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Chapter 1. Random Events and Probability
1.1 Introduction The term Probability refers the study of randomness and uncertainty.
In any situation in which one of a number of possible outcomes may occur, the
theory of probability provides methods for quantifying the chances, or likelihoods,
associated with the various outcomes.
。Tossing a properly balanced coin, it will fall with either a head or a tail showing.
。Measuring the diameters of ball bearing produced by a certain company, the
possible outcomes contain all real numbers in a certain interval.
Definition: A probability is a number between 0 and 1
representing how likely it is that an event will occur.
Probabilities can be:
1. Frequentist (based on frequencies),
2. Subjective: probability represents a person’s degree of belief
that an event will occur,
e.g. I think there is an 80% chance it will rain today, written as
P(rain) = 0.80.
1.2 Sample Spaces and Events
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Definition: A random experiment is an experiment whose
outcome is not known until it is observed.
Definition: A sample space, Ω (or S), is the set of all possible
outcomes of a random experiment.
Definition: A sample point is an element of the sample space.
Experiment: Toss a coin twice and observe the result.
Sample space: Ω = HH,HT, TH, TT
An example of a sample point is: HT
Experiment: Toss a coin twice and count the number of heads.
Sample space: Ω = 0, 1, 2
Experiment: Toss a coin twice and observe whether the two
tosses are the same (e.g. HH or TT).
Sample space: Ω = same, different
Definition: An event is any collection (subset) of outcomes
contained in the sample space. Events will be denoted by capital
letters A,B,C,... . Note:We say that event A occurs if the outcome of the experiment is one of the elements in A.
Definition: An event is said to be simple if it consists of exactly
one outcome and compound if it consists of more than one
outcome.
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Example: Toss a coin twice.
Sample space: Ω = HH,HT, TH, TT
Let event A be the event that there is exactly one head.
We write: A =“exactly one head”
Then A = HT, TH.
Note:A is a subset of Ω, as in the definition. We write A ⊂ Ω.
Definition: Event A occurs if we observe an outcome that is a
member of the set A.
Note: Ω is a subset of itself, so Ω is an event. The empty set,
φ = , is also a subset of Ω. This is called the empty set (or
null event), or the event with no outcomes.
Example: Experiment: throw 2 dice.
Sample space: Ω = (1, 1), (1, 2), . . . , (1, 6), (2, 1), (2, 2), . . . ,
(2, 6), . . . , (6, 6)
Event A = “sum of two faces is 5” = (1, 4), (2, 3), (3, 2), (4, 1)
Example: Consider an experiment in which each of three
automobiles taking a particular freeway exit turns Left(L) or
right (R) at the end of the exit ramp.
The eight possible outcomes that comprise the sample space are
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LLL, RLL, LRL, LLR, LRR, RLR, RRL,and RRR.
Thus, there are eight simple events, among which are E1=LLL
and E5=LRR.
Some compound events include
A=RLL, LRL, LLR=the event that exactly one of the three cars turns right
B=LLL, RLL, LRL, LLR=the event that at most one of the cars
turns right
C=LLL,RRR=the event that all three cars turn in the same direction
Suppose that when the experiment is performed, the outcome is
LLL. Then the simple Event E1 has occurred and so also have
the events B and C (but not A).
Example when the number of pumps in use at each of two
six-pump gas stations is observed, there are 49 possible outcomes, so
there are 49 simple events: E1=(0,0) ,E2= (0,1),…
E49=(6,6). Examples of compound events are
A=(0,0), (1,1), (2,2), (3,3), (4,4), (5,5),(6,6)
=the event that the number of pumps in use is the same for both
B=(0,4), (1,3), (2,2), (3,1), (4,0)
=the event that the total number of pumps in use is four
C=(0,0), (0,1), (1,0), (1,1) =the event that at most one pump is in
use at each station
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1.3 Operations of Set Theory
Definition: Let A and B be events on the same sample space Ω:
so A ⊂ Ω and B ⊂ Ω.
Definition: The complement of event A is written cA (or A ),
and is given by
Experiment: Pick a person in this class at random.
Sample space: Ω = all people in class.
Let event A =“person is male” and
event B =“person travelled by bike today”.
Suppose I pick a male who did not travel by bike. Say whether
the following events have occurred:
1)A 2) B 3) A 4) B
5) BA∪ =female or bike rider or both
6) BA∩ =male and no biker
7) BA∩ =male and biker
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did occur.
1)Yes 2) No 3)No 4) Yes 5) No 6) Yes
7) No 8) Yes.
Chanllenge: Can you express BA∩ using only a ∪ sign?
