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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.
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.
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.
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 n1 ways, followed
by the second n2 ways, and so on until the k th can be performed in
nk ways, then the operation consisting of k steps can be performed by
k n1 × n2 × ... × nk = ∏ ni
i =1 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 = nr .
r 2 times 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,
0000000000.....00
r "0"s and n−1 ""’s The total number of ways is
r +n−1
r
4 = (r + n − 1)!
.
r !(n − 1)! 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 n1 objects of type 1, n2 objects of type 2, ..., nr objects of
type r . (e.g. number of diﬀerent letter arrangements can be formed
using the letters STATISTICS.) Total number of ways is
n!
.
n1 ! × n2 ! × · · · × nr !
Partition of n distinct objects into r distinct groups with speciﬁed
size n1 , n2 , · · · , nr . (e.g. number of ways in dividing a class of 40 into
groups of 10, 10 and 20.) Toal number of ways is
n
n1 , n2 , · · · , nr = n!
.
n1 ! × n2 ! × · · · × nr ! Review: Combinatorial Analysis (cont’d) IV
Partition of n indistinguishable objects into r distinct groups (i.e.
we only concern the number of objects in each group). The problem
is the same as arranging the “0” and “” in, say,
0000000000.....00
n "0"s and r −1 ""’s Total number of ways is
n+r −1
n = (n + r − 1)!
.
n!(r − 1)! Problem 1 Problem 1. A product code of manufacturer consists of 2 letters
followed by 3 digits. The manufacturer only uses the letters A, B, C
and D. For example, AA032 and BD369 are valid product codes. If a
product is selected at random, what is the probability that
1
2
3
4 it begins with the letter A?
the ﬁrst two letters are identical?
the letters are identical and the digits are equal?
the letters are formed by B and D, and the 3digit number is even? Solution to Problem 1 Solution.
1 2
3 4 Since the four letters are equally likely to be the ﬁrst letter,
P(beginning with A) = 1 .
4
4
P(two letters are identical) = 4×4 = 1 .
4
1
P(the digits are equal) = 10×10×10 = 100 .
10
1
1
P(letters are identical and digits are equal) = 100 × 1 = 400 .
4
P(the letters are formed by B and D) = P({B,D} or {D,B}) =
2
1
4×4 = 8 .
5
P(the 3digit number is even) = 10 = 1 .
2
P(the letters are formed by B and D, and the 3digit number is even)
1
1
1
= 8 × 2 = 16 . Problem 2 Problem 2. Three boys and four girls line up randomly in a row.
What is the probability that
1
2
3
4 all boys are on one side?
boys and girls are in alternate positions?
no 2 boys sit next to each other?
the boy named John sits on the right of the girl named Mary (assuming
all boys and girls have diﬀerent names)? Solution to Problem 2 Solution.
1
2
3 4 2
7
=.
3
35
1
P(boys and girls are in alternate positions) = 35 .
The number of ways of arranging in this way is 5 = 10 ( Set the girls
3
as GGGG, put the three boys in to the interspace of the girls). Thus
P(no two boys sit next to each other) = 10 = 2
35
7
6 × 5! 1
P(John sits on the right of Mary)=
=.
7!
7
P(all boys on one side)= 2/ Problem 3
Problem 3. There are 4 red balls, 3 yellow balls, 2 blue balls and 1
white balls. In how many distinct ways can the balls be arranged in a
circle?
Solution.
First we consider the case of arranging the balls in a row. It is a
problem of arrangement of n objects (10 balls), with r
distinguishable types (4 colors).
10
10!
=
4! × 3! × 2! × 1!
4, 3, 2, 1 = 12600 We then understand that for each arrangement of balls in a circle, we
can derive 10 distinct arrangements in a row, each by choosing a
diﬀerent ball to be the ﬁrst in a row. The number of ways the balls
can be arranged in a row is 10 times of that they can be arranged in a
circle. Thus, the answer is:
12600
10 = 1260 Problem 4 Problem 4. (Exam Fall 2008) Three candidates, A, B and C,
participate in an election in which eight voters will cast their votes.
