What Is This Module About?
Probability is a concept we make use of every day. It was an outgrowth of a study on
gambling and games of chance. Today, probability is still used to help people understand
games of chance such as lotto and sweepstakes. But there are certainly many other cases
where probability is of great use.
This module will help you determine what is possible and what is probable in order to
make intelligent predictions or decisions.
This module is made up of three lessons:
Lesson 1—Possibilities
Lesson 2—Basic Concepts of Probability
Lesson 3—Probability of Simple and Compound Events
What Will You Learn From This Module?
After studying this module, you should be able to:
♦
state and apply the fundamental principle of counting;
♦
differentiate permutation from combination;
♦
apply the concepts of permutation and combination to real-life situations;
♦
discuss and illustrate the probability of simple and compound events;
♦
calculate probabilities; and
♦
state and apply the laws of probability in everyday life.
Let’s See What You Already Know
Before you start studying this module, take the following test first to find out how well
you know the topics to be discussed.
Read each problem carefully, then encircle the letter of the correct answer.
1.
A canteen offers a special meal consisting of a meat dish (using one of 5 different
meats), one of four different types of vegetables and one of three different
beverages. In how many ways can a person choose one of these combinations?
a.
2.
60
b. 12
c.
20
d.
15
A telephone system is established in Barangay 707 with the numbers beginning
with 827. How many telephone numbers of the form 827-XXXX are possible?
a. 10000
b. 5040
c. 210
1
d. 1000
3.
If there are 12 contestants in a beauty contest during a town fiesta in Bulacan, in
how many ways can the judges award first and second prizes?
a. 132
4.
b. 25
8.
c. 120
d. 100
b. 336
c.
24
d.
56
A tray of eggs (containing 12 eggs) has two that are defective. If a customer buys
four eggs, what is the probability that not one egg is defective?
a. 0.25
7.
d. 12
A pizza parlor offers 8 different toppings for a pizza. In how many ways can a
customer select three toppings for his/her pizza?
a. 11
6.
c. 24
In how many different ways can a librarian arrange 5 books on a shelf?
a. 5
5.
b. 66
b. 0.42
c. 0.33
d. 0.5
Classify this situation: Jose arranges three of his six reference books on his desk.
This is a problem in _______________.
a.
the fundamental principle of counting
c.
combination
b.
permutation
d.
probability
A bingo caller has a machine that contains 75 balls. The balls are marked in the
following manner:
B1, B2, B3, . . . B15
I16, I17, I18, . . . I30
N31, N32, N33, . . . N45
G51, G52, G53, . . . G60
O61, O62, O63, . . . O75
The balls are mixed and a ball is drawn at random. What is the probability of
drawing a ball that has a B on it?
a. 1/75
9.
c. 1/15
d. 1/60
Barangay Matulungin sold 1500 raffle tickets at P10 each for the construction of
their barangay hall. The first prize is a colored TV set. If Aling Maring buys 5
tickets, what is the probability of her getting the first prize?
a. 1/300
10.
b. 1/5
b. 1/1500
c. 1/5
d. 1/30
If one letter is chosen at random from the word statistics, what is the probability
that the letter is a vowel?
a. 1/10
b. 3/10
c. 7/10
2
d. 2/10
11.
There are 300 houses in a neighborhood. On a certain evening 100 of the
residents there are not at home. What is the probability that when a person
conducting a telephone survey calls on that particular evening, he/she will call a
house in which someone is at home?
a. 1/3
12.
b. 75%
c. 95%
d. 100%
b. 0.08
c. 0.06
d. 0.32
If 80 of 140 moviegoers who watched a certain movie said that they would like to
see the movie again, what is the probability that any moviegoer who saw the said
movie would like to see it again?
a. 0.07
15.
d. 1/300
Lorna was born in October. What is the probability that she was born on the 10th?
a. 0.03
14.
c. 1/100
The owner of a carinderia found that 75% of the customers order a soup with
their meal, 60% order two cups of rice and 40% order both. What is the
probability that a customer will order either a soup or two cups of rice?
a. 60%
13.
b. 2/3
b. 0.01
c. 0.57
d. 0.43
If the probability that a certain person will go out for breakfast is 0.40, what is the
probability that a certain person will not go out for breakfast?
a. 0.6
b. 1
c. 0.16
d. 0.3
Well, how was it? Do you think you fared well? Compare your answers with those in
the Answer Key on page 43 to find out.
