Example_class_1_slides - Example class 1 STAT1301...

Example class 1
STAT1301 Probability and Statistics I
Chan Chi Ho, Chan Tsz Hin & Shi Yun
September 14, 2011

Information about tutors
Tutors: Chan Chi Ho (MW515), Chan Tsz Hin (MW518) and Shi Yun
(MW518)
Emails: [email protected], [email protected], [email protected],
respectively.
Assignment: Put assignments in assignment box outside MW515.
I
Deadline of Assignment 1: Sep. 26th

Information about tutors
Tutors: Chan Chi Ho (MW515), Chan Tsz Hin (MW518) and Shi Yun
(MW518)
Emails: [email protected], [email protected], [email protected],
respectively.
Assignment: Put assignments in assignment box outside MW515.
I
Deadline of Assignment 1: Sep. 26th

Information about tutors
Tutors: Chan Chi Ho (MW515), Chan Tsz Hin (MW518) and Shi Yun
(MW518)
Emails: [email protected], [email protected], [email protected],
respectively.
Assignment: Put assignments in assignment box outside MW515.
I
Deadline of Assignment 1: Sep. 26th

Review: Combinatorial Analysis
Multiplication Principle: If an operation consists of a sequence of k
separate steps of which the first can be performed in n
1
ways, followed
by the second n
2
ways, and so on until the k th can be performed in
n
k
ways, then the operation consisting of k steps can be performed by
n
1
× n
2
× ... × n
k
=
k
∏
i = 1
n
i
ways.

Review: Combinatorial Analysis (cont’d) I
Selection of r from n distinct objects:
1
Ordered selection with replacement: Consider drawing r balls from n
distinct balls. The total number of ways is
n × n × ... × n
| {z }
r
times
= n
r
.
2
Ordered selection without replacement: Consider drawing r balls from
n distinct balls without replacement. The total number of ways is
n × ( n - 1 ) × ( n - 2 ) × ... × ( n - r + 1 ) =
n !
( n - r )!
.

Review: Combinatorial Analysis (cont’d) II
3
Unordered selection with replacement: Consider drawing r balls from n
distinct balls with replacement and without regard to order. This is the
same as placing r identical objects into n distinct containers (why?).
The problem is the same as arranging the “0” and “|” in, say,
00 | 000 || 00 | 0 ||| 00 ..... | 0 || 0
| {z }
r
"0"s and
n - 1
"|"’s
The total number of ways is
r + n - 1
r
=
( r + n - 1 )!
r !( n - 1 )!
.
4
Unordered selection without replacement: Consider drawing r balls
from n distinct balls without replacement and without regard to order.
The total number of ways is
n
r
=
n !
r !( n - r )!
.

Review: Combinatorial Analysis (cont’d) III
Arrangement of n objects, with r distinct types. In other words,
there are n
1
objects of type 1, n
2
objects of type 2, ..., n
r
objects of
type r . (e.g. number of different letter arrangements can be formed
using the letters STATISTICS.) Total number of ways is
n !
n
1
! × n
2
! ×···× n
r
!
.
Partition of n distinct objects into r distinct groups with specified
size n
1
, n
2
, ··· , n
r
. (e.g. number of ways in dividing a class of 40 into
groups of 10, 10 and 20.) Toal number of ways is
n
n
1
, n
2
, ··· , n
r
=
n !
n
1
! × n
2
! ×···× n
r
!
.

Review: Combinatorial Analysis (cont’d) IV
Partition of n indistinguishable objects into r distinct groups (i.e.