The candidate who receives the absolute majority, that is at least ﬁve,
of the votes will win the election. All voters must vote for one and
only one candidate and cannot abstain.
Suppose the voting is done by a secret ballot such that the voter’s
identities are kept conﬁdential.
1 2 3 Show that there are 45 distinct ways to distribute the eight votes
among the three candidates.
Show that there are 10 distinct ways to distribute the eight votes
among the three candidates so that A will win the election.
How many distinct ways are there to distribute the eight votes among
the three candidates so that nobody will win the election? Solution to Problem 4 I
Solution.
1 This is a problem of partitioning n indistinguishable objects (8 votes)
into r distinguishable groups (3 candidates).
8+3−1
8 =
= 2 10!
8! × 2!
45 As we have ﬁxed the result to be A winning the election, we ﬁrst
allocate 5 votes to A and consider the number of ways to distribute the
remaining 3 votes. The problem becomes a partition of n
indistinguishable objects (3 votes) into r distinguishable groups (3
candidates).
3+3−1
3 =
= 5!
3! × 2!
10 Solution to Problem 4 II 3 We can simply deduct the number of ways to distribute the votes for
candidate A, B and C to win respectively from the total number of
ways to distribute the eight votes. This is based on the fact that the
events {A winning}, {B winning}, {C winning} and {nobody winning}
are mutually exclusive and exhaustive.
45 − 10 − 10 − 10 = 15 Problem 5 Problem 5. A total of r identical robots are being lifted from the
ground ﬂoor to the n upper ﬂoors of a building.
1
2 Find the total number of distinct patterns.
A student claims that if each robot is independently and equally
probably allocated to each of the n ﬂoors, then the probability that all
robots end up on the ﬁrst ﬂoor is given by
Number of ways of sending all robots to 1/F
=
Number of distinct exit patterns
Comment on the above statement. Is it correct? n+r −1
r −1 . Solution to Problem 5
Solution.
1 This is a problem of allocating r identical objects into n distinct
containers. Thus the total number of ways is
r +n−1
r 2 . The statement is wrong because, if each robot is independently and
equally probably allocated to each of the n ﬂoors, the distinct patterns
are not equally likely (Why? For example, if there are 3 ﬂoors and 2
robots, what is the probability of all robots going to the 1st ﬂoor?
What is the probability of one robot going to the 1st ﬂoor and the
other robot going to the 2nd ﬂoor?). The correct probability should be
1
.
nr Problem 6 Problem 6. In an experiment k balls are drawn randomly without
replacement from an urn containing 1 red ball, 2 blue balls and 17
yellow balls. All the balls are identical apart from their colours.
1 Suppose 0 ≤ k ≤ 17. Show how you may obtain the number of distinct
combinations of the k balls by considering the identity
(1 + x )(1 + x + x 2 )(1 − x )−1 ≡ (1 − x 2 )(1 − x 3 )(1 − x )−3 . 2 Show that there are six distinct ways of drawing three balls randomly
without replacement from the urn. Solution to Problem 6 Solution.
1 The number of ways of drawing k balls from the urn is equal to the
coeﬃcient of x k in the expression (why? Note that
(1 + x )(1 + x + x 2 )(1 − x )−1 = (1 + x )(1 + x + x 2 )(1 + x + x 2 + x 3 + ...). 2 The number of ways of drawing k balls from the urn is equal to the
coeﬃcient of x k in the above expression.)
Using the expression on the RHS, we get
(1 − x 2 )(1 − x 3 )(1 − x )−3 = (1 − x 2 − x 3 + x 5 )(1 + 3x + 6x 2 + 10x 3 + ...).
From above, we see that the coeﬃcient of x 3 is −1 − 3 + 10 = 6. Problem 7 Problem 7. (Class test Fall 2008) A lady has three rings and eight
distinguishable ﬁngers (excluding the two thumbs).
1 Suppose the three rings are identical.
1 2 3 2 How many distinct ways are there to wear the three rings on the same
ﬁnger?
