X=x), “sample data”, etc. which form the basic notions of engineering reliability. How these
notions, which are sometimes abstractive in nature, all fit together in a real-world setting is a
complicated matter, especially for the beginners in the field.
As in nature, the occurrence of some engineered event is often imperfect; indeed, it may seem to
occur in random; but when it is observed over a large sample or over a long period of time, there
may appear a definitive “mechanism” which causes the event to occur. If the mechanism is exactly
known, the probability for the event to occur can be inferred exactly; if the mechanism is not
known at all, sampling of a set of relevant data (observations) can provide a statistical base from
which at least the nature of the mechanism may become more evident. The latter is, of course,
keenly dependent on the details of data sampling and on how the sample is analyzed.
Example 1-3. The number obtained by rolling a die is a random variable, X. In this case, we know all the possible values of X (the integers from 1 to 6) and the exact mechanism that causes a number to occur. Thus, the associated probability density function f(x) is determined exactly as:
f(x)=1/6, for x = 1, 2,.. 6.
Note, for example, the probability that the number from a throw is less than 3 is given by
F(x<3) = f(1) + f(2) = 1/6+1/6 = 2/6 = 1/3.
Similarly, the probability that the number from a throw is greater than 3 is given by
F(x>3) = f(4) + f(5) + f(6) = 3/6 =1/
The probability that the number from a throw is any number x ≤ 6 is given by
F(x ≤ 6) = f(1)+f(2)+.. F(6) = 1.
Note: In this example, X is known as a discrete random variable, since all the possible values of X are distinct; and the number of all the values is finite. The distribution of f(x) is said to be uniform since f(x)=1/6, for all x = 1, 2,.. 6.
Example 1-4. Now, let us pretend that we do not know any thing about the die. By conducting a sampling test in which the die is rolled N=100 times and each time the integer “x” on the die
Chapter - I Introduction I - 6
i
f(i)
Here, X is also a discrete random variable; but it's probability density function f(i) is not uniform. It is, however, a symmetric function with respect to X=7. For X=7, the theoretical probability is 6/36, the largest among the 11 numbers.
Discussion. Again, if we do a statistical sampling by actually rolling 2 dice N times, we may obtain an estimated f(i) function. In that case, we need a very large sample in order to approach the theoretical f(i) as displayed above.
Central Tendency. In most engineering settings, the random variable X often represents some
measured quantity from identical products; for example, the measured diameters of a lot of
bearing balls off the production line constitute just such a case. In general, the measured values
tend to cluster around the designed value (i.e. the diameter), thus the central tendency. To
evaluate this tendency is clearly important for design or quality control purposes.
The following example illustrates the evaluation of such a tendency:
Example 1-6. A master plumber keeps repair-call records from customers in his service area for 72 consecutive weeks:
71 73 22 27 46 47 36 69 38 36 36 37 79 83 42 43 45 45 55 47 48 60 60 60 49 50 51 75 76 78 31 32 35 85 58 59 38 39 40 40 41 42 42 54 73 53 54 65 66 55 55 56 56 57 49 51 46 54 62 62 54 62 63 64 67 37 58 58 61 62 52 52
Here, let X be the number of repair-calls per week, which seems uniformly random. While one sees that the smallest X value is 22 and the largest is 85, the sample really does not provide a definitive value range for X. Furthermore, since no definitive mechanism(s) could be identified as to how and why the values of X are generated, the true probability distribution of X could never be determined. Hence, instead of looking for the mechanism(s), the sample data can be analyzed in some way to show its central tendency, which may in turn be used to estimate the probability distribution function f(x) for X. Here, we follow a simple procedure as described in the following:
First, we note that there are 72 data values in the sample, roughly within the range from 21 to 90; so we divide this range into 7 equal "intervals" of 10; namely, 21-30, 31-40, 41-50, etc. Second, for each
Chapter - I Introduction I - 7
interval, we count from the sample the number of X values that fall within the interval. For instance, in the first interval (21-30), there are 2 values (22, 27); in the second interval (31-40), there are 13 values (36, 38, 36, 36, 37, 31, 32, 35, 38, 39, 40, 40 and 37); and so on. After all is counted in the 7 intervals, the following result is obtained:
intervals: 21-30 31-40 41-50 51-60 61-70 71-80 81-
X values (^2 13 15 22 11 7 )
in interval:
In this manner, we can already observe that fewer values fall into the lower interval (21-30), or into the upper interval (81-90); but more values fall into the middle intervals, especially in the central interval (51-60). With the above “interval grouping”, we may estimate the probability for X to fall inside the intervals. Instead of treating X, we introduce a new variable I representing the value- intervals; the values of I are the integers 1 to 7, since there are 7 intervals. Thus, the probability density function of I, f(i) can be approximated as follows:
f(1)=2/72 f(2)=13/72 f(3)=15/72 f(4)=22/ f(5)=11/72 f(6)=7/72 f(7)=2/72.
