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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. …….