If all your answers are correct, very good! This shows that you already know much
about the topics in this module. You may still study the module to review what you already
know. Who knows, you might learn a few more new things as well.
If you got a low score, don’t feel bad. This means that this module is for you. It will
help you understand some important concepts that you can apply in your daily life. If you
study this module carefully, you will learn the answers to all the items in the test and a lot
more! Are you ready?
You may now go to the next page to begin Lesson 1.
3
LESSON 1
Possibilities
The simple process of counting plays an important role in our daily life, particularly in
business. One has to count for example when determining the number of votes a certain
candidate in a certain town received or when preparing halu-halo with scoops of different
ingredients. Sometimes, the process of counting is simplified by using mechanical devices
(like a mechanical counter) or by counting indirectly (for instance, by subtracting the serial
numbers of invoices to determine the total number of sales).
You can hardly predict the outcome of a kagawad election unless you know who the
candidates are and you cannot predict very well what songs will be in the list of the top 10
most popular songs unless you know at least which one is in the market. Generally, you
cannot make predictions or decisions unless you know at least what is possible.
After studying this lesson you should be able to:
♦
state and apply the fundamental principle of counting;
♦
differentiate permutations from combinations; and
♦
apply the concepts of permutations and combinations in solving problems in daily
life.
Let’s Study and Analyze
A school canteen cook has to determe how many breakfast meals consisting of a fruit, a
sandwich and a drink are possible if she can select from 4 kinds of fruits (banana, papaya,
mango and pineapple), 3 kinds of sandwiches (cheese, ham and egg sandwiches) and 2 kinds
of drinks (coffee and juice).
4
One method of solving such a problem is through the use of a tree diagram. A tree
diagram consists of a number of “branches” that represent the possible outcomes of the
activity. The possibilities may be read directly from the branches.
Solving the problem through the aid of a tree diagram, you get:
Fruit
Possible Meals
cheese
coffee
juice
(banana, cheese, coffee)
(banana, cheese, juice)
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
coffee
juice
s
s
coffee
juice
s
egg
s
pineapple
coffee
juice
s
s
ham
s
cheese
s
s
coffee
juice
s
s
egg
s
s
s
coffee
juice
s
s
ham
s
s
s
mango
coffee
juice
s
s
s
cheese
coffee
juice
s
s
egg
s
s
s
coffee
juice
s
s
ham
coffee
juice
s
s
s
papaya
coffee
juice
s
s
s
cheese
coffee
juice
s
egg
s
s
s
ham
s
s
s
banana
s
Drink
s
s
Sandwich
(banana, ham, coffee)
(banana, ham, juice)
(banana, egg, coffee)
(banana, egg, juice)
(papaya, cheese, coffee)
(papaya, cheese, juice)
(papaya, ham, coffee)
(papaya, ham, juice)
(papaya, egg, coffee)
(papaya, egg, juice)
(mango, cheese, coffee)
(mango, cheese, juice)
(mango, ham, coffee)
(mango, ham, juice)
(mango, egg, coffee)
(mango, egg, juice)
(pineapple, cheese, coffee)
(pineapple, cheese, juice)
(pineapple, ham, coffee)
(pineapple, ham, juice)
(pineapple, egg, coffee)
(pineapple, egg, juice)
1.
Can you give other possible meals besides the meals listed in the tree
diagram?_________________________________________________________
2.
How many different meals are there?____________________________________
5
To answer the first question on the previous page, let’s prepare another tree diagram
with the type of sandwich in the first column, fruit in the second and drink in the third. The
tree diagram would look like this:
Possible Meals
banana
coffee
juice
(cheese, banana, coffee)
(cheese, banana, juice)
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
coffee
juice
s
s
pineapple
s
s
s
coffee
juice
s
s
mango
coffee
juice
s
s
s
coffee
juice
s
s
s
papaya
egg
coffee
juice
s
s
banana
s
pineapple
s
s
s
coffee
juice
s
s
mango
s
s
s
ham
coffee
juice
s
s
s
papaya
coffee
juice
s
s
banana
coffee
juice
s
pineapple
s
s
coffee
juice
s
mango
s
s
cheese
coffee
juice
s
s
s
papaya
s
Drink
s
Fruit
s
Sandwich
(cheese, papaya, coffee)
(cheese, papaya, juice)
(cheese, mango, coffee)
(cheese, mango, juice)
(cheese,pineapple,coffee)
(cheese, pineapple, juice)
(ham, banana, coffee)
(ham, banana, juice)
(ham, papaya, coffee)
(ham, papaya, coffee)
(ham, mango, coffee)
(ham, mango, juice)
(ham, pineapple, coffee)
(ham, pineapple, juice)
(egg, banana, coffee)
(egg, banana, juice)
(egg, papaya, coffee)
(egg, papaya, juice)
(egg, mango, coffee)
(egg, mango, juice)
(egg, pineapple, coffee)
(egg, pineapple, juice)
Notice that if you compare the tree diagram above with the first tree diagram, you will
find that the possible meal combinations are the same. This means that the order in which
you arrange the diagram does not matter because you will still have the same combinations
of meals.
