CHAPTER 4
4.1
COUNTING RULES
Objective: The student should be able to find the total number of outcomes in a sequence
of events, using fundamental counting rule.
Tree Diagram and Multiplication Principle
Tree Diagram
Example 1: A coin is tossed and a dice is rolled. Find the number of outcomes for the
sequence of events by drawing tree diagram.
Answer: 
Example 2: Find all possible outcomes for the gender of three children in a family.
Answer: 
Example 3: Find all possible outcomes for rolling two dice.
Answer: 
(If the number of outcomes is large, then is not easy to draw a tree diagram. The other
method is multiplication principle)
Multiplication Principle
In a sequence of n events in which the first one has k1 possibilities and
the second one has k2 possibilities and
the third one has k3 possibilities and so forth….
The number of outcomes = k1  k2  k3    kn
(Example 1)
Answer: 
(Example 2)
Answer: 
(Example 3)
Answer: 
Example 4: ( no 12 pg 211) There are 2 major roads from city X to city Y and 4 major
roads from city Y to city Z. How many different trips can be made from city
X to city Z passing through city Y?
Answer: 
( tell that tree diagram can be drawn to solve example 4 because the number of outcome
is small)
Example 5: ( no9 pg 211) How many different 3-digit identification tags can be made if
the digits can be used more than once? If the first digit must be a 5 and
repetitions are not permitted?
Answer: 
( show the student how tedious it is to draw the tree diagram for example 5)
Example 6: (no 5 pg 211) A store manager wishes to display 8 brands of shampoo in a
row. How many ways can this be done?
Answer: 
( tell that understood the number of shampoo for the next one is reduced by one.)
Permutation and Combination
Factorial Notation
5 ! = 5  4  3  2 1
0!=1
n ! = nn  1n  2   1
( student can use calculator)
Permutation is an arrangement of n objects in a specific order. ( order / position is
important)
n
Pr 
n!
n  r !
n objects taking r objects at a time.
Combination is a selection of objects without regard to order ( order/ position is not
important )
n
Cr 
n!
r!n  r !
r objects selected from n objects.
Example 7: Given letters A , B and C.
Arrange 2 letters ( permutation) : AB, BA, AC, CA, BC, CB
( AB and BA are different outcome, even though both having the same letter but the
position of A and B are different, so position is important, considered as different
outcome, and it can be solved using multiplication principle)
use formula : 3 P2  6 outcomes.
Select 2 letters ( combination) : AB , AC , BC
( AB and BA are the same, it does not matter whether select A first then B or select A
second after B, position is not important, because in AB or BA, both letter A and B are
selected, so considered the same)
use formula : 3C2  3 outcomes.
Example 8: Draw 4 people with different height.
(i)How many ways to arrange 3 people for photography session.
Use formula : 4 P3  24 outcomes.
one of the arrangements
another arrangement
A
B
C
B
C
A
(why permutation? even though for both outcomes having the same people but different
position they stand will result different arrangement, so position is important)
Answer: ABC, ACB, BAC, BCA, CAB, CBA, ABD, ADB, BAD, BDA, DBA, DAB,
BCD, BDC, CBD, CDB, DBC, DCB, ACD, ADC, CDA, CAD, DAC, DCA.
( can be solved using multiplication rule )
(ii)How many ways can a committee of three be chosen from four people?
Use formula : 4C3  4 outcomes.
Possible selections : ABC and BCA.
(Why combination? Does position important or not? Selection ABC and BCA are the
same because the same people being selected as the member of the club, so it does not
matter whether A selected first in ABC or the last one in BCA )
Answer: ABC, ABD, BCD, ACD
(iii)How many ways can a president, vice president, and secretary be chosen from four
people?
Use formula: 4 P3 = 24 ways.
(why permutation? and can be solved using multiplication principle)
Example 9:
(i)Four tickets are to be selected from 20 tickets. The holder of the first ticket wins a car,
the second a motorcycle, the third a bicycle and the fourth a skateboard. In how many
different ways can these prizes be awarded?
