ProbabilityProbability

(Kebarangkalian)(Kebarangkalian)

EBB 341

Probability and StatisticsProbability and Statistics

Parameters

population

2,

sample

before observation

after observation

nXXX ,...,, 21

nxxx ,...,, 21data

random variables

Statistical procedure

inference

probability

statistics

Sampling techniques

Definition of probabilityDefinition of probability Probability:

• Likelihood-kemungkinan• Chance-peluang• Tendency-kecenderungan• Trend-gaya/arah aliran

P(A) = NA/N• P(A) = probability of event A• NA = number of successful outcomes of event A

• N = total number of possible outcomes

Example:Example: A part is selected at random from container

of 50 parts that are known have 10 noncomforming units. The part is returned to container. After 90 trials, 16 noncomforming unit were recorded. What is the probability based on known outcomes and on experimental outcomes?

Known outcomes:• P(A) = NA/N = 10/50 = 0.200

Experimental outcomes:• P(A) = NA/N = 16/90 = 0.178

Probability theoremsProbability theorems Probability is expressed as a number

between 0 and 1. (Theorem 1) If P(A) is the probability that an event will

occur, then the probability the event will not occur is 1.0 - P(A) or

P(A) = 1.0 – P(A). (Theorem 2)

P(A) = probability of not event A

Example for Theorem 2:Example for Theorem 2: If probability of finding and error on an

income tax return is 0.04, what is the probability of finding an error-free or conforming return?

P(A) = 1.0 – P(A) = 1.0 -0.04 = 0.96

Probability theoremsProbability theorems For mutually exclusive events, the probability that

either event A or B will occur is the sum of their respective probabilities• P(A or B) = P(A) + P(B). (Theorem 3).

When events A and B are not mutually exclusive events, the probability that either event A or event B will occur is • P(A or B or both) = P(A) + P(B) - P(both).

(Theorem 4). “mutually exclusive” means that accurrence of

one event makes the other event impossible.

ExampleExamplefor Theorem 3:for Theorem 3:

What is the probability of selecting a random part produced by supplier X or by supplier Z?P(X or Z) = P(X) + P(Z)= 53/261 + 77/261 = 0.498

What is the probability of selecting a nonconforming part from supplier X or a conforming part from supplier Z?

Supplier

No. Conforming

No. Nonconforming

Total

X 50 3 53

Y 125 6 131

Z 75 2 77

TOTAL 250 11 261

P(nc X or co Z)

= P(nc X) + P(co Z)

=3/261 + 75/261

= 0.299

Example for Theorem 4:Example for Theorem 4: What is the probability that a randomly

selected part will be from supplier X or a nonconforming unit?

P(X or nc or both)

= P(X) + P(nc) –P(X and nc)

= (53/261) + (11/261) – (3/261)

= 0.234

Probability theoremsProbability theorems Sum of the probabilities of the events of a

situation equals 1.• P(A) + P(B) + … + P(N) = 1.0 (Theorem 5).

If A and B are independent events, then the probability that both A and B will occur is• P(A and B) = P(A) x P(B) (Theorem 6).

If A and B are dependent events, the probability that both A and B will occur is • P(A and B) = P(A) x P(B|A) (Theorem 7).

P(B|A): probability of event B provided that even A has accurred.

Example for Theorem 6 & 7:Example for Theorem 6 & 7: What is probability that 2 randomly selected parts will be

from X and Y? Assume that the first part is returned to the box before the second part is selected (called with replacement).

P(X and Y) = P(X) x P(Y)= (53/261) x (131/261) = 0.102

Assume that the first part was not returned to the box before the second part is selected. What is the probability?P(X and Y) = P(X) x P(Y|X) = (53/261) x (131/260) = 0.102

Since 1st part was not returned, the was a total of 260.

Example:Example:Theorem 7 What is the probability of choosing both parts from Z?

P(Z and Z) = P(Z) x P(Z|Z) = (77/261) (76/260) = 0.086

Theorems 3 and 6-to solve many problems it is necessary to use several theorems:

What is the probability that 2 randomly selected parts (with replacement) will have one conforming from X and one conforming part from Y or Z?P[co X and (co Y or co Z) = P(co X) [P(co Y) + P(co Z)]= (50/261) [(125/261) + (75/261)] = 0.147

Counting of eventsCounting of events Many probability problems, such as

those where the evens are uniform probability distribution, can be solved using counting techniques.

There are 3 counting techniques: Simple multiplication Permutations Combinations

Simple multiplicationSimple multiplication If event A can happen in any a ways and, after it

has occurred, another event B can happen in b ways, the number of ways that both event can happen is ab.

Example.: A witness to a hit and run accident remembered the first 3 digits of the licence plate out of 5 and noted the fact that the last 2 were numerals. How many owners of automobiles would the police have to investigate?• ab = (10)(10) = 100

If last 2 were letters, how many would need to be investigate?• ab = (26)(26) = 676

PermutationsPermutations A permutation is the number of arrangements

that n objects can have when r of them are used.

For example: The permutations of the word “cup” are cup,

cpu, upc, ucp, puc & pcu n = 3, and r = 3

= number of permutations of n objects taken r of them (the symbol is sometimes written as nPr)

n! is read “n factorial” = n(n-1)(n-2) …(1)

)!(

!

rn

nPnr

nrP

Example:Example: How many permutations are there of 5 objects

taken 3 at a time?

