a device used to list all p
ossibilities of a sequenc
e of events in a systemat
1. Suppose a sales rep can travel
from New York to Pittsburgh by p
lane, train, or bus, and from Pitts
burgh to Cincinnati by bus, boat,
or automobile. List all possible w
ays he can travel from New York
A tree diagram can be drawn to
show the possible ways.
• First the salesman can
travel from New York
to Pittsburgh by
• Then the salesman can travel
from Pittsburgh to Cincinnati by
bus, boat, or automobile.
• Next the second branch is paired
up with the first branch in three
• Finally, all outcomes can
be listed by starting at
New York and following
Cincinnati, as shown at
the right end of the tree.
There are nine ways.
2. A coin is tossed and a die is
rolled. Find all possible outcomes
of this sequence of events.
Since the coin can land either
heads up or tails up, and since
the die can land with any one of
six numbers shown face up.
In a sequence of n events in which
the first one has k1 possibilities and
the second event has k2 and the third
has k3, and so forth, the total number
of possibilities of the sequence will be
k1 • k2 • k3 ••••• kn
“And” in this case means to multiply
1. A paint manufacturer wishes
to manufacture several different
paints. The categories include:
green, brown, yellow
kinds of paint can be
made if a person can
select one color, one type,
one texture, and one use?
A person can choose one color and one
type and one texture and one use. Since
there are seven color choices, two type
choices, three texture choices, and two
use choices, the total number of possible
different paint is
2. There are four blood types,
A, B, AB, and O. Blood can
also be Rh+ and Rh-. Finally,
a blood donor can be
classified as either male or
female. How many different
ways can a donor have his or
her blood labeled?
Since there are four possibilities for
blood type, two possibilities for Rh
factor, and two possibilities for
gender of the donor, there are 4 • 2 •
2 or 16, different classification
categories as shown:
When determining the number
of different possibilities of a
sequence of events, one must
know whether repetitions are
1. Find all possible outcomes for the
genders of the children in a family
that has three children.
2. If a baseball manager has five
pitchers and two catchers, how
many different possible pitchercatcher combinations can he field?
3. How many different three-digit
identification tags can be made if the
digits can be used more than once?
If the first digit must be a 5 and
Two other rules that can be used
to determine the total number of
possibilities of a sequence of
events are the permutation rule
and combination rule.
These rules use factorial notation.
The factorial notation uses the
5! Means 5•4•3•2•1
9! = 9•8•7•6•5•4•3•2•1
In order to use the formula in the
permutation and combination rules, a
special definition of 0! is needed. 0! = 1
For any counting n
n! = n(n-1) (n-2)…1
0! = 1
an arrangement of
n object in a
1. Suppose a business owner
has a choice of five locations in
which to establish her business.
She decides to rank each
location according to certain
criteria, such as price of the
store and parking facilities. How
many different ways can she
rank the five locations?
5! = 54321
Different possible rankings.
The reason is that she has five
choices for the first location,
four choices for the second
location, three choices for the
third location, etc.
2. Suppose the business
owner in example no. 1
wishes to rank only the top
three of the five locations.
How many ways can she
Using the multiplication rule, she can
select any one of the five for first choice,
then any one of the remaining four
locations for her second choice, and
finally, any one of the remaining three
locations for her third choice, as shown.
Permutation Rule 1
The number of permutations
of n distinct objects is n!.
In how many ways can a
photographer take pictures
of five ladies in a row?
5! = 5 4 3 2 1 = 120
Permutation Rule 2
The arrangement of n objects in a
specific order using r objects at a
time is called a permutation of n
objects taking r objects at a time. It
is written as nPr, and the formula is
five locations were taken and
then arranged in order; hence,
5P5= = = = 120
(Recall 0! = 1)
In Example 2, three locations
were selected from five
locations, so n=5 and r=3;
5P3 = = = = 60
The notation nPr is used for
P means or = = 360
Although example 1 and 2 were
solved by the multiplication rule,
they can now be solved by
2. How many ways can a
basketball team schedule 3
exhibition games with 3 teams
if they are all available on any
of 5 possible dates?
5P3 = = (5) (4) (3) = 60
1. Two lottery tickets are drawn
from 20 for first and second
prizes. Find the number of
sample points in the space S.