Answer: 20P4  1116280
(why permutation? and can be solved using multiplication principle)
(ii) Four tickets are to be chosen from 20 tickets. The holders of the winning tickets are to
be awarded free trips to the Bahamas. In how many ways can the four winners be
chosen?
Answer: 20C4  4845
(why combination?)
(to use either permutation or combination, we need to ask ourselves: Does the order
matter?)
Example 10 : ( no 19 pg 212 ). How many different 4-letter word can be formed from the
letters in the word DECAGON
Answer: 
( why permutation? and student should know that for this problem it is understood no
repetition, also can be solved using multiplication principle)
Example 11 : (no 25 pg 212) How many different signals can be made by using at least 3
distinct flags if there are 5 different flags from which to select?
Answer: 
( why permutation?)
Example 12: ( no 33 pg 212) How many different tests can be made from a test bank of
20 questions if the test consists of 5 questions?
Answer: 
( why combination?)
Example 13: (no 48 pg 212) An advertising manager decides to have an ad campaign in
which 8 specials calculators will be hidden at various locations in a
shopping mall. If he has 17 locations from which to pick, how many
different possibilities can he choose?
Answer: 
(why combination?)
Example 14: (no 37 pg 212). There are 7 women and 5 men in a department . How many
ways can a committee of 4 people be selected? How many ways can this
committee be selected if there must be 2 men and 2 women on the
committee? How many ways can this committee be selected if there must
be at least 2 women on the committee?
Answer: 
Involving Permutation and Combination
Example 15: A committee of seven consisting of a chairman, a vice chairman, a
secretary, and four other members is to be chosen from a class of 20 students. In how
many ways can this committee be chosen?
Distinguishable Permutations
If a set of n objects consists of n1 objects of the first kind
n2 objects of the second kind
n3 objects of the third kind and so on
the number of ways =
n!
n1!n2!n3!  nk !
Example 16: How many ways can the letters of the word DAD be arranged?
Answer: DAD, ADD, DDA
only 3 outcomes
(show that formula n Pr doesn’t work because 3 P3  6 outcomes. So conclude that n Pr
formula for n distinct objects)
Example 17:
How many ways can 2 black marbles and 2 white marbles be arranged in a row?
Answer:
Number of ways =
4!
 6 outcomes
2!2!
Together Problem
Example 18 : Four employees at a company picnic are to stand in a row for a group
photograph. I how many ways can this be done if
(i)
Siti and Ani insist on standing next to each other?
(ii)
Ani and Siti refuse to stand next to each other?
(i)
( SA treat as one person because they have to be together)
No of ways = 3  2 1  6 ways to arrange meaning that
if SA together in the first position then there are two ways: SABC , SACB
If SA together in the middle then there are two ways : BSAC , CSAB
If SA together at the end then there are two ways : BCSA , CBSA
But then SA and AS are different so the answer
Answer : 3  2  1  2 = 12 ways
(SABC , SACB , BSAC , CSAB , BCSA , CBSA , ASBC , ASCB , BASC , CASB ,
BCAS , CBAS )
(ii) No of ways to arrange 4 people = 4  3  2  1 = 24 ways
( in 24 ways there are only 12 ways that there are together. Meaning that the rest of
outcomes they are not together).
Answer : 24 – 12 = 12 ways
Exercise
1. A company has 2844 employees. Each employee is to be given an ID number that
consists of one letter followed by two digits. Is it possible to give each employee a
different ID number using this scheme?
2. In how many ways can five students be seated in a row of five chairs if Jack
insists on sitting in the first chair?
3. Standard automobile license plates in California display a nonzero digit, followed
by three letters, followed by three digits. How many different standard plates are
possible in this system?
4. In how many ways can three pizza toppings be chosen from 12 available
toppings?
5. A combination lock has 60 different positions. To open the lock, the dial is turned
to a certain number in the clockwise direction, then to a number in the
counterclockwise direction, and finally to a third number in the clockwise
direction. If the successive numbers in the combinations cannot be the same, how
many different combinations are possible?
6. In how many ways can seven students from a class of 30 be chosen for a field trip
if Jack must go on the field trip?