A bar chart for the above is constructed as shown below; it is termed a “histogram” for the sample data considered:
10 20 30 40 50 60 70 80 90 calls
f(i)
fitted f(x)
f(x)
i^1 2 3 4 5 6
The above bar chart displays some important features for the weekly repair-calls. Namely, it suggests that the most probable number of the repair calls occurs at i=4, the 4th value-interval; or 50-60 calls per week. Secondly, the shape of the histogram provides another clue as to the form of the estimated probability distribution function, f(x).
Note that the "fitted" f(x) shown in the figure is just a qualitative illustration; details of sample fitting will be further discussed in Chapter III.
Discussions. We note that the “histogram” obtained above is not unique. For one may take more or fewer intervals value intervals in the range from 21 to 90. In either case, one may obtain a somewhat different histogram for the same sample; often, one may even draw a quite different conclusion for the problem under consideration. This aspect in handling sample data will be examined further in later chapters.
Sampling Errors. As has been seen, most engineering-related events are unlike rolling dice.
Rather, the mechanisms that generate the random variable X are not exactly known. Moreover,
Chapter - I Introduction I - 9
Summary.
As a beginner, it is useful to be conceptually clear about the meaning of some of the key terminologies introduced in this chapter, and to distinguish their differences and interrelations:
- The random variable X is generated to occur as an event, by some mechanisms that are or are not known. When it occurs, X may assume a certain real value, denoted by x; and x can be any one value inside a
particular range; say, 0 ≤ x< ∞.
- In one occurrence, there is a chance (probability) that X=x; that chance is denoted by f(x); here X=x means X equals exactly x.
- Since x denotes any value in the range of X values, f(x) is treated mathematically a distribution function over the value range. Thus, f(x) is the mathematical representation of X; f(x) has several properties and these will be discussed in Chapter II.
- In one occurrence, the probability that X ≤ x is denoted by P{x|X ≤ x}, or simply F(x); here, F(x) is the
sum of all f(x) where X ≤ x. Similarly, P{x|X>x} denotes the sum of all f(x) where X>x; sometimes, it is
also denoted by R(x) and/or by 1-F(x).
- In this course, we sometimes mix the use of the symbols between P{x|X ≤ x} and F(x); between P{x|X>x} and R(x) and 1-F(x), etc. This can be a source of confusion at times.
- If the exact mechanism that generates X is known, it is theoretically possible that the exact value-range x, along with the theoretical f(x) is also known; if the mechanism is not known exactly, one can only rely upon statistical samples along with a proper statistical analysis methodology in order to determine f(x).
- A certain quality of an engineered product can be treated as a random variable (X); x is then the measure of that quality in one such product picked in random. Often than not, the exact mechanisms which generate the product quality (X) are not completely known; hence, sampling of the quality and statistical analysis of samples become essential in determining the associate f(x) function for X.
- Engineering reliability is a special case where the random variable X represents time-to-failure, such as service life-time of a light bulb; for obvious reasons, it is necessary to obtain the time-to-failure probability f(t) for, say, the light bulb.
Assigned Homework.
1.1 Let the random variable X be defined as the product of the two numbers when 2 dice are rolled.
- List all possible values of X by this mechanism;
- Determine the theoretical probability function, f(x);
- Sketch a bar-chart for f(x), similar to Example I-5;
- Compute F(25), explain the meaning of F(25);
- Compute R(15), explain the meaning of R(15);
- Show that the sum of all possible value of f(x) equals to one.
[Partial answer: there are 18 values for X; f(6)=1/9; f(25)=1/36]
1.2 A coin-bag contains 3 pennies, 2 nickels and 3 dimes. If 3 coins are to be taken from the bag each time, their sum is then a random variable: X.
Chapter - I Introduction I - 10
- List all the possible values of X by this mechanism;
- Determine the associated probability distribution f(x);
- Plot the distribution in a graphical bar-chart;
- Show that the sum of all possible value of f(x) equals to one.
[Partial answer: there 56 combinations in drawing “three coins”; but only 9 difference values; $0.20 and 0.25 are among them]
1.3 (Optional; for extra effort) Let the random variable X be the sum of the three numbers when 3 dice are rolled.
- Complete the theoretical probability distribution f(x) for X;
- Show your results in a bar-chart;
- Comment on the form of the bar-charts obtained by rolling 1, 2 and 3 dice, respectively.