Answer: .)( BABA ∪=∩
Venn diagrams are generally useful for up to 3 events, although
they are not used to provide formal proofs. For more than 3
events, the diagram might not be able to represent all possible
overlaps of events. (This was probably the case for our transport
Venn diagram.)
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Properties of union, intersection, and complement
The following properties hold.
Distributive laws:
For any sets A, B, and C:
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Definition: Two events A and B are mutually exclusive, or
disjoint, if
This means events A and B cannot happen together. If A happens,
it excludes B from happening, and vice-versa.
Note: Does this mean that A and B are independent?
No, quite the opposite. A EXCLUDES B from happening, so B
depends strongly on whether or not A happens.
Definition: Any number of events nAAA ,,, 21 are mutually
exclusive if every pair of the events is mutually exclusive: ie.
Definition: A partition of the sample space Ω is a collection of
mutually exclusive events whose union is Ω.
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.
form a partition of A.
We will see that this is very useful for finding the probability of
event A.
This is because it is often easier to find the probability of small
‘chunks’ of A (the partitioned sections) than to find the whole
probability of A at once. The partition idea shows us how to add
the probabilities of these chunks together: see later.
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1.4 Axioms, Interpretations, and properties of Probability
Given an experiment and a sample space, the objective of
probability is to assign to each event A a number P(A),called the
probability of the event A, which will give a precise measure of
the chance that A will occur. To ensure that the probability
assignment will be consistent with our intuitive notions of
probability, all assignments should satisfy the following axioms
(basic properties) of probability.
Definition: A probability P is a rule (function) which assigns a
positive number to each event, and which satisfies the following
axioms:
As a direct consequence of the axioms we have the following
properties for P.
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Theorem Let A and B be events. Then,
6. If 1 2, , , nA A A are n arbitrary events in Ω, then
11
1 1 11
( ) ( ) ( ) ( ) ( 1) ( )n n
ni i i j i j k n
i i j n i j k ni
P A P A P A A P A A A P A A−
= ≤ < ≤ ≤ < < ≤=
= − + − + −∑ ∑ ∑ 7. If 1 2, , , nA A A is a finite sequence of mutually exclusive
events in Ω ( Φ=ji AA , ji ≠ ), then
Examples of basic probability calculations
Example. A die is loaded in such a way that the number 1 is twice
as likely to occur as other numbers. Find the probability of the event
E that a number less than 3 occurs in a single toss.
Solution. The sample space is S=1,2,3,4,5,6,
by the assumption, P(2),P(3),…,P(6) are equal and P(1)=2P(2).
Since the sum of P(1),P(2),…,P(6) is 1, so we have P(1)=2/7,
P(2)=P(3)=…=P(6)=1/7 . Thus, P(E)=P(1)+P(2)=3/7.
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Example. 300 Australians were asked about their car preferences in
1998. Of the respondents, 33% had children. The respondents were asked
what sort of car they would like if they could choose any car at all. 13%
of respondents had children and chose a large car. 12% of respondents did
not have children and chose a large car.
Find the probability that a randomly chosen respondent:
(a) would choose a large car;
(b) either has children or would choose a large car (or both).
First formulate events:
Next write down all the information given:
(a) Asked for P(L).
(b) Asked for P(L∪C).
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Respondents were also asked their opinions on car reliability and
fuel consumption. 84% of respondents considered reliability to be of
high importance, while 40% considered fuel consumption to be of
high importance.
Formulate events:
R = “considers reliability of high importance”,
F = “considers fuel consumption of high importance”.
Information given: P(R) = 0.84, P(F) = 0.40.
(c) 16.084.01)(1)( =−=−= RPRP
(d) We can not calculate P(R∩F) from the information given.
(e) Given the further information that 12% of respondents
considered neither reliability nor fuel consumption to be of high
importance, find P(R∪F) and P(R∩F).
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Probability that respondent considers either reliability or fuel consumption, or both, of high importance.
Probability that respondent considers BOTH reliability AND fuel consumption of high importance.
Example. (Student Enrollment) Among a group of 200 students, 137
students are enrolled in a mathematics class, 50 students are enrolled in a history class, and 124 students are enrolled in a music class. Furthermore, the number of students enrolled in both the mathematics and history classes is 33, the number enrolled in both the history and music classes is 29, and the number enrolled in both the mathematics and music classes is 92. Finally, the number of students enrolled in all three classes is 18. We shall determine the probability that a student selected at random from the group of 200 students will be enrolled in at least one of the three classes.
Solution. Let A1 denote the event that the selected student is enrolled in the mathematics class,
A2 denote the event that the selected student is enrolled in the history class,
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A3 denote the event that the selected student is enrolled in the music class.
