Probability
Sample Space = {(H,1),(H,2),(H,3),(H,4),(T,1),(T,2),(T,3),(T,4)}
Tree Diagram
Chapter 1 Introduction
1
2
4
3
1
2
3
4
1
2
3
4
H
Counting Rules
Theorem: Basic Principle of Counting
In an operation consists of n steps in which the first
step can be done is k1 ways and the second step has
k2 ways and the third has k3, and so forth, then the
whole
h l operation
i can be
b done
d
in
i k1  k2  k3  …  kn
ways.
T
2
1
x
4
(H,1)
(H,2)
(H,3)
(H,4)
(T,1)
(T,2)
(T,3)
(T,4)
2
Tree Diagram
What is the sample space for casting two
(1,1)
dice experiment? Die 2
1
2
Die 1
3
1
4
2
5
6
3
(1,2)
(1,3)
(1,4)
(1 5)
(1,5)
(1,6)
4
5
2 x 4 = 8 possible outcomes (coin & wheel)
6 X 6 = 36 possible outcomes (2 dice)
6
k1  k2
6
3
Example:
6
Permutations
= 36 outcomes
4
B A S E
How many different four-letter code
words can be formed by using the four
letters in the word “BASE” without
repeating use of the same letter?
There are three shirts of different colors,
two jackets of different styles and five
pairs of pants in the closet. How many
ways can you dress yourself with one
shirt, one jacket and one pair of pants
selected from the closet?
k1 = 4
k2 = 3
k3 = 2
k4 = 1
4·3·2·1 = 24
Sol: 3 x 2 x 5 = 30 ways
K1 x k2 x k3
x
5
6
Probability - 1
Probability
Permutations
Theorem
How many ways can 8 people be arranged
to sit at a round dinner table facing a main
entrance?
The number of permutations of n
distinct objects in n!
H
A
B
G
8! = 40320
C
F
D
E
7
Ordered Sampling B A S E
How many different four-letter code words
can be formed by using the four letters
selected from letters A through Z and the
p
y
same letter can not be used repeatedly
(sampling without replacement taking an
ordered sample of size 4)?
8
Theorem: Permutation Rule
The number of possible permutations
of r objects from a collection of n
distinct objects is
k1 = 26 k2 = 25 k3 = 24 k4 = 23
n
26·25·24·23 = 358,800
Pr 
n!
( n  r )!
Order does count!
9
Permutations
n
Pr 
n!
( n  r )!
How many ways can a four-digit code be
formed by selecting 4 distinct digits from
nine digits, 1 through 9, without
repeating use of the same digit?
9P4 =
9!
(9-4)!
=
9!
5!
=
9·8·7·6·5!
10
Circular Permutations
How many circular permutations are there
for 4 persons playing bridge game?
A
D
B
4!/4 = 3! = 6
5!
C
= 9·8·7·6 = 3024
11
12
Probability - 2
Probability
Combination Rule
Theorem: Circular Permutations
The number of possible permutations
of r objects arranged in a circle from
a collection of n distinct objects is
(n – 1) !.
Theorem: Combination Rule
The number of possible combinations of r
objects from a collection of n distinct
objects is
n Cr
Order does count!

n
n!
  
r!(n  r )!  r 
Binomial Coefficient
Order does not count!
13
Combinations
n Cr

n!
r!(n  r )!
Combinations
How many ways can a committee be formed
by selecting 3 people from a group 10
candidates?
n = 10,, r = 3
10!
10!
10·9·8·7!
=
=
10C3 =
3!·(10-3)!
3!·7!
3!·7!
10 ·9 ·8
=
3!
14
n Cr

n!
r!(n  r )!
How many ways can a combination of 4 distinct
digits be selected from nine digits, 1 through 9?
9C4
= 120
=
9!
4!·(9-4)!
=
9·8·7·6
4·3·2·1
=
9!
9·8·7·6·5!
=
4!·5!
4!·5!
= 126
ABC, ACB, BAC, BCA, CAB, CBA are the same combination.
15
Combinations
n Cr

n!
r!(n  r )!
Distinguishable Permutations
How many ways can 6 distinct numbers be
selected from a set of 47 distinct
numbers?
n = 47,
47 r = 6
47C6
16
47!
47·46·45·44·43·42·41!
=
6!·41!
6!·41!
47·46·45·44·43·42
=
6!
How many different 4 letter code words
can be formed by rearranging the 4 letters
in the word, DEED?
4!
4x3x2x1
=
=6
2!2!
2x2
=
= 10,737,573
17
18
Probability - 3
Probability
Distinguishable Permutations
Distinguishable Permutations
Theorem: Let a set of n objects of
two types, r of one type, and n – r of
other type. The number of
distinguishable permutations of these n
objects is nCr .
Example: How many possible outcomes could it
be if a coin is tossed 8 times and getting 2
heads and 6 tails?
nCr
n
8
8!
 28
=   = 8C2 =   
r
 
 2  2! 6!
Theorem: Let a set of n objects of s
types, n1 of one type, and n2 of the 2nd
type, … and, ns the s-th type. The number
of distinguishable permutations of these
n objects is
n


n!

 
n1!n2 ! ... ns !  n1 , n2 ,..., ns 
Multinomial Coefficient
19
Distinguishable Permutations
20
Partitions of Distinct Objects
Theorem: The number of ways in which
a set of n distinct objects can be
partitioned into k subsets with n1 objects
in tthe
e first
st subset, a
and
d n2 objects in tthe
e
2nd subset, … and, nk objects in the k-th
subset is
Example: For Christmas decoration,
three different colors of light bulbs are
used to line up is a row with 3 yellow, 5
green, and 6 red light bulbs. How many
different ways can these light bulbs be
arranged?
n


n!

 
n1!n2 ! ... nk !  n1 , n2 ,..., nk 
 14 
14 !

 
= 168168
3
,
5
,
6
3
!
 5! 6!


21
22
Theorem: For any positive integers
n and r = 0, 1, 2, …, n,
n  n 
   

r  n  r
Theorem:
k

Theorem: For any positive integers
n and r = 0, 1, 2, …, n1,
r 0
 m  n   m  n 
 

  
 r  k  r   k 
 n   n  1  n  1

  
   
 r   r   r  1
23
24
Probability - 4