7. In how many ways can a president, vice president, and secretary be chosen from a
class of 20 females and 30 males if the president must be a female and the vice
president a male?
8. A group of 22 aspiring thespians contains ten men and twelve women. For the
next play the director wants to choose a leading man, a leading lady, a supporting
male role, a supporting female role, and eight extras consists of three women and
five men. In how many ways can the cast be chosen.
9. Five letter “ words” are formed using the letters A, B, C, D, E, F, G. How many
such words are possible for each of the following conditions?
i)
no condition is imposed
ii)
no letter can be repeated in a word
iii)
each word must begin with the letter A
iv)
the letter C must be in the middle
v)
the middle letter must be a vowel.
10. From a group of 30 contestants, six are to be chosen as semifinalists, then two of
those are chosen as finalists, and then the top prize is awarded to one of the
finalists. In how many ways can these choices be made in sequence?
11. Three-digit numbers are formed using the digits 2 , 4, 5, and 7, with repetition of
digits allowed. How many such numbers can be formed if
i)
the numbers are less than 700
ii)
the numbers are even
iii)
the numbers are divisible by 5
12. Twelve dots are drawn on a page in such away that no three are collinear. How
many straight lines can be formed by joining the dots?
13. A five-person committee consisting of students and teachers is being formed to
study the issue of student parking privileges. Of those who have expressed an
interest in serving on the committee, 12 are teachers and 14 are students. In how
many ways can the committee be formed if at least one student and one teacher
must be included?
14. In how many different ways can the letters of the word ELEEMOSYNARY be
arranged?
15. In how many different ways can five red balls, two white balls, and seven blue
balls be arranged in a row?
16. In how many ways can four men and four women be seated in a row of eight seats
if
i)
the women are to be seated together
ii)
the men and women are to be seated alternately by gender.
17. In how many ways can ten students be arranged in a row for a class picture if
John and Jane want to stand next to each other, and Mike and Molly also insist on
standing next to each other?
18. In how many ways can the ten students be arranged if Mike and Molly insist on
standing together but John and Jane refuse to stand next to each other?
19. In how many ways can a committee of four be chosen from a group of ten if two
people refuse to serve together on the same committee?
4.2
SAMPLE SPACES AND PROBABILITY
Objective: The student should be able to determine sample spaces and find the
probability of an event.
Probability : the chances of an event occurring.
Probability experiment: flipping coin, rolling a die and so on
Outcome : the result of the probability experiment.
Sample space, S : set of all possible outcomes of a probability experiment
Example : Toss a coin : S = { H , T }
Roll a die : S = { 1 , 2 , 3 , 4 , 5 , 6 }
Two children in a family : S = { BB , BG , GB , GG } ( can use tree diagram )
Event : a set of outcomes of a probability experiment.
Example 1 : ( no 12 pg 180) A die is rolled. List the outcomes for the event of getting an
odd number.
Answer :
S={1,2,3,4,5,6}
Let E : event getting an odd number
E={1,3,5}
Probability of event , PE  
n E 
nS 
Probability Rule
Let S = { a , b , c }. S is sample space
(i) P( S ) = 1 ; P (a) + P (b) + P (c ) = 1
(ii) Pa   0 , Pb  0 , Pc  0
Let E be any event then 0  P( E )  1 . If
(iii) P (E ) = 0  event E cannot occur
(iv) P(E ) = 1  event E certain to occur.
Example 2 : Roll a die. Find probability of getting no 9.
Let E : getting no 9 = { }
P(E) = 0
Example 3: Roll a die. Find the probability of getting a number less than 7?
Let E : getting a number less than 7 = { 1 , 2 , 3 , 4 , 5 , 6 }
P(E ) = 1
( do not discuss question on a deck of 52 cards )
Example 4: ( no 13 pg 180) If two dice are rolled one time, find the probability of
getting these result:
i) a sum of 6
ii) doubles
iii) a sum of 7 or 11
iv) a sum greater than 9
v) a sum less than or equal to 4
Answer: 
Example 5: ( no 21 pg 181) A couple has three children. Find each probability
i) of all boys
ii) of all girls or all boys
iii) of exactly two boys or two girls
iv) of at least one child of each gender.