[There are 216 possible outcomes in rolling 3 dice; they provide only 16 values, from 3 to 18; f(10)=27/216; f(13)=21/216; one die gives a uniform f(x); 2 dice yield a bi-linear f(x);... ]
1.4 In Example 1-6, the exact mechanism that generates repair calls (X) is not known; but the sample provided can be used to gain some insight into the probability distribution function f(x).
- Now, by using the class-interval = 6 calls instead of 10 calls, Re-do a histogram for the sample;
- Discuss the difference between your histogram and the one obtain in Example 1-5.
1.5 (Optional; for extra effort) The Boeing 777 is designed for a mean service life of 20 years in normal use. Let the service life distribution be given by f(t), where t is in years; and the form of f(t) looks like the one shown in Example 1-2.
- Sketch f(t) as a function of t (years); and locate the design life (20 years) on the t-axis;
- If a B-777 has been in serve for 10 years already, what is the chance that the craft is still fit to fly until for another 6 years?
[This is a case of "conditional" probability]
Chapter- II Fundamentals II - 2
Example 2-1: In tossing a coin, the head will or will not appear; we know that the probability for the head to appear is P{X}= p =1/2, and that for the head not to appear is P{X’}= q =1- p =1/2.
Similarly, in rolling a die, let X be the event that the number “1” occurs. Here, we also know that P{X} = p =1/6, and the probability that “1” will not occur is P{X’} = q = 1- p = 5/6.
In the above, we know the exact mechanisms that generate the occurrence of the respective random variable X. In most engineering situations, one can determine P{X} or p from test samples instead.
Example 2-2. In a QC (quality control) test of 500 computer chips, 14 chips fail the test. Here, we let X be the event that a chip fails the QC test; and from the QC result, we estimate using (2.1):
P{X} = p ≅ n/N = 14/500 = 0.028.
Within the condition of the QC test, we say that the computer chip has a probability of failure p = 0.028, or a survivability of q = 0.972.
Discussion. In theory, (2.1) is true only when N → ∝. The p= 0.028 value obtained above is based on
a sample of size N=500 only. Hence, it is only an estimate and we do not know how good the estimate is. There is a way to evaluate the goodness of the estimate; and this will be discussed in Chapter III.
Combination of Two Events. Suppose that two different events X and Y can possibly occur in
one situation; if the respective probabilities are P{X} and P{Y}. The following defined:
P{X ∩ Y} = the probability that both X and Y occur; and
P{X ∪ Y} = the probability that either X, or Y, or both occur.
X ∩ Y is termed the intersection of X and Y; X ∪ Y is termed the union of X and Y; a graphical
representation of these two cases is shown by means of the Venn diagrams:
X Y X Y
X ∩Y X^ ∪Y
In each diagram, the outline square area is 1x1, representing the total probability; the circles X
Chapter- II Fundamentals II - 3
and Y represent the probabilities of the respective events to occur. The shaded area on the left is
X ∩ Y, in which both X and Y occur; the shaded area on the right is X ∪ Y, in which either X or
Y or both occur. The union is mathematically expressible as:
P{X ∪ Y} = P{X}+ P{Y} - P{X ∩ Y} (2.4)
which can be inferred from the Venn diagram.
Note that the blank area outside the circles in each case represents the probabilities of “non
event", that is neither X nor Y will occur, P{X ∪ Y}'= 1 - P{X ∪ Y}.
Independent Events. If the occurrence of X does not depend on the occurrence of Y, or vice
versa, X and Y are said to be mutually independent. Then,
P{X ∩ Y} = P{X}∗ P{Y} (2.5)
Expression (2.5) is an axiom of probability and it cannot be shown on Venn diagram.
Example 2-3. In rolling two dice, let the occurrence of #1 in the first die be X and that in the second is Y. In this case, occurrence of Y does not depend on that of X; and we know P{X}=P{Y}=1/6. It follows from (2.5) and (2.4), respectively, that
P{X ∩ Y}= #1 appears in both dice = (1/6)(1/6) = 1/36.
P{X ∪ Y}= #1 appears in either or both dice = 1/6+1/6-1/36=11/36.
Discussion: The fact that P{X ∪ Y}=11/36 can also be found as follows: there are in all 11 possible
combinations in which #1 will appear in either or both dice - (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (3,1), (4,1), (5,1), (6,1), while there are a total of 36 possible outcomes.
Conditional Probability. If the occurrence of X depends on the occurrence of Y, or vice versa,
then X and Y are said to be mutually dependent. We define
P{X/Y} = probability of X occurrence, given the occurrence of Y;
P{Y/X} = probability of Y occurrence, given the occurrence of X.
