1 / 25

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

2 / 25

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.

3 / 25

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

4 / 25

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

5 / 25

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

6 / 25

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.)

7 / 25

Properties of union, intersection, and complement

The following properties hold.

Distributive laws:

For any sets A, B, and C:

8 / 25

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.

9 / 25

.

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.

10 / 25

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.

11 / 25

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.

12 / 25

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).

13 / 25

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).

14 / 25

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,

15 / 25

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

16 / 25

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

17 / 25

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?

18 / 25

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. )

19 / 25

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

20 / 25

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,

21 / 25

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

22 / 25

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

23 / 25

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

24 / 25

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

25 / 25

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,