module 4 - fundamentals of probability

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.