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Additional Important Properties of Sets
__
A
∪A =S
__
__
S = Φ,
Φ= S
__ ___
__
( A)=A
De Morgan’s Theorems:
(A ∪ B) = A
1)
(A
2)
B
∩ B) = A ∪ B
Cardinality of A, |A|, is defined as the number of elements in the set A.
Example:
|{a,b,c}| =3, |{ }| =1, while | | = 0,
| R | = , where R = set of all real numbers.
Sequence of Sets
Consider sequence A1, A2, ..., An.
^A = {x | x ∈ A for i = 1,2,……n} = A
……. ∩A = {x | x ∈A
Similarly, _A = A ∩A
We define
i
i
i
1
Example:
1∪
2
n
1
A 2 ∪ ….∪An
and A2 and …..An}
Let Bm = {1, 2, ..., m}. Then,
Bi ⊂ Bk .......i < k
B1 = {1}, B2 = {1,2}, B3 {1, 2,3}, etc
Partitions: A sequence A1, A2,
i) A i
∩A j = φ
*A
n
ii)
An is said to be a partition of the sample space, or universal set, S, if
1
i ≠j
=S
i =1
Example:
Let S = real numbers
1) (-4, 5], (5, 4) is a partition
2) ., A-2, A-1, A0, A1, A2, ... is an infinite partition, Am =( m, m+1] = {m ≤ x < m+1}
Example: Twenty electric motors are pulled from an assembly line and inspected for defects.
Eleven are free of defects, eight have defects on the exterior finish, and three have
defects in their assembly and will not run. Write symbolic notations for:
a)
b)
c)
d)
Set of motors with both types of defects;
The set of motors with at least one type of defect;
The set of motors with no defects;
The set of motors with exactly one type of defect.
Let A be the set of motors with assembly defects and let B be the set with finish defects.
Solution:
a) The motors with both defect types must be in A and B. So this event is A 1 B.
Since only nine motors have defects, and A contains three motors, B contains eight, then
there are 11 defects, and two must be in A 1 B.
b) The motors having at least one type must be in A or B, hence, A c B
because eleven have no defect, nine must have at least one type defect.
__
c) Eleven have no defects, e.g.
__
__ _ _ __
AB = A*B
_
_
d) The sets are those elements in A and not in B, or B and not in A, ˆ (A1B)
From the Venn diagram we see that 7 motors fall into this event.
c (B1A).
Example :
Given three arbitrary events, E, F and G. Find expressions for the following:
a) Only G occurs – G 1 E c 1 F c
b) At least one event occurs – E c G c F
c) At most one event occurs – (Ec1 Gc 1 Fc) c(E1 Fc1 Gc) c (Ec1 F1 Gc)c (Ec 1 Fc1G)
d) At most two events occur – (E F G) c
e) All three events occur - EFG
Example:
Determine whether the following data are consistent. In a group of 1000 people, there are
a) 306 people who can swim
b) 335 people over 30 years of age
c) 480 people who jog
d)
43 swimmers who jog
e)
f)
g)
h)
145 joggers over 30 years of age
79 swimmers over 30 years of age
22 swimmers and joggers over 30 years of age
125 non-swimmers and non joggers under 30 years of age
Solution:
There is a total of 1000, and there are 3 groups;
A = 306 swimmers
B = 335 over 30 years of age
C = 480 joggers
We know that (A c B c C c)=(A+B+C) c =125. Also, (A+ B+ C) + (A+B+Cc) =1000
So!! A + B + C = 875
We also know the Number in (A + B + C) = #A + #B + #C -#AB - #AC - +#ABC
= 306 + 335+480-43-145-79+22
= 876
Hence, the data is not consistent!!
So! The concept of random experiments can be illustrated as follows. Consider a group of Random
Experiments;
Random Experiments:
1)
2)
3)
4)
5)
6)
7)
Tossing a coin once.
Tossing a coin three times.
Observing the life span of a person.
The number of busy trunk lines in a telephone exchange.
Random noise in an electrical communication system.
Noting if a component is functioning properly.
Execution time of a program.
The possible results of the random experiment are called outcomes:
1)
2)
3)
4)
5)
6)
7)
Define:
Heads, Tails.
(H,H,H), (H,H,T), (H,T,H), etc.
72.5 years, 46.3 years.
7, 10, 15.
2.2 volts, 7.1 volts.
Properly functioning, improperly functioning.
5 minutes, 70 seconds.
The totality of the possible outcomes of the experiment is called the sample space of the
experiment and it will be denoted by S.
We characterize the space S for each experiment as:
1)
S = {H,T}
2)
S = {(H,H,H), (H,H,T), (H,T,H), ...}
3)
S={X}X>0}
4)
S = {0, 1, 2, 3, ..., N}
S = 2
S = 23 = 8
|S| = ∞
S = N + 1
5)
6)
7)
S = S = 2
S = R = real numbers
S = {properly, improperly}
S = {X|X>0}
|S| = ∞
We now can define a Sample Space, S, as a mutually exclusive, collectively exhaustive, listing of all
possible outcomes of a random experiment. Subsets of S that are mutually exclusive and collectively
exhaustive are called event spaces. A collection of events can be mutually exclusive, collectively
exhaustive, both, or neither.