To answer question 2, we will count the number of different meal combinations. We
have a total of 24 meal combinations.
Recall the given facts—you have 4 fruits, 3 sandwiches and 2 drinks to choose from.
Multiplying, we get 4 × 3 × 2 = 24.
The answer is 24 which is equal to the number of meal combinations we arrived at
using the tree diagram.
6
Hence another way of determining the number of different ways an event can happen is
to multiply the number of ways one event can occur by the number of ways the second event
can occur and the number of ways the third event can occur.
In the given example of breakfast meal combinations, we have the following:
Event
Number of Ways It Can Occur
Fruits
first
4
Sandwiches
second
3
Drinks
third
2
Multiplying the number of ways each event can occur will give us: 4 × 3 × 2 = 24.
Let’s Learn
What we have just shown is the use of the fundamental principle of counting (FPC). In
general, the principle can be stated as: If an activity has n different outcomes, a second
1
activity has n different outcomes and a third activity has n different outcomes, then the
2
3
first, second and third activities performed together have n × n × n different outcomes.
1
2
3
This concept may be extended if there are other events to follow. Thus, you would have
n × n × n × . . . n outcomes, where k is the number of activities. The list of all
1
2
3
k
outcomes is called sample space while the elements are called sample points. Sample
space is usually denoted by S.
EXAMPLE 1
A bus service has 2 buses and 5 drivers. In how many different ways can a driver and
a bus be assigned to a job?
What is the sample space n1?__________ buses
What is the sample space n2?__________ drivers
7
a.
Using a tree diagram, you get:
s
s
driver C
s
(bus 1, driver C)
s
driver D
s
(bus 1, driver D)
s
driver E
s
(bus 1, driver E)
s
(bus 2, driver A)
s
(bus 2, driver B)
s
(bus 2, driver C)
s
(bus 2, driver D)
s
(bus 2, driver E)
driver A
driver B
driver C
s
(bus 1, driver B)
s
driver B
s
(bus 1, driver A)
s
driver A
s
bus 2
Possible Job
s
bus 1
Driver
s
Bus
driver D
s
driver E
How many jobs are possible? ____________
If you answered 10, you are right. This is because following the FPC, we have 2
events; the first event will have 2 occurrences and the second event will have 5 occurrences.
So there are 2 × 5 = 10 possible assignments of drivers and buses.
Let’s have another example:
EXAMPLE 2
In a recently concluded election, the following criteria were used for
choosing a senatorial candidate. A candidate must be: (1) male or
female Filipino, (2) from Luzon, Visayas or Mindanao and (3) may
belong to the LP, NP, GAD, UNIDO or PPC. How many
different kinds of candidates are possible using these three sets of
criteria?
SOLUTION
Using the FPC, we have:
Sex
2
Area
×
Political Party
3
×
8
5
Possibilities
=
30 distinct
candidates
Let’s Try This
NFE learners in Manilag Learning Center are required to undergo medical checkup at
the barangay health center once a year. Find the number of ways in which a learner may be
classified if the categories include (1) blood type ( AB, A, B or O), (2) sex (male or female)
and (3) blood pressure (high, normal or low).
Compare your answer with the one in the Answer Key on page 43.
Let’s Study and Analyze
Five friends—Mario, Carlo, Rhea , Ana and Joy—took a jeepney in going home from a
learning session. There were exactly five empty seats in the vehicle. In how many ways
could they seat themselves?