Answer: 
Equally Likely Events : events that have the same probability of occurring
Example 6 : Toss a coin S = { H , T }
;
P (H ) =
1
2
, P(T)=
1
2
Roll a die S = { 1 , 2 , 3 , 4 , 5 , 6 } ; P(1) = P(2) = P(3) = P(4) = P(5) = P(6)=
1
6
Complementary Events, E : set of outcomes in the sample space that not included in
the outcomes of event E.
Example 7 : Roll a die : S = { 1 , 2 , 3 , 4 , 5 , 6 }
Let E be the event of getting odd number , E = { 1 , 3 , 5 }.Therefore , E  { 2 , 4 , 6 }
In Venn Diagram
S
E
5
6
1
3
2
4
P(S) = 1 ; PE   PE   1 ;
PE   1  PE 
Example 8: ( no 18 pg 181 )Sixty-nine percent of adults favor gun licensing in general.
Choose one adult at random. What is the probability that the selected adult
does not believe in gun licensing?
Answer: 
Empirical Probability: relies on the actual experience to determine the likelihood of
outcomes
P(E) =
f
n
Example 9: A researcher asked 25 people if they liked the taste of a new soft drink. The
responses were classified as “yes” , “no” .
Response
Yes
No
Frequency
16
9
Relative
Frequency
0.64
0.36
1.00
The probability that a person who responded yes is
16
25
( if the survey was not done then the probability of selecting a person who liked the taste
1
is , assumes equally likely event that is chances of selecting a person who liked or did
2
not like the taste is 50-50 percent)
Example 10 : ( no 17 pg 181) In a college class of 250 graduating seniors, 50 have jobs
waiting, 10 are going to medical school, 20 are going to law school, and 80
are going to various other kinds of graduate schools. Select one graduate at
random
i) What is the probability that the student is going to graduate school?
ii) What is the probability that the student is going to medical school?
iii) What is the probability that the student will have to start paying back his
deferred student loans after 6 months (i.e. does not continue in school)?
Exercise
1. The wheel spinner shown here is spun twice. Find the sample space, and then
determine the probability of the following events.
i)
an odd number on the first spin and an even number on the second spin ( 0 is
considered even)
ii)
a sum greater than 4
iii)
even number on both spins
iv)
a sum that is odd
v)
the same number on both spins
2. In a survey, 16 percent of American children said they use flattery to get their parents
to buy them things. If a child is selected at random, find the probability that the child said
he or she does not use parental flattery?
2.3
THE ADDITION RULES FOR PROBABILITY
Objective: The student should be able to find the probability of compound events, using
the addition rules.
A  B  event A or B or both A and B occurs
A  B  both event A and B occurs
A
B
A
A B
B
A B
P A  B  P A  PB  P A  B
Mutually Exclusive Event : two events cannot occur at the same time or they have no
outcomes in common.
Example 1: (i)Roll a die: Get a number greater than 3 and get a number less than 3
A : number greater than 3 = { 4 , 5, 6 }
B: number less than 3 = { 1 , 2 }
Event A and B  A  B = {} no outcome in common. Therefore mutually exclusive
event. ( or we can say that event A and event B can not occur at the same time meaning
that impossible to have a number which greater than 3 and at the same time less than 3)
(ii) Select one student. Get a female or male.
A : female B: male . Event A and B are mutually exclusive. ( For a student: Event A
and B cannot occur at the same time meaning that impossible for a student who is female
and at the same time the student is male)
( for not mutually exclusive event: syllabus covers up to two events)
P A  B  P A  PB
when two events A and B are mutually exclusive.
A
B
A  B  {}
For more than two mutually exclusive events:
P A  B  C   P A  PB  PC 
A
B
C
Example 2: ( no. 9 pg 187) At a particular school with 200 male students, 58 play
football, 40 play basketball, and 8 play both. What is the probability that a
randomly selected male student plays neither sport?