= 60

In the licence plate example, suppose the witness further remembers that the numerals were not the same

1.2

1.2.3.4.5

)!35(

!553

P

901...7.8

1...8.9.10

)!210(

!10102

P

CombinationsCombinations If the way the objects are ordered in

unimportant. “cup” has 6 permutations when 3 objects

are taken 3 at a time. There is only one combinations, since the

same 3 letters are in different order.

FormulaFormula The formula for combination:

where

= number combinations of n object taken r at a time.

)!(!

!

rnr

nC nr

nrC

Discrete Probability Discrete Probability DistributionsDistributions

Typical discrete probability distributions: Hypergeometric Binomial Poisson

Discrete probability Discrete probability distributionsdistributions

Hypergeometric - random samples from small lot sizes.• Population must be finite• Samples must be taken randomly without

replacement Binomial - categorizes “success” and

“failure” trials Poisson - quantifies the count of discrete

events.

HypergeometricHypergeometric Occurs when the population is finite and

random sample taken without replacement The formula is constructed of 3 combinations

(total combinations, nonconforming combinations, and conforming combinations):

Nn

DNdn

Dd

C

CCdP

)(

P(d) = prob of d nonconforming units in a sample of size n. N = number of units in the lot (population) n = number of unit in the sample. D = number nonconforming in the lot d = number nonconforming in the sample N-D = number of conforming units in the lot n-d = number of conforming units in the sample

= Combinations of all units

= combinations of nonconforming units

= combinations of conforming units

NnC

DdC

DNdnC

ExampleExample A lot of 9 thermostats located in a container has 3

nonconforming units. What is probability of drawing one nonconforming unit in a random sample of 4?

N = 9, D = 3, n = 4 and d = 1

LotSample

nonconformingunit

conforming unit

Similarly, P(0) = 0.119, P(2) = 0.357, P(3) = 0.048. P(4) is impossible- only 3 nc units. The sum probability: P(T) = P(0) + P(1) + P(2) + P(3)

= 0.119 + 0.476 + 0.357 + 0.048 = 1.000

476.0

)!49(!4

!9)!36(!3

!6

)!13(!1

!3

)1( 94

3914

31

C

CCP

Nn

DNdn

Dd

C

CCdP

)(

““or less” & “or more” probabilityor less” & “or more” probability Some solutions require an “or less” or “or

more” probability.

P(2 or less) = P(2) + P(1) + P(0)

P(2 or more) = P(T) – P(1 or less)

= P(2) + P(3) + …

BinomialBinomial This is applicable to the infinite number of items or

have steady stream of items coming from a work center.

The binomial is applied to problem that have attributes, such as conforming or nonconforming, success or failure, pass or fail.

Binomial expansion:

nnnnn qqpnn

qnppqp

...2

)1()( 221

p = prob. an event such as a nonconform q = 1-p = prob. nonevent such as conform n = number of trials or the sample size

Since p = q, the distribution is symmetrical regardless of the value of n.

When p q, the distribution is asymmetrical.

In quality work p is the proportion or fraction nonconforming and usually less than 0.15.

Binomial for single termBinomial for single term

P(d)= prob. of d nonconforming n = number of sample d = number nonconforming in sample po = proportion(fraction) nc in the population qo = proportion(fraction) conforming (1-po) in the

population

dno

do qpdnd

ndP

)!(!

!)(

ExampleExample A random sample of 5 hinges is selected from

steady stream of product, and proportion nc is 0.10.

What is the probability of 1 nc in the sample? What is probability of 1 or less? What is probability of 2 or more?

qo = 1-po = 1.00 - 0.10 = 0.90

328.0)9.0()1.0()!15(!1

!5)1( 151

P

590.0)9.0()1.0()!05(!0

!5)0( 50

P

P(1 or less) = P(0) + P(1) = 0.918

P(2 or more) = P(2) + P(3) + P(3) + P(4)

= P(T) – P(1 or less) = 0.082

PoissonPoisson The distribution is applicable to

situations:• that involve observations per unit time

(eg. count of car arriving at toll in 1 min interval).

• That involve observations per unit amount (eg. count nonconformities in 1000 m2 of cloth) .

PoissonPoisson Formula for Poisson distribution

where c=count or number npo=average count, or average number e=2.718281

onpc

o ec

npcP

!

)()(

PoissonPoisson Suppose that average count of cars that

arrive a toll booth in a 1-min interval is 2, then calculations are:

003.0!7

)2()7(012.0

!6

6)2()6(

036.0!5

)2()5(090.0

!4

)2()4(180.0

!3

)2()3(

271.0!2

)2()2(271.0

!1

)2()1(135.0

!0

)2()0(

27

2

25

24

23

22

21

20

ePeP

ePePeP

ePePeP

PoissonPoisson The probability of zero cars in any 1-min interval is

0.135. The probability of one cars in any 1-min interval is

0.271. The probability of two cars in any 1-min interval is

0.271. The probability of three cars in any 1-min interval is

0.180. The probability of four cars in any 1-min interval is

0.0.090. The probability of five cars in any 1-min interval is

0.036. …….