It follows the axiom (2.5) that
P{X ∩ Y}= P{X/Y}∗ P{Y}
= P{Y/X}∗ P{X} (2.6)
Example 2-4. Inside a bag, there are 2 red and 3 black balls. The probability for drawing a red ball out is: P(X}=2/5 and that to draw another red ball from the rest of the balls in the bag is: P{Y/X}=1/4. Thus, to draw both red balls consecutively, the probability is:
P{X ∩ Y}=P{Y/X} ∗ P{X}=(1/4)(2/5)=1/10.
Example 2-5. An electrical system is protected by two circuit breakers that are arranged in-series. When an electrical surge passes through, at least one breaker must break in order to protect the system; if both do not break, the system would be damaged.
In a QC test of the breakers individually, the probability for the breaker not to break is P{X}=0.02;
Chapter- II Fundamentals II - 5
P{X 1 ∩ X 2 ∩ X 3... ∩ XN} = P{X 1 }∗ P{X 2 }∗ P{X 3 }∗ ∗ ∗ P{XN} (2.9)
Note that (2.9) represents the probability that all N events occur.
Similarly, X' 1 , X' 2 , X' 3 ,.. X'N are the respective non-events; and their intersection is:
P{X' 1 ∩ X' 2 ∩ X' 3... ∩ X'N}=P{X' 1 }∗P{X' 2 }∗P{X' 3 }∗ ∗ ∗ P{X'N} (2.10)
And the above is the probability that none of the N events occur.
As for the union of the N-events P{X 1 ∪ X 2 ∪ X 3... ∪ XN}, it represents the probability that
one or more or all of the N events occur. Since P{one or more or more events occur}+ P{none
occurs} = 1, we can write:
P{X 1 ∪ X 2 ∪ X 3... ∪ XN}+[P{X' 1 }∗ P{X' 2 }∗ P{X' 3 }∗ ∗ P{X'N}]= 1 (2.11)
Note that (2.11) is the total probability of all possible outcomes; thus, it is a unity.
The terms P{X'i}, i=1,2,...N in (2.11) can be replaced by
P{X'i} = 1 - P{Xi}; i=1,2,..N (2.12)
Hence, the union of all the N-events can be expressed in the following alternate form:
P{X 1 ∪ X 2 ∪ X 3... ∪ XN}=1-[1-P{X 1 }] ∗ [1-P{X 2 }] ∗ ∗ ∗ [1−P{XN}] (2.13)
A Special case: if P{Xi} = p for all i=1,N, then the intersect in (2.9) becomes:
P{X 1 ∩ X 2 ∩ X 3... ∩ XN} = p N;
And the union in (2.13) becomes:
P{X 1 ∪ X 2 ∪ X 3... ∪ XN} = 1- (1- p )N
The example below illustrates such a special case.
Example 2-6: A structural splice consists of two panels connected by 28 rivets. QC finds that 18 out of 100 splices have at least one defective rivet. If we assume defective rivets occur independently and the probability of being defective is p, what can we say about the quality of the rivets?
Here, let Xi, i=1,28 be the event that the ith rivet is found defective in one randomly chosen splice; and P{Xi}= p. The probability for one or more or all rivets to be found defective in one splice is the
union of all Xi, i=1,28: P{X 1 ∪ X 2 ∪ X 3... ∪ X 28 }.
But, QC finds the probability of a splice to have at least one defective rivet is 18/100; hence,
P{X 1 ∪ X 2 ∪ X 3... ∪ X 28 } = 1-[1- p ]^28 = 0.18.
Chapter- II Fundamentals II - 6
Solving, we obtain p = 0.0071. We can say that about 7 out of 1000 rivets may be found defective
Discussion. In this example, QC rejection rate of the splice (0.18) is given but the probability of being defective of a single rivet is not. By using the definitions of intersect and union of multiple events, we can estimate the probability of being defective for a single rivet, p.
Conversely, if p is given, we can use the same relations to estimate the rejection rate of the splice.
II-2 The Binomial Distribution.
A special case of N-events leads to the so-called “ binomial distribution ”; it is also referred to as
the “ Bernoulli Trials ”. Specifically, if the events X 1 , X 2 ,.. XN are independent as well as
statistically identical :
P{Xn}= p and P{X'n} =1- p = q ; for n = 1,2,...N.