The total number of sample
20P2 = = (20) (19) = 380
Permutations that occur by
arranging objects in a circle are
called circular permutations. Two
circular permutations are not
corresponding objects in the two
arrangements are preceded or
followed by a different object as
we proceed in a clockwise direction.
For example, if 4 people are
playing bridge, we do not have
a new permutation if they all
considering 1 person in a fixed
position and arranging the
other 3 in 3! Ways, we find that
arrangements for the bridge
Permutation Rule 3
The number of permutations of n
distinct objects arranged in a circle
In how many ways can eight
persons be seated in a round
Permutation Rule 4
The number of distinct permutations
of n things of which n1 are one kind,
n2 of a second kind, … , nk of a kth
How many different ways can 3
red, 4 yellow, and 2 blue bulbs
be arranged in a string of
Christmas tree light with 9
The total number of distinct
Permutation Rule 5
The number of ways of partitioning
a set of n objects into r cells with
n1 elements in the first cell, n2
elements in the second, and so on,
Where n1 + n2 + … + nr = n.
How many ways can 7 people
be assigned to 1 triple and 2
The total number of possible
partitions would be
= = 210
The difference between
permutation and combination
is that in combination, the
order of arrangement of the
objects is not important: by
contrast, order is important
in a permutation.
interested in the number of ways of
selecting r objects from n without
regard to order. These selections are
called combinations. A combination
creates a partition with 2 cells, one cell
containing the r objects selected and
the other cell containing the n-r
objects that are left.
without regard to
1. Given the letters A, B, C, and
D, list the permutations and
combinations for selecting two
The listing follow:
¤ Note that in permutations, AB is
combinations, AB is the same as BA,
so only AB is listed. (Alternatively BA
could be listed instead of AB.)
¤ The elements of a combination are
usually listed alphabetically.
¤ Combinations are used when the order
or arrangement is not important, as
in the selecting process.
The number of combinations of r
objects selected from n objects
is denoted by nCr and is given by
combinations, denoted by , is
usually shortened to , since the
number of elements in the
second cell must be n – r.
2. How many combinations of
four objects are there taken
two at a time?
Since this is a combination
problem, the answer is
4C2 = = = = 6
Notice that the formula for nCr is
Which is the formula for permutation,
With an r! in the denominator. This
r! divides out the duplicates from the
number of permutations, as shown in
Example 2. For each two letters, there
are two permutations but only one
combination. Hence, dividing the
number of permutations by r!
eliminates the duplicates. This result
can be verified for other values of n
and r. Note: nCn = 1.
From 4 Republicans and 3
Democrats find the number of
committees of 3 that can be
formed with 2 Republicans
The number of ways of
selecting 2, Republicans from 4
nCr = = 6.
The number of ways of
selecting 1 Democrat from 3 is
nCr = = 3.
So 4C2 ● 3C1 = 6 ●3 = 18.
We human beings are fond of gambling. When we
gamble, we bet money. Some of us do not just bet money.
We bet our life. We indulge in excessive drinking of
alcoholic drinks. We smoke excessively or even we tend to
drive exceeding the normal speed. We don’t care about
the risks when we are involved in these activities because
we don’t understand the concept of probability. On the
other hand, some of us may fear activities that involve
little risk to health or life because these activities have
been sensationalized by press and the media.
So at this point, we will learn the concept of
probability. Its meaning and how it is computed.
The theory of probability grew out of the
of various games of chance using coins, dice and cards.
So these devices will be used as
as a general concept
can be defined as
the chance of an
* Processes such as flipping a coin,
rolling a die, or drawing a card from a
* A probability experiment is a chance
process that leads to well-defined
results called outcomes.
* An outcome is the result of a single
trial of a probability experiment.
* A trial means flipping a coin once,
rolling a die once, or the like. When a
coin is tossed, there are two possible
outcomes: head or tail. In the roll of a
single die, there are six possible
outcomes: 1,2,3,4,5, or 6. In any
experiment, the set of all possible
outcomes is called the sample space.
* A sample space is the set of all
possible outcomes of a probability
Toss one coin
Roll a die
Answer a true-false
Toss two coins
coins are tossed, there are four
possible outcomes, as shown in the
fourth experiment above. Both coins
could fall heads up. Both coins could fall
tails up. Coin 1 could fall heads up
coin 2 tails up.
1. Find the sample space for
rolling two dice.