To solve the problem, we must determine the value of )( 321 AAAP ∪∪ .
From the given numbers,
,20018)(
,20092)(,
20029)(,
20033)(
,200124)(,
20050)(,
200137)(
321
313221
321
=∩∩
=∩=∩=∩
===
AAAP
AAPAAPAAP
APAPAP
It follows from Property (6),
6. If 1 2, , , nA A A are n arbitrary events in Ω, then
11
1 1 11
( ) ( ) ( ) ( ) ( 1) ( )n n
ni i i j i j k n
i i j n i j k ni
P A P A P A A P A A A P A A−
= ≤ < ≤ ≤ < < ≤=
= − + − + −∑ ∑ ∑
8/7200/175)()(
)()()()()()(
32131
3221321321
==∩∩+∩−
∩−∩−++=∪∪AAAPAAP
AAPAAPAPAPAPAAAP
1.5 Probabilities from combinatorics: equally likely outcomes
Sometimes, all the outcomes in a discrete finite sample space
are equally likely. This makes it easy to calculate probabilities.
If
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Example: For a 3-child family, possible outcomes from oldest to
youngest are:
Let 1 2 8 , , , p p p be a probability distribution on Ω . If every
baby is equally likely to be a boy or a girl, then all of the 8
outcomes in Ω are equally likely, so 1 2 81 .8
p p p= = = =
Event A contains 4 of the 8 equally likely outcomes, so event A
occurs with probability 4 1( ) .8 2
P A = =
Counting equally likely outcomes
The number of permutations, n
rP , is the number of ways of
selecting r objects from n distinct objects when different
orderings constitute different choices.
The number of combinations, nrC , is the number of ways of
selecting r objects from n distinct objects when different
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orderings constitute the same choice.
Then
Use the same rule on the numerator and the denominator
When ( ) ,# outcomes in A# outcomes in
P A =Ω
we can often think about the
problem either with different orderings constituting different
choices, or with different orderings constituting the same choice.
The critical thing is to use the same rule for both numerator and
denominator.
Example: (a) Tom has five elderly great-aunts who live together in a tiny
bungalow. They insist on each receiving separate Christmas cards, and threaten to disinherit Tom if he sends two of them the same picture. Tom has Christmas cards with 12 different designs. In how many different ways can he select 5 different designs from the 12 designs available? Order of cards is not important, so use combinations. Number of ways of
selecting 5 distinct designs from 12 is
b) The next year, Tom buys a pack of 40 Christmas cards, featuring 10 different pictures with 4 cards of each picture. He selects 5 cards at random to send to his great-aunts. What is the probability that at least two of the great-aunts receive the same picture?
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Total number of outcomes is
(Note: order mattered above, so we need order to matter here too.)
So
Thus P(A)=P(at least 2 cards are the same design)
=1 ( ) 1 0.392 0.608.P A− = − =
Example: What is the probability p that at least two people in a
group of k people ( 365≤k ) will have the same birthday, that is,
will have been born on the same day of the same month but not
necessarily in the same year. Solution. (We assume that the birthdays of the k people are unrelated (in particular, we assume that twins are not present) and that each of the 365 days of the year is equally likely to be the birthday of any person in the group. In particular, we ignore the fact that the birth rate actually varies during the year and we assume that anyone actually born on February 29 will consider his birthday to be another day, such as March 1.) Since there are 365 possible birthdays for each of k people,
the sample space S will contain k365 outcomes (all of which will be equally probable).
The number of outcomes in S for which all k birthdays will be different is kP ,365
(since the first person’s birthday could be any one of the 365 days, the second person’s birthday could then be any of the other 364 days, and so on. )
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Hence , the probability that all k persons will have different birthdays is
kkP
365,365 .
The probability p that at least two of the people will have the same birthday is
kkk
kP
p365)!365(
)!365(1365
1 ,365
−−=−= .
Numerical values of this probability p for various values of k are given in the following table.
The probability p that at least two people in a group of k people will have the same birthday k p 5 0.027
10 0.117 15 0.253 20 0.411 22 0.476 23 0.507 25 0.569 30 0.706 40 0.891 50 0.970 60 0.994
The calculation in this example illustrates a common technique for solving probability problems. If one wishes to compute the probability of some event A, it might be more straightforward to calculate P(AC) and then use the fact that P(A)=1- P(AC). This idea is particularly useful when the event A is of the form “at least n things happen” where n is small compared to how many things could happen.
