Probability (Kebarangkalian)

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Probability(Kebarangkalian)

• EBB 341

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Probability and Statistics
random variables
probability
population
sample
Parameters
Sampling techniques
before observation
after observation
data
statistics
inference
Statistical procedure
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Definition 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

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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

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Probability 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

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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

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Probability 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.

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Examplefor Theorem 3
Supp lier No. Conforming No. Noncon forming Total
X 50 3 53
Y 125 6 131
Z 75 2 77
TOTAL 250 11 261

• 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?

P(nc X or co Z) P(nc X) P(co Z) 3/261
75/261 0.299
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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

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Probability 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(BA) (Theorem 7).
• P(BA) probability of event B provided that even
  A has accurred.

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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(YX) (53/261) x
  (131/260) 0.102
• Since 1st part was not returned, the was a total
  of 260.

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Example

• Theorem 7
• What is the probability of choosing both parts
  from Z?
• P(Z and Z) P(Z) x P(ZZ) (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?
• Pco X and (co Y or co Z) P(co X) P(co Y)
  P(co Z)
• (50/261) (125/261) (75/261) 0.147

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Counting 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

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Simple 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

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Permutations

• 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)

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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

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Combinations

• 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.

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Formula

• The formula for combination
• where
• number combinations of n object taken r at a
  time.

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Discrete Probability Distributions

• Typical discrete probability distributions
• Hypergeometric
• Binomial
• Poisson

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Discrete probability distributions

• 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.

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Hypergeometric

• 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)

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• 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

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Example

• 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

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• 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

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or 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)

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Binomial

• 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

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• 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.

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Binomial 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

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Example

• 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

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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
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Poisson

• 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) .

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Poisson

• Formula for Poisson distribution
• where ccount or number
• npoaverage count, or average number
• e2.718281

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Poisson

• Suppose that average count of cars that arrive a
  toll booth in a 1-min interval is 2, then
  calculations are

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Poisson

• 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.
• .