Since each die can land in six
different ways, and two dice are
rolled, the sample space can be
represented by a rectangular
array. The sample space is
list of pairs of numbers in the
Outcomes when two dice are rolled.
Since there are four suits
(hearts, clubs, diamonds, and
spades) and 13 cards for each
suit (ace through king), there are
52 outcomes in the sample
An event consists of the outcomes of a
An event can be one outcome or more than
one outcome. For example, if a die is rolled and
a 6 shows, this result is called an outcome, since
it is a result of a single trial. An event with one
outcome is called a simple event. The event of
getting an odd number when a die is rolled is
called a compound event, since it consists of
three outcomes or three simple events. In
general, a compound event consists of two
or more outcomes or simple events.
THREE BASIC TYPES OF
1. Classical probability
2. Empirical or relative frequency
3. Subjective probability
- uses sample spaces to determine the
numerical probability that an event will
happen. One does not actually have to
perform the experiment to determine that
probability. Classical probability is so
named because it was the first type of
mathematicians in the 17th and 18th
* Classical probability assumes that all
outcomes in the sample space are equally
likely to occur.
* Equally likely events are events that have
the same probability of occurring
Formula for Classical Probability
The probability of any event E is
This probability is denoted by
This probability is called
classical probability, and it uses
the sample space S.
Rounding Rule for Probabilities
Probabilities should be expressed as
reduced fractions or rounded to two
or three decimal places. When the
probability of an event is an
extremely small decimal, it is
permissible to round the decimal to
the first nonzero digit after the
1. For a card drawn from an
ordinary deck, find the
probability of getting a king.
Since there are 52 cards in a
deck and there are 4 kings,
P(king) = = .
2. If a family has three children,
find the probability that all
children are girls.
The sample space for the gender
of children for a family that has
three children is BBB, GBB, BBG,
BGB, GGG, GGB, GBG, and BGG.
Since there is one way in
possibilities for all three
children to be girls,
3. A card is drawn from an ordinary deck.
Find these probabilities.
a. Of getting a jack.
b. Of getting the 6 of clubs.
c. Of getting a 3 or a diamond.
a. There are 4 jacks and 52 possible
P(jack) = =
b. Since there is only one 6 of clubs, the
probability of getting a 6 of clubs is
P(6 of clubs) =
c. There are four 3s and 13
diamonds, but the 3 of diamonds is
counted twice in this listing. Hence,
there are 16 possibilities of drawing
a 3 or a diamond, so
P(3 of diamond) = =
There are four basic probability rules.
These rules are helpful in solving
probability problems, in understanding the
nature of probability, and in deciding
if your answers to the problems
Probability Rule 1
The probability of any event E is a number (either
a fraction or decimal) between and including 0
and 1. This is denoted by 0 < P(E) < 1
Rule 1 states that probabilities cannot be negative
or greater than one.
Probability Rule 2
If an event E cannot occur (i.e., the event contains
no members in the sample space), the probability
When a single die is rolled, find the probability of
getting a 9
Since the sample space is 1,2,3,4,5, and 6, it is
Probability Rule 3
If an event E is certain, then the probability of
In other words, if P(E) = 1, then the event E is
certain to occur.
When a single die is rolled, what is the
probability of getting a number less than 7?
Since all outcomes 1,2,3,4,5, and 6, are less
than 7, the probability is
P(number less than 7) = = 1
The event of getting a number less than 7 is
Probability Rule 4
The sum of the probabilities of the
outcomes in the sample space is 1.
In the roll of a fair die, each outcome in
the sample space has a probability of
1/6. Hence, the sum of the probabilities
of the outcomes is as shown.
complementary events. When a die is
rolled, for instance, the sample space
consists of the outcomes 1,2,3,4,5, and
6. The event E of getting odd numbers
consists of the outcomes 1,3, and 5. The
event of not getting an odd number is
called the complement of event E, and it
consists of the outcomes 2,4, and 6.
The complement of an event E is the set of
outcomes in the sample space that are not
included in the outcomes of event E. The
complement of E is denoted by Ē.