Example. Tossing a Coin. Suppose that a fair coin is to be tossed 10 times, and it is
desired to determine (a) the probability p of obtaining exactly three heads and (b) the probability p’ of obtaining three or fewer heads. Solution. The total possible number of different sequences of 10 heads and tails is 210. (it may be assumed that each of these sequences is equally probable.) The number of these sequences that contains exactly three heads will be equal to the number of different arrangements that can be formed with three heads and seven tails. Here are some of those arrangements: HHHTTTTTTT HHTHTTTTTT HHTTHTTTTT TTHTHTHTTT, etc. Each such arrangement is equivalent to a choice of where to put the 3heads among the
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10 tosses, so there are
3
10such arrangements. The probability of obtaining exactly
three heads is then
.1172.023
10
10 =
=p
(b) using the same reasoning as in part (a), the number of
sequences in the sample space that contain exactly k heads
(k=0,1,2,3) is
k
10. Hence , the probability of obtaining three heads or fewer
heads is
1719.02176
212045101
23
102
101
100
10
' 101010 ==+++
=
+
+
+
=p .
Example. (Sampling without replacement.) Suppose that a class contains 15 boys and 30 girls, and that 10 students are to be selected at random for a special assignment. We shall determine the probability p that exactly three boys will be selected. Solution. The number of different combinations of the 45 students that might be
obtained in the sample of 10 students is
1045
, the statement that the 10 students are
selected at random means that each of these
1045
possible combinations is equally
probable. Therefore, we must find the number of these combinations that contain exactly three boys and seven girls. When a combination of three boys and seven girls is formed, the number of different combinations in which three boys can be selected from the 15 available boys is
3
15 , and the number of different combinations in which seven girls can be selected
from the 30 available girls is
730
. Since each of these combinations of three boys
can be paired with each of the combinations of seven girls to form a distinct sample,
the number of combinations containing exactly three boys is
730
315
. Therefore,
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the desired probability is
=
1045
730
315
p
Example. (Playing Cards) Suppose that a deck of 52 cards containing four aces is
shuffled thoroughly and the cards are then distributed among four player so that each player receives 13 cards. We shall determine the probability that each player will receive one ace.
Solution. The number of possible different combinations of the four positions in the
deck occupied by the four aces is
4
52. If each player is to receive one ace, then there must
be exactly one ace among the 13 cards that the first player will receive and one ace among each of the remaining three groups of 13 cards that the other three players will receive. In other words, there are 13 possible positions for the ace that the first player is to receive, 13 other possible positions for the ace that the second player is to receive, and so on. Therefore, among the
4
52possible combinations of the positions for the four aces, exactly 134 of these combinations
will lead to the desired result. Hence, the probability p that each player will receive one ace is
1055.0
452134
=
=p
Example. (Lottery Tickets) In a lottery game, six numbers from 1 to 30 are
drawn at random from a bin without replacement, and each player buys a ticket with six different numbers from 1 to 30. If all six numbers drawn match those on the player’s ticket, the player wins. We assume that all possible draws are equally likely.
One way to construct a sample space for the experiment of drawing the winning
combination is to consider the possible sequences of draws. That is, each outcome consists of an ordered subset of six numbers chosen from the 30 available numbers. There are P30,6=30!/24! such outcomes. With this sample spaces S , we can calculate probabilities for events such as
A=the draw contains the numbers 1,14,15,20,23, and 27 B=one of the numbers drawn is 15 C=the first number drawn is less than 10
P(A)= 00000168.0!30!24!6!6
6,30
==P
, P(B)= 2.03066
6,30
5,29 ==×P
P
P(C)= 3.01039
6,30
5,29 ==×P
P
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1.6 Multinomial Coefficients We learn how to count the number of ways to partition a finite set into more than two disjoint subsets. The generalization is useful when outcomes consist of several parts selected from a fixed number of distinct types. We begin with a fairly simple example that will illustrate the general ideas of this section. Example. (Choosing Committees) Suppose that 20 members of an organization are to be divided into three committees A, B, C in such a way that each of the committees A and B is to have eight members and committee C is to have four members. We shall determine the number of different ways in which members can be assigned to these committees. Notice that each of the 20 members gets assigned to one and only one committee. Solution. To form committee A, we must choose eight out of 20 members , and this can be done in
820
ways. Then to split the remaining 12 members into committee B and C there are
8
12ways to do it. Every choice of committee A can be paired with every one of the splits of the
remaining 12 members into committees B and C. Hence the number of assignments into three committees is the product of the numbers combinations for the two parts of the assignment,
820
8
12= 150,355,62
!4!8!8!20
!4!8!12
!12!8!20
==
In general, suppose that n distinct elements are to be divided into k different groups ( 2≥k ) in such a way that, for j=1,2,…,k, the jth group contains exactly nj elements, where n1+ n2+…+ nk =n. It is desired to determine the number of different ways in which the n elements can be divided into the k groups. The n1 elements in the first group can be selected from the n available elements in
1nn
different ways. After the n1 elements in the first group have been selected, the n2 elements in
the second group can be selected from the remaining n-n1 elements in
−
2
1
nnn
different ways.
