Transcript Probability (Kebarangkalian) - USM :: Universiti Sains

Probability
(Kebarangkalian)
EBB 341
Probability and Statistics
population
sample
random
variables
X 1 , X 2 ,..., X n
Sampling
techniques
Parameters
,  2
before observation
after observation
x1 , x2 ,..., xn
data
inference
Statistical
procedure
probability
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.
Example
for Theorem 3:

No.
Conform
ing
No. Noncon
forming
Total
X
50
3
53
125
6
131
75
2
77
250
11
261
What is the probability of
selecting a random part Y
produced by supplier X
or by supplier Z?
Z
P(X or Z)
= P(X) + P(Z)
= 53/261 + 77/261
= 0.498

Supp
lier
What is the probability of
selecting a
nonconforming part from
supplier X or a
conforming part from
supplier Z?
TOTAL
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.
•

If A and B are independent events, then the
probability that both A and B will occur is
•

P(A) + P(B) + … + P(N) = 1.0 (Theorem 5).
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:

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:
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 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
!
n
n = 3, and r = 3
P 
r
n
r
P
(n  r )!
= 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?
5!
5.4.3.2.1
P 

= 60
(5  3)!
2.1
5
3

In the licence plate example, suppose the witness
further remembers that the numerals were not the
same
10!
10.9.8...1
P 

 90
(10  2)! 8.7...1
10
2
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:
n!
C 
r!(n  r )!
n
r
where
n
r
C = 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):
D
d
N D
nd
N
n
C C
P(d ) 
C
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







N
n
C
= Combinations of all units
CdD
= combinations of nonconforming units
N D
n d
= combinations of conforming units
C
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? Lot
Sample
nonconforming
unit

N = 9, D = 3, n = 4 and d = 1
conforming unit
D
d
N D
nd
N
n
C C
P(d ) 
C
3!
6!
C13C4913 1!(3  1)! 3!(6  3)!
P(1) 

 0.476
9
9!
C4
4!(9  4)!




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:
n(n  1) n  2 2
n
( p  q)  p  np q 
p q  ... q
2
n
n
n 1






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
n!
d nd
P( d ) 
po qo
d!(n  d )!





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
5!
1
51
P(1) 
(0.1) (0.9)  0.328
1!(5  1)!
5!
0
5
P(0) 
(0.1) (0.9)  0.590
0!(5  0)!
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
c
(npo ) npo
P (c ) 
e
c!



where c=count or number
npo=average count, or average number
e=2.718281
Poisson

Suppose that average count of cars that
arrive a toll booth in a 1-min interval is 2,
then calculations are:
(2) 0 2
P(0) 
e  0.135
0!
(2) 3 2
P(3) 
e  0.180
3!
(2)6 2
P(6) 
e  0.012
6!
(2)1 2
(2) 2 2
P(1) 
e  0.271 P(2) 
e  0.271
1!
2!
(2) 4 2
(2) 5 2
P(4) 
e  0.090 P(5) 
e  0.036
4!
5!
(2) 7 2
P (7 ) 
e  0.003
7!
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.
…….