Answer: 
Example 3: The probability that a customer selects a pizza with mushroom or
pepperoni is 0.55, and the probability that the customer selects mushrooms
only is 0.32. If the probability that he or she selects pepperoni only is 0.17,
find the probability of the customer selecting both items.
Answer: 0.06
Example 4: (no 3 pg 187) A furniture store decides to select a month for its annual sale.
Find the probability that it will be April or May. Assume that all months
have equal probability of being selected.
Answer: 
Example 5: (no 25 pg 189) An urn contains 6 red balls, 2 green balls, 1 blue ball, and 1
white ball. If a ball is drawn, find the probability of getting a red or a white
ball.
Answer: 
Example 6: (no 12 pg 188) At a used-book sale, 100 books are adult books and 160 are
children’s books. Seventy of the adult books are nonfiction while 60 of the
children’s books are nonfiction. If a book is selected at random, find the
probability that it is
i)
fiction
ii) not a children’s nonfiction
an adult book or a children’s nonfiction.
iii)
Answer: 
Example 7: (no 15 pg 188) A grocery store employs cashiers, stock clerks, and deli
personnel. The distribution of employees according to marital status is
shown here.
Marital Status
Married
Not Married
Cashiers
8
5
Stock clerks
12
15
Deli personnel
3
2
If an employee is selected at random, find these probabilities:
i) the employee is a stock clerk or married
ii) the employee is not married
iii) the employee is a cashier or is not married.
Answer: 
4.4 THE MULTIPLICATION RULES AND CONDITIONAL PROBABILITY
Objective: The student should be able to find the probability of compounds events, using
multiplication rules.
Multiplication rule: used to find the probability of two or more events that occur in
sequence. For example toss a coin and then a die is rolled.
Independent Events : Two events A and B are independent if event A occurs does not
affect the probability of B occurring.
Example: (i) rolling a die and getting a 6 , and then rolling a second die and getting a 3
(ii) There are 7 blue balls and 3 red balls, select a blue ball, replace and then
select a red ball.
P(3) =
1
6
P(R)=
1
6
6
1
6
3
3
10
3
PB 
PR 
B
R
3
10
 PB  R
R
P B  R   P B   P R  
7 3
21
 
10 10 100
(The probability of getting number 3 still the same, the first event of getting number 6
does not affect the probability of the second event i.e. getting number 3)
(The probability of getting red ball still the same, the first event of getting blue ball
does not affect the probability of the second event i.e. getting red ball.)
P A  B  P A  PB
If event A and B are independent.
Example: A coin is tossed twice. Find probability of getting HH.
Answer:
Method 1: S = { HH , TH , HT , TT }
P(HH) =
1
.
4
Method 2: Event getting head for the first toss and event getting head for the second
toss are independent event ( because the first event does not affect the probability of
the second event)
P(H) =
1
2
1
2
1
2
H
H
T
T
P(H  H ) = PH   PH  =
1 1 1
 
2 2 4
P( A  B  C  D)  P A  PB  PC   PD
If A, B, C and D are independent
events
Dependent event: Two events are said to be dependent when the occurrence of the
first event changes the probability of the occurrence of the second event.
Example: There are 7 blue balls and 3 red balls, select a blue ball, not replacing it and
then select a red ball.
Answer:
B
P B  
P R  
6/9
7
3
10
R  PB  R
3/ 9
B
10
7/9
R
B
2/9
R
P B  R   P B   P R \ B  
Therefore for independent event
7 3 21
 
10 9 90
PR \ B  PR
Conditional probability, P( B \ A) : probability that event B occurs after / given that
/ if it is known event A has already occurred.
P A  B  P A  PB \ A
If A and B are dependent.
rearrange
P  B \ A 
P( A  B)
P  A
P(A)
A
P(B)
B
P(C\A)
C
P(D\A)
D
P(C\B)
P(D\B)
C
D
Example: In a statistics class there are 18 juniors and 10 seniors; 6 of the seniors are
females, and 12 of the juniors are males. If a student is selected at random, find the
probability that the student is a female given that she is a junior
Answer:
F
M
J
6
12
S
6
4
(we want to find probability of selecting a female student with the condition that she
is a junior. So if we ignore the condition, selecting a female student, then there are
two possibilities either she is a junior or a senior, but what the question want is that
with the condition that she is a junior, in order to fulfill the condition that she is a
junior, so we will let the condition event occur first, then out of 18 juniors there 6
female, so the chance that the student is a female given that she is a junior is 6/18)
F
J
S
We just want to find the chances of this event to
occur.