Let us assume that they boarded the jeepney in the sequence given above, that is, Mario
first, followed by Carlo, Rhea, Ana and Joy. When Mario boarded the jeepney, there were
five seating choices; when Carlo got inside, there were four choices left; for Rhea, three
choices left; for Ana, two choices left and Joy will have no choice but take the last
unoccupied seat. Therefore, by FPC, the number of possible arrangements is:
5 × 4 × 3 × 2 × 1 = 120
9
Let’s Learn
In mathematics, we call the product of a succession of positive integers from 1 to the
greatest value such as 5 × 4 × 3 × 2 × 1 a factorial. Generally a factorial is symbolized by
n! where n is the greatest integer in the set of factors. Thus, in the example, the factorial 5 ×
4 × 3 × 2 × 1 is symbolized by 5!, read as “five factorial.”
n! = n × (n – 1) × (n – 2) × . . . . × 2 × 1
By definition, 1! = 1 and 0! = 0
Other rules of counting deal with the number of arrangements of items considering the
positions of the items.
Take a look at the example below:
If you are given the following four letters of the English alphabet:
A
B
E
T
How many different three-letter arrangements can be made with these letters, given that
each can be used only once?
a.
Using a tree diagram, you get
1st letter
2nd letter
s
s
s
s
s
s
s
s
s
B
A
10
s
s
A
E
s
s
B
E
s
B
A
s
s
E
s
s
s
A
T
s
s
B
s
s
s
T
B
T
s
s
s
A
A
T
s
s
T
s
s
s
s
s
B
s
s
s
E
s
s
s
A
A
E
s
s
E
s
s
E
T
s
s
T
B
E
s
s
s
B
B
T
s
s
s
A
E
T
s
T
s
A
3-letter arrangement
s
s
s
s
E
s
s
B
3rd letter
ABE
ABT
AEB
AET
ATB
ATE
BAE
BAT
BTA
BTE
BEA
BET
EAB
EAT
EBA
EBT
ETB
ETA
TAB
TAE
TBA
TBE
TEA
TEB
Each of the arrangements is different from the others. The letters are in a definite order
and the order in which the letters are arranged is important. When you have a group of things
arranged in a definite order, you have a permutation. A permutation is the number of ways
by which the elements of a given set are arranged in a definite order.
A key word in the definition of permutation is the word order. In determining the
number of possible permutations in a given situation, order counts.
For instance, the arrangement AB is different from the arrangement BA. Therefore,
rearranging the order of two objects creates a new permutation.
Permutations may use all the given elements of a given set (as we did in the previous
example) or only a certain number of these elements.
When there are n items to be arranged all at a time, and where none of these items are
repeated, nPn = n!
where left subscript n = total number of items; and
right subscript n = number of items taken at a time.
Suppose you have n things and you want to count the possible arrangements using r of
them at a time. The formula is
n
Pr =
n!
(n − r )!
where n! is read as n factorial and is equal to n (n – 1) (n – 2) …1 or the
product of all the numbers from n down to 1.
Let’s Try This
In how many ways may the positions of president and vice-president be given to 5
nominees?
How will we apply the formula in finding the number of permutations to this problem?
What is n?
What is r?
Compare your answer with the one below:
n
Pr =
n!
(n − r )!
Substituting the given values in the formula, you get
P =
5 2
5!
5! 5 × 4 × 3 × 2 × 1
= =
= 20
(5 − 2 )! 3!
3 × 2 ×1
Therefore there are 20 different permutations.
11
Try this one:
In a singing contest with 8 contestants, in how many ways can a judge award the first,
second and third prizes?
In using the formula for finding the number of permutations of n objects taken r
at a time, that is,
n
Pr , what is n? __________ What is r? __________
Using n = 8 and r = 3 in the formula, you have
n
Pr =8 P3 =
8!
8!
=
(8 − 3)! 5!
8 × 7 × 6 × 5 × 4 × 3 × 2 ×1
5 × 4 × 3 × 2 ×1
P = 8×7×6
P = 336
P=
Hence, there are 336 different ways of awarding the first, second and third
prizes.
In some cases where the items are not all distinct (meaning some items are the same)
then the number of permutations will be reduced. The formula is:
n
P(x1, x2 ... xn ) =
n!
x1! x2!... xn !
where: x1 are items of one kind
x2 are items items of another kind
. . . xn are similar items of other kinds
n is the number of items
Let’s Try This
1.
How many distinct permutations are possible with the letters in the word
mathematics?
SOLUTION
How many letters are there in the word? __________
How many ms are there? __________
How many ts are there? __________
How many letter as are there? __________
The remaining letters appear only once. These are e, h, i, c and s.