Find the complement of each event.
a. Rolling a die and getting a 4.
b. Selecting a letter of the alphabet and
getting a vowel.
c. Selecting a month and getting a month
that begins with a J.
d. Selecting a day of the week and
getting a weekday.
a. Getting a 1,2,3,5, and 6
b. Getting a consonant
c. Getting February, March,
April, May, August,
November, and December
d. Getting a Saturday and
The outcomes of an event and the
outcomes of the complement make
up the entire sample space. For two
coins are tossed, the sample space
is HH, HT, TH and TT. The
complement of “getting all heads” is
complement of the event “all heads”
is the event “getting at least one
Since the event and its complement make up
the entire sample space, it follows that the sum
of the probability of the event and the
probability of its complement will equal to 1.
That is, P(E) + P(Ē) = 1. In the previous example,
E = all heads, or HH, and let Ē = at least one
tail, or HT, TH, TT. Then P(E) = and P(Ē) = ;
hence, P(E) + P(Ē) = + = 1.
The rule for complementary events can be
stated algebraically in three ways.
Rule for Complementary Events
P(Ē) = 1 – P(E) or P(E) = 1 – P(Ē) or P(E) + P(Ē) = 1
Stated in words, the rule is: If the
probability of an event or the probability of
its complement is known, then the other can
be found by subtracting the probability from
1. This rule is important in probability theory
because at times the best solution to a
problem is to find the probability of the
complement of an event and then subtract
from 1 to get the probability of
the event itself.
If the probability that a person lives
in an industrialized country of the
world is , find the probability that a
person does not live in an
P (not living in an industrialized
= 1 – P (living in an industrialized
The difference between classical and empirical
probability is that classical probability assumes
that certain outcomes are equally likely while
empirical probability relies on actual experience to
determine the likelihood of outcomes. In empirical
probability, one might actually roll a given die 6000
times and observe the various frequencies and use
these frequencies to determine the probability of
an outcome. Suppose, for example, that a
researcher asked 25 people if they liked the taste
of a new soft drink. The responses were classified
as “yes” “no” or “undecided.” The results were
a frequency distribution, as
Probabilities now can be compared
for various categories. For example,
the probability of selecting a person
who liked the taste is , or , since 15
out of 25 people in the survey
Formula for Empirical Probability
Given a frequency distribution, the
probability of an event being in a given
P(E) = =
This probability is called
and is based on observation
1. In the soft-drink survey just described, find
the probability that a person responded “no.”
P(E) = =
2. In a sample of 50 people, 21 had type O
blood, 22 had type A blood, 5 had type B blood,
and 2 had type AB blood. Set up a frequency
distribution and find the following probabilities:
a. A person has type O blood.
b. A person has type A or type B blood.
c. A person has neither type A nor type O
d. A person does not have type AB blood.
a. P (O) = =
b. P (A or B) = + =
c. P (neither A nor O) = + =
d. P(not AB) = 1 - = =
- uses a probability value based on an educated guess
or estimate, employing opinions and inexact
In subjective probability, a person or group makes
an educated guess at the chance that an event will
occur. This guess is based on the person’s experience
and evaluation of a solution. For example, a
sportswriter may say that there is a 70% probability
that the ADMU Blue Eagle will win the UAAP next
year. A physician might say that on the basis of her
diagnosis, there is 30% chance that the patient will
need an operation. A seismologist might
is an 80% probability that an earthquake will occur
in a certain area.
Two events are mutually exclusive if they
cannot occur at the same time.
In another situation, the events of getting
a 4 and getting a 6 when a single card is
drawn from a deck are mutually exclusive
events, since a single card cannot be both a 4
and a 6. On the other hand, the events of
getting a 4 and getting a heart on a single
draw are not mutually exclusive, since one can
select the 4 of hearts when drawing a single
card from an ordinary deck.
1. Determine which events are mutually
exclusive and which are not when a single die is
a) Getting an odd number and getting an even
b) Getting a 3 and getting an odd number.
c) Getting an odd number and getting a number
less than 4.
d) Getting a number greater than 4 and getting a
number less than 4.
e) The events are mutually exclusive, since the
first event can be 1,3, or 5, and the second
event can be 2,4, or 6.
c) The events are not mutually exclusive since the first
event can be 1,3, or 5, and the second can be 1,2, or 3.
Hence, 1 and 3 are contained in both events.
d) The events are mutually exclusive, since the first
event can be 5 or 6, and the second event can be 1,2,
2. Determine which events are mutually exclusive and
which are not when a single card is drawn from a deck.
a) Getting a 7 and getting a jack.
b) Getting a club and getting a king.
c) Getting a face card and getting an ace.
d) Getting a face card and getting a spade.