How many distinct ways are there to wear two rings on the same ﬁnger
and the third ring on a diﬀerent ﬁnger?
How many distinct ways are there to wear the three rings on three
separate ﬁngers? Now suppose the three rings are distinguishable. Using part (a) or
otherwise, ﬁnd the total number of distinct ways to wear the three
rings on the lady’s eight ﬁngers without any restriction. Solution to Problem 7 Solution.
1 Since the rings are identical,
1 2 3 2 Number of distinct ways to wear the three rings on the same ﬁnger =
number of distinct ﬁngers = 8.
First we choose a ﬁnger to wear 2 rings, then we choose another ﬁnger
to wear 1 ring. Thus the number of ways = 8 × 7 = 56.
The number of ways to wear the three rings on three separate ﬁngers is
8
= 56.
3 Since the rings are distinguishable, the total number of distinct ways is
8 × 3! + 56 × 3! + 56 × 3! = 720. Problem 8 Problem 8. Mary has 5 dresses, 3 skirts, 4 blouses, 3 pairs of shoes,
and 2 hats. She always wears shoes and either a dress or a blouse and
a skirt. She may or may not wear a hat.
1
2 3 How many diﬀerent combinations can she wear?
Suppose Mary can aﬀord to buy either an extra dress or an extra hat
(but not both). Which should she buy if she decides to maximize the
number of diﬀerent combinations that she can wear?
Suppose Mary’s brown shoes do not match her pink or blue dress, and
that the blue hat does not match her yellow blouse. How many
matching combinations can she wear? Solution to Problem 8
Solution.
1
2 3 Total number of combinations = 3 × (2 + 1) × (5 + 3 × 4) = 153.
Extra dress: total number of combinations
= 3 × (2 + 1) × (6 + 3 × 4) = 162.
Extra hat: total number of combinations
= 3 × (3 + 1) × (5 + 3 × 4) = 204.
Therefore she should buy an extra hat.
Total number of combinations with brown shoes and pink or blue dress
= 1 × 3 × 2 = 6.
Total number of combinations with blue hat and yellow blouse
= 3 × 1 × (1 × 3) = 9.
Since the two events “brown shoes and pink or blue dress” and “blue
hat and yellow blouse” are mutually exclusive, there is no double count
in the above two countings. Therefore the matching combinations she
can wear is
153 − 6 − 9 = 138. Problem 9 Problem 9 (a more diﬃcult question). Suppose there are N boys and
M girls sitting around a table, and the chairs are all diﬀerent.
1
2 How many diﬀerent ways to sit them?
How many ways to sit them such that there are k boys with a girl
sitting to his right (assume M ≥ k )? Solution to Problem 9 I
Solution
1
2 (M + N )!.
We can count the total number of combinations using the following
steps:
1 2 3
4 Suppose there is a boy called John. Line up the boys with John in
front. The number of ways to line up the N − 1 boys is (N − 1)!.
Choose k boys from the N boys and mark them. The number of ways
N
is
.
k
Line up the M girls. The number of ways is M !.
Divide the M girls (which are already all lined up) into k groups
M −1
without disturbing the order. The number of ways is
(why?
k −1
Consider for example there are 4 girls lined up in a row G1 G2 G3 G4 , how
many ways to separate them into 3 groups without disturbing the
order?). Put the ﬁrst group on the right of the ﬁrst marked boy, and
put the 2nd group on the right of the 2nd marked boy, etc. Solution to Problem 9 II 5 Choose a sit for John. People queuing behind John occupy seats to
John’s right one by one anticlockwise. There are M + N ways.
Total number of ways
= the product of the number of ways in (i), (ii), (iii), (iv), (v)
=(N − 1)!
=N !M ! N
k
N −1
k −1 M! M −1
k −1
M −1
k −1 (M + N )
M +N
k ...
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This note was uploaded on 01/16/2012 for the course STAT 1301 taught by Professor Smslee during the Fall '08 term at HKU.
 Fall '08
 SMSLee
 Statistics, Probability

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