Module 4
Fundamentals of Probability
Module 4
Fundamentals of Probability
OutlineOutline Definition of Probability Theorems of Probability Counting of Events Discrete Probability Distributions Continuous Probability Distribution
When you have completed this chapter you should be able to: Define probability using the frequency definition. Know the seven basic theorems of probability. Identify the various discrete and continuous
probability distributions.
Learning ObjectivesLearning Objectives
• Calculate the probability of non-conforming units occurring using the Hyper-geometric,
Binomial and Poisson distributions.• Know when to use the Hyper-geometric, Binomial and Poisson distributions.
Likelihood, chance, tendency, and trend. The chance that something will happen. Examples:
1. If a Nickel is tossed, the probability of a head is ½ and the probability of the tail is ½.
2.When a die is tossed on the table, the probability of one spot is 1/6, the probability of two spots is 1/6,.....3.We are drawing a card from a deck of cards. The probability of a spade is 13/52.
Definition of ProbabilityDefinition of Probability
The area of each distribution is equal to 1.
The are under the normal distribution curve, which is
a probability distribution, is equal to 1.
The total probability of any situation will be equal to 1.
The probability is expressed as a decimal (the
probability of a head is 0.5).
An event is a collection of outcomes (six-sided die
has six possible outcomes).
When the number of outcomes is known or when the number of outcomes is found by experimentation:
P(A) = NA/N
where:P(A) = probability of an event A occurring to 3 decimal places
NA=number of successful outcomes of event A N= total number of possible outcomes
Example:
A part is selected at random from a container
of 50 parts that are known to have 10 non-
conforming units. The part is returned to the
container and a record of the number of trials
and the number non-conforming is
maintained. After 90 trials, 16 non-conforming
units were recorded. What is the probability
based on known outcomes and on
experimental outcomes.
Solution
For known outcomes
P(A) = NA/N =10/50 = 0.200
For experimental outcomes
P(A) = NA / N = 16 /90 = 0.178
The probability calculated using known
outcomes is the true probability, and the
one calculated using experimental
outcomes is different due to the chance
factor.
For an infinite situation (N), the definition
would always lead to a probability of zero.
In the infinite situation the probability of
an event occurring is proportional to the
population distribution. .
Theorem 1
Probability is expressed as a number
between 1 and 0, where a value of 1 is a
certainty that an event will occur and a value
of 0 is a certainty that an event will not occur.
Theorems of ProbabilityTheorems of Probability
Theorem 2
If P(A) is the probability that event A will
occur, then the probability that A will
not occur is:
P(not A) = 1- P(A)
Example
If the probability of finding an error on
an quality inspection form is 0.04, what
is the probability of finding an error free
or conforming form.
Answer
i.e. 1- 0.040 = 0.960
One EventOut or Two
or More Events
MutuallyExclusive
Theorem 3
Not MutuallyExclusive
Theorem 4
Two or More EventOut or Two
or More Events
Independent
Theorem 6
Dependent
Theorem 7
When to use Theorems 3,4,6 and 7
Mutually exclusive means that the occurrence of one event makes the other event impossible
Theorem 3
If A and B are two mutually exclusive events (the
occurrence of one event makes the other event
impossible), then the probability that either
event A or event B will occur is the sum of their
respective probabilities:
P(A or B) = P(A) +P(B)
This is the “additive law of probability”.
Supplier Number
conforming
Number non-
conforming
TOTAL
X 50 3 53
Y 125 6 131
Z 75 2 77
TOTAL 250 11 261
ExampleTable below shows inspection results of components from 3 suppliers X, Y and Z.
Calculate the probability of selecting a random part produced by supplier X or by supplier Z.
Answer P(X or Z) = P(X) + P(Z)
498.0261
77
261
53
Theorem 4
If event A and event B are not mutually
exclusive, then the probability of either
event A or event B or both is given by:
P(A or B or both) = P(A) +P(B) – P(both)
Events that are not mutually exclusive have
some outcomes in common
Example
Using previous table, calculate probability that
a randomly selected part will be from supplier
X or a non-conforming unit.
Answer: P(X) + P(nc) – P(X and nc)
234.0261
3
261
11
261
53
Theorem 5
The sum of the probabilities of the
events of a situation is equal to 1.000
P(A) + P(B) + …..+ P(N) = 1.000
Theorem 6
If A and B are independent events (one where
its occurrence has no influence on the
probability of the other event or events), then
the probability of both A and B occurring is the
product of their respective probabilities:
P(A and B) = P(A) X P(B)
Example
Use previous table to calculate the probability
that 2 randomly selected parts will be from
supplier X and supplier Y, assuming that the
first part is returned to the box before the
second part is selected.
Answer: P(X and Y) = P(X) x P(Y)
102.0261
131
261
53
Theorem 7
If A and B are dependent events, the
probability of both A and B occurring is the
probability of A and the probability that if A
occurred, then B will occur also:
P(A and B) = P(A) X P(B\A)
P(B\A) is defined as the probability of event
B, provided that event A has occurred.
Example
Use previous table to calculate the probability
that 2 randomly selected parts will be from
supplier X and supplier Y, assuming that the first
part was from supplier X and was not returned
to the box before the second part was selected.
Answer: P(X and Y) = P(X) x P(Y/X)
102.0260
131
261
53
1.Simple multiplication
If an event A can happen in any of a ways
or outcomes and, after it has occurred,
another event B can happened in b ways
or outcomes, the number of ways that
both events can happen is ab.
Counting of EventsCounting of Events
2. Permutations A permutation is an ordered arrangement of a set of
objects.
Where n is total number of objectsr = number of objects selected out of the total
numberExample: The word “cup”…… cup, cpu, upc, ucp, puc,
and pcu. . i.e. 3 objects in a set arranged in groups of 3n=3 r=3 to give 6 permutations
!