Only the events in parts a and c are mutually
Addition Rule is used when the events are
Addition Rule 1
When two events A and B are mutually
exclusive, the probability that A or B will
P(A or B) = P(A) + P(B)
1. A restaurant has 3 pieces of apple pie, 5
pieces of cherry pie, and 4 pieces of pumpkin
pie in its dessert case. If a customer selects
a piece of pie for dessert, find the
probability that it will be either cherry or
Since there is a total 12 pieces of pie,
P(cherry or pumpkin) = P(cherry) +
The events are mutually exclusive.
2. At a political rally, there are 20
Republicans, 13 Democrats, and 6
Independents. If a person is selected at
random, find the probability that he or she
is either a Democrat or Independent.
P(Democrat or Independent) =
Addition Rule 2
If A and B are not mutually exclusive, then
P(A or B) = P(A) + P(B) – P(A and B)
Note: This rule can be also be used when the
events are mutually exclusive, since P(A and B)
will always equal 0. However, it is important to
make a distinction between the two situations.
1. In a hospital unit there are eight nurses
and five physicians. Seven nurses and
three physicians are females. If a
person is selected, find the probability that
subject is a nurse or a male.
The probability is
P(nurse or male) = P(nurse) + P(male) – P(male
nurse) = + - =
1.Determine whether the following events are
a.Roll a die: Get an even number, and get a
number less than 3.
b.Roll a die: Get a prime number, and get an
c.Roll a die: Get a number greater than 3,
and get a number less than 3.
d.Select a student in your class: The student
has blond hair, and the student has blue
e.Select a student in your college: The
student is a sophomore, and the student is
a business major.
2. On New Year’s Eve, the probability of a
person driving while intoxicated is 0.32,
the probability of a person having a
driving accident is 0.09, and the
probability of a person having a driving
accident while intoxicated is 0.06. What is
the probability of a person driving while
intoxicated or having a driving accident?
P(intoxicated or accident) = P(intoxicated)
+ P(accident) – P(intoxicated and
= 0.32 + 0.09 – 0.06 =
3. 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 an equal probability of being selected.
4. At a convention there are seven
mathematics instructors, five computer
science instructors, three statistics
instructors, and four science instructors. If an
instructor is selected, find the probability of
getting a science instructor or a math
5. An automobile dealer has 10 Fords, 7
Ferrari, and 5 Benz on his used-car lot. If a
person purchases a used car, find the
6. On a small college campus, there are five English
professors, four mathematics professors, two science
professors, three psychology professors, and three
history professors. If a professor is selected at random,
find the probability that the professor is the following:
a. An English or psychology professor
b. A mathematics or science professor
c. A history, science, or mathematics professor
d. An English, mathematics, or history professor
7. The probability of a California teenager owning a
surfboard is 0.43, of owning a skateboard is 0.38, and
of owning both is 0.28. If a California teenager is
selected at random, find the probability that he or she
owns a surfboard or skateboard.
8. On any given day, the probability of a tourist visiting
Indian Caverns is 0.80 and of visiting the Safari Zoo is
0.55. The probability of visiting both places on the
9. A single card is drawn from a deck. Find
the probability of selecting the following:
a.A 4 or a diamond
b.A club or a diamond
c.A jack or a black card
10. In a statistics class there are 18 juniors
and 10 seniors: 6 of the senior are females,
and 12 of the juniors are males. If a student
is selected at random, find the probability
of selecting the following:
d.A junior or a female
e.A senior or a female
f. A junior or a senior
11. A woman’s clothing store owner buys
from three companies: A, B, and C. The
most recent purchases
If one item is selected at random, find the
a.It was purchased from company A or
is a dress.
b.It was purchased from company B or
Two events A and B are independent if the fact that A
occurs does not affect the probability of B occurring.
Multiplication Rule 1
When two events are independent, the
probability of both occurring is
P(A and B) = P(A) * P(B)
1. A coin is flipped and a die is rolled. Find
the probability of getting a head on the
coin and a 4 on the die.
P(head and 4) = P(head) * P(4) = * =
A card is drawn from a deck and replaced; then a
second card is drawn. Find the probability of getting a
queen and then ace.
3. An urn contains three red balls, two blue balls, and
five white balls. A ball is selected and its color noted.
