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Probability(Kebarangkalian)

- EBB 341

Probability and Statistics

random variables

probability

population

sample

Parameters

Sampling techniques

before observation

after observation

data

statistics

inference

Statistical procedure

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

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 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

- 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 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.

Examplefor 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

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

75/261 0.299

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 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.

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.

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

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

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

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)

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

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.

Formula

- The formula for combination
- where
- number combinations of n object taken r at a

time.

Discrete Probability Distributions

- Typical discrete probability distributions
- Hypergeometric
- Binomial
- Poisson

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.

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)

- 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

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

- 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

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)

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

- 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 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

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

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

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) .

Poisson

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

Poisson

- Suppose that average count of cars that arrive a

toll booth in a 1-min interval is 2, then

calculations are

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. - .

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