( Don’t confused with the question of finding probability that the student is a junior
and a female, PJ  F  )
J
S
F
 P( J  F )
both events occur , from table 6/28 or use the formula
PJ  F   PJ   PF  if event junior and female
independent or
PJ  F   P( J )  P( F / J ) if event junior and female
dependent. ( does PF   PF / J  ?, mathematically
not equal, so definitely dependent event).
Example 1: (no. 19 pg 202) In a class containing 12 men and 18 women, 2 students are
selected at random to give an impromptu speech. Find the probability that
both are women.
Answer: 
( solving problem selecting 2 objects/ people is the same as select one by one without
replacing)
Example 2: ( no 2 pg 202) If 18% of all Americans are underweight, find the probability
that if three Americans are selected at random, all will be underweight.
Answer: 
(solving problem selecting 3 objects/ people is the same as select one by one without
replacing. Explain why for a small sample selected from a large population, the
probability is considered to remain the same meaning using independent event formula)
Example 3: (no 26 pg 203) At a small college, the probability that a student takes
physics and sociology is 0.092. The probability that a student takes sociology
is 0.73. find the probability that the student is taking physics, given that he or
she is taking sociology.
Answer: 
Example 4: (no 32 pg 203). In a pizza restaurant, 95% of the customers order pizza. If
65% of the customers order pizza and a salad, find the probability that a customer who
orders pizza will also order a salad.
Answer: 
Example 5: ( no 36 pg 203). The medal distribution from the 2000 summer Olympic
Games is shown in the table
United States
Russia
China
Australia
Others
Gold
39
32
28
16
186
Silver
25
28
16
25
205
Bronze
33
28
15
17
235
Choose one medal winner at random.
i) Given that the winner was from the United States, find the probability
that the winner won the gold medal
ii) Find the probability that the winner was from the United States, given
that she or he won a gold medal.
iii) Are the events “medal winner is from United States” and “gold medal
was won” independent? Explain.
Answer: 
Probabilities for “At least”
Example 6: ( no 7 pg 202). At a local university 54.3% of incoming first-year students
have computers. If three students are selected at random, find the following
probabilities
i)
none have computers
ii)
at least one has a computer
iii)
all have computers
Answer: 
Example 7: (no 40 pg 204) A lot of portable radios contains 15 good radios and 3
defective ones. If two are selected and tested, find the probability that at least
one will be defective.
Answer: 
Example 8: (no 51 pg 204). If a die is rolled 3 times, find the probability of getting at lest
one even number.
Answer: 
Example 9: (no 21 pg 203). A manufacturer makes two models of an item: model I,
which accounts for 80% of unit sales, and model II, which accounts for 20%
of unit sales. Because of defects, the manufacturer has to replace (or
exchange) 10% of its model I and 18% of its model II. If a model is selected
at random, find the probability that it will be defective.
Answer: 
( explain that in example 9, select a model , find the probability that it will be defective,
so it could be from model I or model II) .
Venn Diagram
I
II
D
PI  D
PII  D
PD  PI  D  P( II  D)  PI   PD \ I   PII   PD \ II 
Tree Diagram:
D
 PI  D  PI   PD \ I 
I
II
D  PII  D  PII   PD \ II 
PD  PI  D  P( II  D)  PI  PD \ I   PII  PD \ II 
Example 10: (no. 25 pg 203). Urn 1 contains 5 red balls and 3 black balls. Urn 2 contains
3 red balls and 1 black ball. Urn 3 contains 4 red balls and 2 black balls. If
an urn is selected at random and a ball is drawn, find the probability it will
be red.
Answer: 
4.5 PROBABILITY AND COUNTING RULES
Objective: Student should be able to find the probability of an event, using the counting
rules.