Example.
Four red balls and four black balls labeled one through four are placed into two urns,
respectively. An experiment consists of drawing one red and one black ball. Let Rn and Bn
be the events, “the red and black balls have value n, n = 1, 2, 3, 4.” For each of the
following, determine whether the list of events is M.E., C.E. (collectively exhaustive), a
sample space or an event space.
a). List of events (R 1 B1;)(R1 B2 )(B2, )(B4, )(R2)
#
#
#
From the diagram we see the events are not mutual exclusive and not completely exhaustive.
Hence are not a sample space nor an event space.
b) The events (R1 B1), (R1 B2), (R1) c, (B1 ª B2)c , (B1 «N2c), R4
These events are completely exhaustive, but not mutually exclusive. Hence not a sample space
nor event space.
.
We are ready now to define a Probability Space. Let there be a sample space, S, with a giveN
field, ℑ=P(s). Then the pair (S,ℑ) is called a probability pair.
Assume that a number, or probability measure can be assigned to the events in the sample space
S. We write P(A) for any AŒF. We write P(s) for s Œ S.
Then, a Probability Space exists. P is said to be a probability measure for the probability pair,
(S, ℑ, P).
Combinations & Permutations
Combinatorics
Let (S, ℑ, P) be a probability space. Suppose S is finite but large and that
P (s) =
1
S
The problem is said to be combinatorial. It follows that for any event A.
P (s) =
A
S
To calculate P(A) it suffices to obtain A and S.
Counting Methods
Pairs: With m elements a1, a2, ..., am and n elements b1, ..., bn it is possible to form mn
ordered pair (ai, bJ).
Example:
Let one set of elements be the four suits of cards, the other set of elements be the thirteen
face values. There are therefore 4 x 13 = 52 combinations, or cards.
Multiplets: Given n1 elements a 1, a2, ..., an1 and n2 elements b1, b2, ..., bn2, etc., up to nr
elements x1, ..., xnr, it is possible to form n1, n2, ... nr ordered r-tuplets , (aj1, bj2, ..., xjr).
Thus, r successive selections (decisions) with exactly nk choices possible at the kth step can
produce a total of n1, n2, ..., nk different outcomes.
Examples
1) People classified according to sex, marital status, and profession. If there are seventeen
professions, then we have 2 x 2 x 17 = 68 classes in all.
2) Placing balls in a cell amounts to choosing a cell for each ball. With r distinguishable
balls we have r independent choices, and therefore r balls can be placed into n cells in nr
different ways.
3) r tosses of a die. The number of possible outcomes is S = 6 r. Let A = “no ace turns
up.” Then A = 5 r. Thus
()
5 r
6
Ordered Samples
Consider a set or “population” of n elements a 1, a2, ..., an. Any ordered arrangement of r elements is
called an ordered sample, of size r drawn from our population.
Two procedures:
i)
ii)
Sampling with replacement
Sampling without replacement
With replacement, the number of possible outcomes is S = nr.
Without replacement, we have n possible choices for the first element, n-1 for the second, n-2 for the
third, etc., so that there are S = n (n - 1)(n - 2) ... (n - r + 1) = (n) r choices in all. If r = n, the sample
represents a permutation of all the elements: S = n!
Example
In sampling, the probability for any fixed element to be included is equal to one minus the
probability that the element is not included. Let A - “element not included.”
Without replacement:
|A|=(n-1) r
|S|=(n) r
∴
P( A ) =1 −
( n − 1) r
(n)r
With replacement, we have S = nr, A = (n - 1)r so that
 n - 1
P (A ) = 1 - 

 n 
r
Example: Consider random sample of size r with replacement. Let A = “no element appears
twice in the sample. AS.
|S| = n r
|A|=(n)
P (A ) =
(n ) r
nr
Illustration:
• Let population consist of ten digits 0, 1, ..., 9 and consider five samples. Let A =
“all digits are different.”
P (A ) =
(1 0 ) 5
= 0 .3 0 2 4
105
• If n distinguishable balls are randomly placed into n cells, the probability that
each cell will be occupied equals n!/nn.
• The birthdays of r people form sample size r from a population of size 365.
If we let A = “ the birthdays are all different”, then;
P (A ) =
| A | (3 6 5 ) r 1
=
≤ fo r all r > 2 3
|S|
365r
2
As a side note, to evaluate factorials numerically, we use Stirling’s Formula:
n! ≅ 2π nn+1/ 2e − n
Then,
P ( A) =
(365)r =
r
365
r = 23
365!
1
=
*
(342 )! 36523
365!
1
*
and, when we let
(365 − r )! 365 r
Using Stirling’s Formula: n! ≅
2π (365)
e −365
1
* −342 *
342 .5
e
(365)23
2π (342 )
365.5
=
=
(365)342.5 * e − 23 = 0.36997978 = 0.4927
0.75091883
(342)342.5
2π n
n+ 1
2 −n
e