Then it is replaced. A second ball is selected and its
color noted. Find the probability of each of the following.
a. Selecting two blue balls
b. Selecting a blue ball and then white ball
c. Selecting a red ball and then a blue ball
a. P(blue and blue) = P(blue) * P(blue) = * = =
b. P(blue and white) = P(blue) * P(white) = * = =
c. P(red and blue) = P(red) * P(blue) = * = =
4. Approximately 9% of men have a type of
color blindness that prevents them from
distinguishing between red and green. If 3
men are selected at random, find the
probability that all of them will have this
type of red-green color blindness.
P(C and C and C) = P(C) * P(C) * P(C)
When the outcome or occurrence
of the first event affects the
outcome or occurrence of the
second event in such a way that
the probability is changed, the
events are said to be dependent.
Multiplication Rule 2
When two events are dependent, the
probability of both occurring is,
P(A and B) = P(A) * P(B/A)
1.In a shipment of 25 microwave ovens, 2 are
defective. If two ovens are randomly selected
and tested, find the probability that both are
defective if the first one is not replaced after it
has been tested.
P(D1 and D2) = P(D1) * P(D2/D1) = * = =
2. The World Wide Insurance Company
found that 53% of the residents of a city
had homeowner’s insurance with the
company. Of these clients, 27% also had
automobile insurance with the company.
If a resident is selected at random, find
the probability that the resident has both
homeowner’s and automobile insurance
with the World Wide Insurance Company.
P(H and A) = P(H) * P(A/H) = (0.53)(0.27) =
3. Box 1 contains two red balls and one blue
ball. Box 2 contains three blue balls and one
red ball. A coin is tossed. If it falls heads up,
box 1 is selected and a ball is drawn. If it falls
tails up, box 2 is selected and a ball is drawn.
Find the probability of selecting a red ball.
With the use of tree diagram, the sample
space can be determined. First, assign
probabilities to each branch. Next, using the
multiplication rule, multiply the possibilities
for each branch.
Finally, use the addition rule, since
a red ball can be obtained from box
1 or box 2.
P(red) = + = + =
The sum of all probabilities will
always be equal to 1.
The conditional probability of an event B in
relationship to an event A was defined as the
probability that event B occurs after event A has
The conditional probability of an event can be
found by dividing both sides of the equation for
multiplication rule 2 by P(A), as shown:
P(A and B) = P(A) • P(B/A)
Formula for Conditional Probability
The probability that the second event
B occurs given that the first event A has
occurred can be found by dividing the
probability that both events occurred by
the probability that the first event has
occurred. The formula is
1. A box contains black chips and white chips. A
person selects two chips without replacement. If the
probability of selecting a black chip and white chip is ,
and the probability of selecting a black chip on the
first draw is , find the probability of selecting the white
chip on the second draw, given that the first chip
selected was a black chip.
B = selecting a black chip
P(W/B) = =
= = • =
W = selecting a white
1. A box contains 24 transistors, four of which are
defective. If four are sold at random, find the following
a. Exactly two are defective.
b. None is defective.
c. All are defective.
d. At least one is defective.
There are 24C4 ways to sell four transistors, so the
denominator in each case will be 10,626.
a. Two defective transistors can be selected as 4C2 and
two nondefective ones as 20C2
P(exactly 2 defectives) = = =
b. The number of ways to choose no defective is
P(no defective) = = =
c. The number of ways to choose four defectives
from four is 4C4, or 1. Hence,
P(all defectives) = =
d. To find the probability of at least one defective
transistor, find the probability that there are no
defective transistors, and then subtract that
probability from 1
P(at least 1 defective) = 1 – P(no defectives)
=1- =1- =
A binomial experiment is a probability
experiment that satisfies the following four
1. Each trial can have only two outcomes or
outcomes that can be reduced to two
outcomes. These outcomes can be
considered as either success or failure.
2.There must be a fixed number of trials.
3.The outcomes of each trial must be
independent of each other.
4. The probability of success must remain
the same for each trial.
* A binomial experiment and its results give rise
to a special probability distribution called the
* The outcomes of a binomial experiment and
the corresponding probabilities of these
outcomes are called a binomial distribution.
* In binomial experiments, the outcomes are
usually classified as successes or failures. For
example, the correct answer to a multiplechoice item can be classified as a success, but
any of the other choices would be incorrect
and hence classified as a failure.
