probability

2. 2.1.1 Random Experiment
Definition
An experiment that can result in different
outcomes, even though it is repeated in the same
manner every time, is called a random experiment.
2.1.2 Sample Spaces
Definition
The set of all possible outcomes of a random
experiment is called the sample space. This is
often denoted by the symbol S.

3. Two types of sample spaces:
Discrete Continuous
It consists of a finite or
countable infinite set of
outcomes.
It contains an interval
(either finite or infinite) of
real numbers.
Tree Diagrams
Sample spaces can also be described graphically with tree diagrams.
 When a sample space can be constructed in several steps or stages,
we can represent each of the n1 ways of completing the first step as a
branch of a tree.
 Each of the ways of completing the second step can be represented
as n2 branches starting from the ends of the original branches, and so
forth.

4. Example 1
Each message in a digital communication is classified as to
whether it is received within the time specified by the system
design. If three message are classified, use a tree diagram to
represent the sample space of possible outcomes.
Each message can either be received on time or late. The
possible results for three messages can be displayed by eight
branches in the tree diagrams shown in Fig. 2.1.

5. Solution
Figure 2.1 Tree diagram for three messages.

6. 2.1.3 Events
Definition
An event is a subset of the sample space of a
random experiment.
The union of the two events A and B, denoted by the symbol
is the event containing all the outcomes that belong to A
or B or both
Union
,A B

7. Intersection
The intersection of the two events A and B, denoted by the
symbol is the events containing all the outcomes in both
A and B.
,A B
Complement
The complement of an event A with respect to S is the subset
of all outcomes of S that are not in A. We denote the
complement of A by the symbol .A
Mutually Exclusive
Two events A and B are mutually exclusive, if
that is, A and B have no outcomes in common. In particular, A
and A’ are mutually exclusive.
,A B  

8. Venn Diagrams
Figure 2.2 Venn diagrams.

9. Example 2
In light-dependent photosynthesis, light quality refers to the
wavelengths of light that are important. The wavelength of a
sample of photosynthetically active radiations (PAR) is measured
to the nearest nanometer. The red range is 675 – 700 nm and
the blue range is 450 – 500 nm. Let A denote the event that PAR
occurs in the red range and let B denote the event that PAR
occurs in the blue range. Describe the sample space and indicate
each of the following events:
(a) (b) (c) (d) A BA BA B

10. Solution
Let w ~ the wavelength of photosynthetically active radians (PAR)
The sample space
A ~ PAR occurs in the red range
B ~ PAR occurs in the blue range
(a)
(b)
(c)
(d)
 0,1,2,w w 
 675,676,...,700nmA w w 
 450,451,...,500nmB w w 
 450,451,...,500,675,676,...,700nmA B w 
A B  

11. 2.1.4 Counting Techniques
Multiplication Rules
If an operation can be described as a sequence of k steps, and
if the number of ways of completing step 1 is n1, and
if the number of ways of completing step 2 is n2 for each way of
completing step 1, and
if the number of ways of completing step 3 is n3 for each way of
completing step 2, and so forth,
the total number of ways of completing the operation is
1 2 kn n n  

12. Example 3
In a design of a casing for a gear housing, we can use four
different types of fasteners, three different bolt lengths, and three
different bolt locations. How many different designs are possible?
Solution
From the multiplication rule,
different designs are possible.1 2 3 4 3 3 36n n n     

13. Example 4
Sam is going to assemble a computer by himself. He has the
choice of ordering chips from two brands, a hard drive from four,
memory from three, and an accessory bundle from five local
stores. How many different ways can Sam order the parts?
Solution
Since and there are
different ways to order the
parts.
1 2 3 4 2 4 3 5 120n n n n       
1 2 32, 4, 3,n n n   4 5,n 

14. Permutation
The number of permutation of n different elements is where!n
   ! 1 2 2 1n n n n       
Permutation of Subsets
The number of permutation of subsets of r elements selected from a
set of n different elements is
     
 
!
1 2 1
!
n
r
n
P n n n n r
n r
         


15. Example 5
A printed circuit board has eight different locations in which a
component can be placed. If four different components are to be
placed on the board, how many different designs are possible?
Solution
Each design consists of selecting a location from the eight
locations for the first component, a location from the remaining
seven for the second component, a location from the remaining
six for the third component, and a location from the remaining
five for the fourth component. Therefore,
different designs are possible.8
4
8!
8 7 6 5 1680
4!
P      

16. Permutation of Similar Objects
The number of permutation of objects of which
are of one type, are of a second type, … , and are of rth type is
1 2 3
!
! ! ! !r
n
n n n n
1 2 rn n n n    1n
2n rn

17. Example 6
Consider a machining operation in which a piece of sheet metal
needs two identical-diameter holes drilled and two identical-size
notches cut. We denote a drilling operation as d and a notching
operation as n.
In determining a schedule for a machine shop, we might be
interested in the number of different possible sequences of the
four operations.
The number of possible sequences for two drilling operations and
two notching operations is
4!
6
2!2!


18. Example 7
A part is labeled by printing with four thick lines, three medium
lines, and two thin lines. If each ordering of the nine lines
represents a different label, how many different labels can be
generated by using this scheme?
Solution
The number of possible part labels is
9!
1260
4!3!2!


19. Example 8
In the design of an electromechanical product, seven different
components are to be stacked into a cylindrical casing that holds
12 components in a manner that minimizes the impact of shocks.
One end of the casing is designated as the bottom and the other
end is the top.
(a) How many different designs are possible?
(b) If the seven components are all identical, how many
different designs are possible?
(c) If the seven components consist of three of one type of
component and four of another type, how many different
designs are possible? (more difficult)

20. Solution
(a) Every arrangement of 7 locations selected from the 12 comprises
different designs. Therefore,
designs are available.
(b) Every subset of 7 locations selected from 12 comprises a new design.
Therefore, the number of different designs is:
 
12
7
12!
3,991,680
12 7 !
P  

12!
792
7!5!


21. Solution
(c) Three locations for the first component selected. Therefore,
designs.
Then, four components selected from the nine remaining location:
designs.
Therefore, from the multiplication rule, the number of designs is
12!
220
3!9!

9!
126
4!5!

220 126 27,720 

22. Combinations
The number of combinations, subsets of size r that can be selected
from a set of n elements, is denoted as and
 
!
! !
n
r
n n
C
r r n r
 
  
 
n
rC

23. Example 9
A printed circuit board has eight different locations in which a
component can be placed. If five identical components are to be
placed on the board, how many different designs are possible?
Solution
Each design is a subset of size five from the eight locations that
are to contain the components. The number of possible design is
8!
56
5!3!


24. Example 10
A bin of 50 manufactured of parts contains three defective parts
and 47 nondefective parts. A sample of six parts is selected from
50 parts without replacement. That is, each part can only be
selected once and the sample is a subset of the 50 parts. How
many different samples are there of size six that contain exactly
two defective parts?

25. Solution
A subset containing exactly two defective parts can be formed by first
choosing the two defective parts from the three defective parts. This step
san be completed in
different ways
Then, the second step is to select the remaining four parts from the 47
acceptable parts in the bin. The second step can be completed in
different ways
Therefore, from the multiplication rule, the number of subsets of size six
that contain exactly two defective items is
3 3!
3
2 2!1!
 
  
 
47 47!
178,365
4 4!43!
 
  
 
3 178,365 535,095 