( )!nr
nP
n r
3.Combinations If the way the objects are ordered is
unimportant, then we have a combination:
!
!( )!nr
nC
r n r
Example: The word “cup” has 6 permutations when the 3 objects are taken 3 at a time. There is only one combination, since the same three letters are in different order. n=3 and r= 3 to give C =1
Hyper-geometric Probability Distribution1. Occurs when the population is finite
and the random sample is taken without replacement.
2. The formula is constructed of 3 combinations (total, nonconforming, and conforming):
( )D N Dd n d
Nn
C CP d
C
Discrete Probability Distributions
where P(d) = probability of d nonconforming units in a sample of size nNCn = combinations of all unitsDCd = combinations of nonconforming unitsN-DCn-d = combinations of conforming unitsN = number of units in the lot (population)n = number of units in the sampleD = number of nonconforming units in the lotd = number of nonconforming units in the sampleN – D = number of conforming units in the lotn – d = number of conforming units in the sample
Example
A random sample of 4 insurance claims is
selected from a lot of 12 that has 3 non-
conforming units. Using the hyper-geometric
distribution, calculate the probability that the
sample will contain exactly;
a) 0 non-conforming units
b) 3 non-conforming units
Nn
DNdn
Dd
C
CCdP
)(
255.0)0(124
94
30
124
31204
30
C
CC
C
CCP
018.0)3(124
91
33
124
31234
33
C
CC
C
CCP
a) 0 non-conforming units is;
where N = 9, n = 4, D = 3 and d = 0
b)3 non-conforming units where N = 9, n = 4, D = 3 and d = 3 is;
Example
Suppose we randomly select 5 cards
without replacement from an ordinary
deck of playing cards. What is the
probability of getting exactly 2 red cards
(i.e., hearts or diamonds)?
Solution: This is a hyper-geometric experiment in which we know the following: N = 52; since there are 52 cards in a deck. D = 26; since there are 26 red cards in a deck. n = 5; since we randomly select 5 cards from the deck. d = 2; since 2 of the cards we select are red. We plug these values into the hyper-geometric formula as follows:[ DCd ] [ N-DCn-d ] / [ NCn ] = [ 26C2 ] [ 26C3 ] / [ 52C5 ] = [ 325 ] [ 2600 ] / [ 2,598,960 ] = 0.32513Thus, the probability of randomly selecting 2 red cards is 0.32513.
Binomial Probability Distribution
1. It is applicable to discrete probability problems
that have an infinite number of items or that
have a steady stream of items coming from a
work center.
2. It is applied to problems that have attributes such
as conforming and nonconforming.
3. It is applicable provided the two possible
outcomes are constant and the trials are
independent
1 2 2( 1)( ) .........
2n n n n nn n
p q p np q p q q
It corresponds to successive terms in the binomial expansion, which is
Where p = probability of an event such as conforming unitq = 1 – p = probability of a non-event such as a conforming unit.
Distribution of the number of tails for an infinite number of tosses of 11 coins
3. For graph behind, since p=q, the distribution
is symmetrical regardless of the value of n,
however, when p is not equal to q, the
distribution is asymmetrical.
4. In quality work p is the portion or fraction
nonconforming and is usually less than 0.15
5. In most case in quality work, we are not
interested in the entire distribution, only in
one or two terms of the binomial expansion.
0 0!
( )!( )!
d n dnP d p q
d n d
The binomial formula for a single term is
where P(d) = proportion of d nonconforming unitsn = number in the sampled = number conforming in the samplepo =proportion (fraction) nonconforming in the
populationqo = proportion conforming (1 –po) in the population
5. As the sample size gets larger, the shape of the curve will become symmetrical even though p is not equal to q.
6. It requires that there be two and only two possible outcomes (C, NC) and that the probability of each outcome does not change.
7. The use of the binomial requires that the trials be independent.
8. It can be approximated by the Poisson when Po≤0.10 and nPo≤5.
9. The normal curve is an excellent approximation when Po is close to 0.5 and n/N>̳� 0.10
Example
Using binomial distribution, find the probability
of obtaining 2 or less non-conforming
components in a sample of 9 when the lot is
15% non-conforming.
dn
odo qpdnd
ndP
!!
!
232.085.015.0!09!0
!90 90
P
368.085.015.0!19!1
!91 81
P
260.085.015.0!29!2
!92 72
P
859.0260.0368.0232.0210 PPP
Poisson Probability Distribution
1. It is applicable to many situations that
involve observations per unit of time.
2. It is also applicable to situations involving
observations per unit amount.
3. In each of the preceding situations, there
are many equal opportunities for the
occurrence of an event.
4. The Poisson is applicable when n is quite large and Po is small.
5. When Poisson is used as an approximation to the binomial, the symbol c has the same meaning as d has in the binomial and hyper-geometric formulas..
00( )( )
!
cnpnp
P c ec
6. When nPo gets larger, the distribution approaches symmetry.
7. The Poisson probability is the basis for attribute control charts and for acceptance sampling.
9. It is used in other industrial situations, such as accident frequencies, computer simulation, operations research, and work sampling.
10.Uniform (generate a random number table), Geometric, and Negative binomial (reliability studies for discrete data).
11.The Poisson can be easily calculated
using Table C.
12.Similarity among the hyper-geometric,
binomial, and Poisson distributions
can exist.
Normal Probability Distribution
1. When we have measurable data.
2. The normal curve is a continuous probability
distribution.
3. Under certain condition the normal
probability distribution will approximate the
binomial probability distribution.
Continuous Probability DistributionsContinuous Probability Distributions
4. The Exponential probability distribution is used in reliability studies when there is a constant failure rate.
5. The Weibull distribution is used when the time to failure is not constant.