Notation for the Binomial Distribution
P(S) The symbol for the probability of
P(F) The symbol for the probability of
p the numerical probability of success
q The numerical probability of failure
P(S) = p and P(F) = 1 – p = q
n The number of trials
X The number of successes
Note that 0 < X < n.
probability of a success in a binomial experiment can be
computed with the ff. formula.
Binomial Probability Formula
In a binomial experiment, the probability of
exactly X successes in n trials is.
P(X) = px qn-X
1. A coin is tossed three times. Find the
probability of getting exactly two heads.
HHH, HHT, HTH, THH, TTH, THT, HTT,
This answer is or 0.375.
Looking at the problem in Example 1 from
the standpoint of a binomial experiment, one
can show that it meets the four
1. There are only two outcomes for each
trial, heads or tails.
2. There is a fixed number of trials (three).
3. The outcomes are independent of each
other (the outcome of one toss in no way
affects the outcome of another toss).
4. The probability of a success (heads)
is in each case.
In this case, n = 3, X = 2,p = . Hence, substituting in
the formula gives
P(2 heads) = ( ( = = 0.375
Which is the same answer obtained by using the
The same example can be used to explain the
formula. First, note that there are three ways to get
exactly two heads and one tail from a possible eight
ways. They are HHT, HTH, and THH. In this case,
then, the number of ways of obtaining two heads from
three coin tosses is 3C2, or 3. In general, the number
of ways to get X successes from n trials without
regard to order is
2. If a student randomly guesses at five
multiple-choice questions, find the
probability that the student gets exactly
three correct. Each question has five
In this case n = 5, X = 3, and p = , since
there is one chance in five of guessing a
correct answer. Then,
P(3) = ( = 0.05
Recall that in order for an experiment to be
binomial, two outcomes are required for each
trial. But if each trial in an experiment has more
than two outcomes, a distribution called the
multinomial distribution must be used. For
example, a survey might require the responses
of “approve,” “disapprove,” or “no opinion.” In
another situation a person may have a choice of
one of five activities for Friday night, such as a
movie, dinner, baseball game, play, or party.
Since these situations have more than two
possible outcomes for each trial, the binomial
distribution cannot be used to
The multinomial distribution can
be used for such situations if the
probabilities for each trial remain
constant and the outcomes are
independent for a fixed number of
trials. The events must also be
Formula for the Multinomial
If X consists of events E1, E2, E3, ….., Ek, which
have corresponding probabilities p1, p2, p3, …… pk
of occurring, and X1 is the number of times E1 will
occur, X2 is the number of times E2 will occur X3
is the number of times E3 will occur, etc., then
the probability that X will occur is
P(X) = P1X1 P2X2 …….. PkXk
Where X1 + X2 + X3 + … + Xk = n,
and P1 + P2 + P3 … + Pk = 1
1. In a large city, 50% of the people
choose a movie, 30% choose dinner and a
play, and 20% choose shopping as a
leisure activity. If a sample of five people
is randomly selected, find the probability
that three are planning to go to a movie,
one to a play, one to a shopping mall.
n = 5, X = 3, X = 1, X = 1, P = 0.50, P = 0.30,
and P3 = 0.20. Substituting in the formula gives
P(X) = (0.50)3 (0.30)1 (0.20)1 = 0.15
Again, note that the multinomial
distribution can be used even
though replacement is not done,
provided that the sample is small
in comparison with the population.
2. In a music store, a manager found
that the probabilities that a person buys
zero, one, or two or more CDs are 0.3,
0.6, and 0.1, respectively. If six
customers enter the store, find the
probability that one won’t buy any CDs,
three will buy one CD, and two will buy
two or more CDs.
Thus the multinomial distribution is similar to
the binomial distribution but has the advantage
of allowing one to compute probabilities when
there are more than two outcomes for each trial
in the experiment. That is, multinomial
distribution is a general distribution, and the
binomial distribution is a special case of the
When sampling is done without replacement,
the binomial distribution does not give exact
probabilities, since the trials are not
independent. The smaller the size of the
population, the less
accurate the binomial
probabilities will be.