Then, we can always write:
P{Xn}+ P{X'n} = p + q = 1 for n=1,2,...N (2.14)
It follows that
( q + p )N^ = 1 (2.15)
Note that (2.15) is a binomial of power N; upon expansion, we have
CNo q N+CN 1 q N-1 p +CN 2 q N-2 p^2 +.. + CNi q N-i p i +.. + CNN p N^ = 1 (2.16)
where
CNi = (N!) / [(N-i)! i!] i = 0, 1, 2,... N. (2.17)
It turns out that each term in the binomial expansion (2.16) has a distinct physical meaning:
CNo q N^ = q N^ is the probability that none of the N-events ever occur: f(i =0);
CN 1 q N-1 p is the probability that one of the N-events occur: f(i =1);
CNN p N^ = p N^ is the probability that all of the N-events occur: f(i =N); and
In particular,
CNi q N-i p i are the probabilities that i out of N events occur: f(i).
The expression
Chapter- II Fundamentals II - 8
no no yes no = pq^3 no no no yes = pq^3 f(2): yes yes no no = p^2 q^2 yes no yes no = p^2 q^2 yes no no yes = p^2 q^2 no yes yes no = p^2 q^2 no yes no yes = p^2 q^2 no no yes yes = p^2 q^2 f(3): yes yes yes no = p^3 q yes yes no yes = p^3 q yes no yes yes = p^3 q no yes yes yes = p^3 q f(4): yes yes yes yes = p^4
The total probability of all outcomes is thus:
f(0)+f(1)+f(2)+f(3)+f(4) = q^4 + 4 q^3 p +6 q^2 p^2 +4 q p^3 + p^4 = ( q + p )^4 = 1
Again, it follows the binomial distribution.
Discussion: For engineering products, a system or a single component may be under repeated and statistically identical demands. Say, in each demand, the failure probability is p =1/6 while that for non-failure is q =5/6. Then, the probability that failure of the system (or component) occurs at least once in 4 repeated demands is given by:
f(1)+f(2)+f(3)+f(4) = 1- f(0) = 1 - q^4 = 1 - (5/6)^4 = 51.8%
The above result can be obtained by applying (2.18) to (2.20) directly.
The Pascal Triangle. The coefficients in the binomial expansion of power N in (2.17) can be
easily represented by a geometrical construction, known as the Pascal Triangle, if N is not very
large:
1 N=
1 1 N=
1 2 1 N=
1 3 3 1 N=
1 4 6 4 1 N=
1 5 10 10 5 1 N=
1 6 15 20 15 6 1 N=
1 7 21 35 35 21 7 1 N=
1 8 28 56 70 56 28 8 1 N=
Example 2-9. Suppose the probability of a light bulb being burnt out is p whenever the switch is turned on. In a hallway, 8 light bulbs are controlled by one switch. Compute f(0), f(1),.. , f(8) when the switch is turned on.
Chapter- II Fundamentals II - 9
Using the Pascal Triangle, for N=8, we can quickly write:
f(0) = q^8 ; f(1) = 8 q^7 p ; f(2) = 28 6 p^2 ; f(3) = 56 q^5 p^3 ; f(4) = 70 q^4 p^4 ; f(5) = 56 q^3 p^5 ; f(6) = 28 q^2 p^6 ; f(7) = 8 q p^7 ; f(8) = p^8.
The Poisson Distribution. While the Pascal triangle becomes cumbersome to use when N
is large, say N>20, there is a simple expression for (2.18) when N is large and p is small (say
N>20 and p <<1):
f(i) = (N p )i exp[-N p ]/(i!) i = 0,1,2,.. N (2.21)
Expression (2.21) is known as the Poisson Distribution. It is an approximation of the binomial
distribution (2.18) when N is large and p small.
Example 2-10: Suppose that the probability for a compressor to pass a QC test is q =0.9. If 10 compressors are put through the QC test, compute the various probabilities of failure: f(i), i=0,1,..
In this case, N=10 and p =0.1; the binomial distribution (2.18) and the simplified Poisson distribution (2.21) will be used; the following are the respective results:
f(0) f(1) f(2) f(3) f(4) f(5) f(6), f(7), f(8), f(9), f(10)
by (2.18) 0.349 0.387 0.194 0.057 0.011 0.0015 ≈ 0
by (2.21) 0.368 0.368 0.184 0.061 0.015 0.0031 ≈ 0
Discussion: The exact binomial distribution (2.18) gives a maximum at f(0)=0.349 and f(1)=0.387; the Poisson approximation (2.21) gives f(0)=f(1)=0.368. For the rest, the two distributions are rather close.
The Poisson approximation would yield better results if N is larger or p smaller. In this example, the N (=10) value is not large enough and p (=0.1) is not small enough, thus the difference.