Hence, the total number of different ways of selecting the elements fro both the first group and the
second group is
1nn
−
2
1
nnn
. It follows from the preceding explanation that, for each
j=1,2…,k-2 after the first j groups have been formed, the number of different ways in which the nj+1 elements in the next group (j+1) can be selected from the remaining n-n1-…-nj elements is
−−
+1
1
j
j
nnnn
.After the elements of group k-1 have been selected, the remaining nk elements
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must then form the last group. Hence, the total number of different ways of dividing the n elements into the k groups is
1nn
−
2
1
nnn
−−
3
21
nnnn
−−
−
−
1
21
k
k
nnnn
=!!!
!
21 knnnn
,
Where the last formula follows from writing the binomial coefficients in terms of factorials.
Definition. (Multinomial Coefficients).
The number !!!
!
21 knnnn
, which we shall denote by
knnnn
,,, 21 ,is called a multinomial
coefficient. The name multinomial coefficient derives from the appearance of the symbol in the multinomial theorem.
Theorem (Multinomial Theorem). For all numbers kxxx ,,, 21 and each positive
integer n,
knk
nn
k
nk xxx
nnnn
xxx
2121
2121 ,,,
)( ∑
=+++ , where the summation extends
over all possible combinations of nonnegative integers knnn ,,, 21 such that
nnnn k =+++ 21 .
Example. Choosing Committees. In above example, we see that the
solution obtained there is the same as the multinomial coefficient fro
which n=20, k=3, n1=n2=8 and n3=4, namely,
.150,355,62!4)!8(
!204,8,8
202 ==
.
Example. Suppose that 12 dice are to be rolled. We shall determine the
probability p that each of the six different numbers will appear twice.
Solution. Each outcome in the sample space S can be regarded as an ordered
sequence of 12 numbers, where the ith number in the sequence is the outcome of the ith roll. Hence, there will be 612 possible outcomes in S. The number of these
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outcomes that would contain each of the six numbers 1, 2, .., 6 exactly twice will be equal to the number of different possible arrangements of these 12 elements. This number can be determined by evaluating the multinomial coefficient for which n=12,
k=6, and n1=n2=…=n6=2.
Hence the number of such outcomes is
6)!2(!12
2,2,2,2,2,212
=
, the required probability p is p= 0034.0
6)!2(!12
126 = .
Example. A deck of 52 cards contains 13 hearts. Suppose that the cards
are shuffled and distributed among four players A, B, C, and D so that
each player receives 13 cards. We shall determine the probability p that
player A will receive six hearts, player B will receive four hearts, player
C will receive two hearts, and player D will receive one heart.
Solution. The total number N of different ways in which the 52 cards can be distributed among
the four players so that each player receives 13 cards is
4)!13(!52
13,13,13,1352
=
=N
We must now calculate the number M of ways of distributing the cards so that each player receives the required number of hearts. The number of different ways in which the hearts can be distributed to players A, B, C, and D so that the numbers of hearts they receive are 6, 4, 2, and 1,
respectively, is !1!2!4!6
!131,2,4,6
13=
.
Also, the number of different ways in which the other 39 cards can then be distributed to the four players so that each will have a total of 13 cards is
!12!11!9!7!39
12,11,9,739
=
.
Therefore, the required probability p is .00196.0)
)!13(!52(
)!12!11!9!7
!39!1!2!4!6
!13(
4
===p
There is another approach to this problem. The number of possible different combinations of the
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13 positions in the deck occupied by the hearts is
1352
. If player A is to receive six hearts, there
are
6
13 possible combinations of the six positions these hearts occpy among the 13 cards that
A will receive. Similarly, if player B is to receive four hearts, there are
4
13 possible
combinations of their positions among the 13 cards that B will receive. There are
2
13 possible
combinations for player C, and there is
1
13 possible combinations for player D. Hence,
=
1352
113
213
413
613
p ,
Which produces the same value as the one obtained by the first method of solution.
Hw. P15/7,11
P21/1,3,7,11
P25/1,3,6,8
P32/1,3,5,7,9
P41/7,9,13,17
P45/1,3,7,9
P50/1,2,3,5,7,
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