For example, suppose a committee
of four people is to be selected from
seven women and five men. What is
the probability that the committee
will consist of three women and one
To solve this problem, one must
find the number of ways a committee
of three women and one man can be
selected from seven women and five
men. This answer can be found by
Next, find the total number of ways
a committee of four people can be
selected from 12 people. Again, by
the use of combinations, the answer
12C4 = 495
Finally, the probability of getting a
committee of three women and one
man from seven women and five men
P(X) = =
The results of the problem can be
generalized by using a special
probability distribution called the
hypergeometric distribution is a
distribution of a variable that has two
outcomes when a sampling is done
Formula for the Hypergeometric
Given a population with only two types of
objects (females and males, defective and
nondefective, successes or failures, etc.), such
that there are a items of one kind and b items of
another kind and a + b equals the total
population, the probability P(X) of selecting
without replacement a sample of size n with X
items of type a and n –X items of type b is
The basis of the formula is that
there are aCX ways of selecting the
first type of items, bCn-X ways of
selecting the second type of items,
and (a+b)Cn ways of selecting n items
from the entire population.
1. Ten people apply for a job as assistant
manager of a restaurant. Five have
completed college and five have not. If
the manager selects three applicants at
random, find the probability that all
three are college graduates.
Assigning the values to the variable gives
a = 5 college graduates
b = 5 nongraduates
and n – X = 0. Substituting in the formula
2. A recent study found that four out of
nine houses were underinsured. If five
houses are selected from the nine
houses, find the probability that exactly
two are underinsured.
In this problem
P(X) = = =
1. When two dice are rolled, find the probability of
a. A sum of 6 or 7
b. A sum greater than 8
c. A sum less than 3 or greater than 8
d. A sum that is divisible by 3
e. A sum of 16
f. A sum less than 11
2. The probability that Sam will be accepted by the
college of his choice and obtain a scholarship is 0.35. If
the probability that he is accepted by the college is 0.65,
find the probability that he will obtain a scholarship
given that he is accepted by the college.
3. The probability that John has to work overtime and it
rains is 0.028. John hears the weather forecast, and
there is a 50% chance of rain. Find the probability that
geology professor has 5 silicates, 7 pyrites and 8
carbonates in a rock collection. He picks 6 rocks at
random for a student to analyze. Find the probability
that the professor picked 2 silicates, 1 pyrite and 3
carbonates. Express your answer in lowest fraction.
If there are 20 rocks in all, the number of ways picking 6 of these is
= = 38,760
The number of ways of picking 2 silicates, 1 pyrite and 3
Hence, the probability of picking these rocks is or approximately
Russian roulette is played by two players as
follows. One bullet is placed into empty cylinder
of a six-gun and the cylinder is spun. The first
player holds the gun to his head and pulls the
trigger. If he does not die, he wins the game. If he
dies, then the second player takes his turn by
spinning the cylinder, putting the gun to his head,
and pulling the trigger. If he dies the first player
wins. If he does not die, then the game is played
again. What is the probability that the second
player will die?
Probability of the second player will die is (5/6) (1/6)
Hint: Compute the probability that the first player
A student elects to solve a problem about
compound interest using either arithmetic
methods, algebraic methods or calculus
methods with corresponding probabilities of
2/9, 4/9 and 1/3. With each of these methods,
the respective probabilities of obtaining the
correct answer are 9/10, 7/8 and 3/4 .
Determine the probability that the student
obtains the correct answer.
P(correct) = (choosing arithmetic and being
correct OR choosing algebra and being
correct OR choosing calculus and being
Debra is taking two college entrance exams, in English
and in Math. The probability that she will pass the
English exam is 0.75. The probability that she will fail
the Math is 0.20. The probability that she will pass
both exams is 0.65. What is the probability that Debra
will pass either the Math or the English exam?
P(A or B) = P(A) + P(B) – P(A and B)
P(English or Math) = P(English) + P(Math) – P(Math
P(English) = 0.75
P(not Math) = 0.20
P(Math) = 1 – 0.20 = 0.80
P(English or Math) = 0.75 + 0.80 – 0.65
= 1.55 – 0.65 = 0.90
In how many ways can a committee of 5
be formed with 3 seniors and 2 juniors
as its members if there are 6 seniors
and 5 juniors to choose from?
C3 * 5C2 = 20 * 10 = 200
“We cannot hope that many children will
learn MATHEMATICS unless we find a way to
share our enjoyment and show them its
beauty as well as its utility.”