Using the given formula,
12
11
2.
11!
2!2!2!
11 × 10 × 9 × 8 × 5 × 7 × 6 × 5 × 4 × 3 × 2 × 1 399168
=
=
(2 × 1)( 2 × 1)( 2 × 1)
64
= 4989600
P(2, 2, 2 ) =
In how many ways can 3 red bulbs, 4 yellow bulbs and 3 green bulbs be arranged
on a Christmas tree using 10 bulb sockets?
SOLUTION
Applying the formula for the number of permutations when the
objects are not all alike, you get
10
3.
10!
3!4!3!
10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
=
(3 × 2 × 1)( 4 × 3 × 2 × 1)(3 × 2 × 1)
10 × 9 × 8 × 7 × 6 × 5
=
(3 × 2 × 1)( 3 × 2 × 1)
151200
=
36
= 4200
P(3, 4,3) =
How many different permutations can be formed from the word Philippines?
SOLUTION
How many letters are there in the word Philippines? __________
How many different letters are there? __________
How many ps are there? __________
How many is are there? __________
Substituting these numbers in the formula, you get:
11
11!
3!3!
11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
=
(3 × 2 × 1)(3 × 2 × 1)
11 × 10 × 9 × 8 × 7 × 6 × 5 × 4
=
3 × 2 ×1
6652800
=
6
= 1108800
P(3,3) =
Therefore, there are 1108800 different permutations.
With permutations, one is very particular in counting the arrangements. Let us consider
a slightly different type of choice.
13
Let’s Study and Analyze
Suppose there are 3 beginner ballroom dancers—Lolit, Fe and Jenny. The dance
instructor (DI) will teach them individually. In other words, the three girls will each
become a partner of the DI and of each other. In how many ways could a double pair be
formed?
This type of problem is different from the one involving permutations because when a
pair is formed, it does not matter which person is chosen first and which person is chosen
second. Hence, the order of choice is not important. There are 12 different arrangements.
They are:
DI, Lolit
Lolit, DI
Fe, DI
Jenny, DI
DI, Fe
Lolit, Fe
Fe, Lolit
Jenny, Fe
DI, Jenny
Lolit, Jenny
Fe, Jenny
Jenny, Lolit
If you eliminate the duplicate pairs, you will have:
DI, Lolit
Lolit, Fe
DI, Fe
Lolit, Jenny
DI, Jenny
Fe, Jenny
Hence, there are 6 different pairs that can be formed.
The problem you have just solved is a combination problem. A combination is a
distinct group of objects where the arrangement or order is not taken into consideration.
Committees, groups, teams and sets are good examples of combinations. You are only
concerned with who is in a committee, not with who is first, who is second and so on.
It is not always possible to list all the possible combinations of n objects taken r at a
time. There is a formula used to determine the number of combinations of n things taken r
at a time. That is,
C=
n!
(n − r )! r!
read as “the number of combinations of n things taken r at a time.”
Let us try to apply this formula to the case of the 3 ballroom dancers and the dance
instructor. Here, you are actually finding the number of combinations of 4 dancers taken 2 at
a time. Using the formula, you get
n = 4 and r = 2
C=
n!
(n − r )! r!
C=
4!
(4 − 2 )!2!
14
4!
2!2!
4 × 3 × 2 ×1
=
(2 × 1)(2 × 1)
12
=
2
=6
C=
There are 6 possible pairs.
Let’s See What You Have Learned
Solve the following problems. Write your answers on a separate sheet of paper.
1.
List all the different possible outcomes when a coin and a die are tossed. How
many outcomes are there?
2.
In how many ways can 5 candidates be listed on a ballot?
3.
Aling Emma has a sari-sari store on Calle 107. Every Saturday she makes an
inventory of her stocks and places her order the following Monday. She can place
her order by telephone, by sending somebody through a tricycle or by herself
requesting in each case that her order be confirmed by telephone or by sending
somebody. Draw a tree diagram to show the various ways she can place and
confirm her order.
4.
Mang Nardo wants to buy a new television set. When he arrives at the appliance
store, he finds out that there are 6 different television brands available in 3 colors
and 5 sizes. How many choices does he have?
5.
A carinderia offers a promo package in which a buyer can order any one of 6
different viands, any one of 4 vegetable dishes and any one of the 5 kinds of drinks
for P85 only. How many different choices does a buyer have?
6.
A store has prepared 5 different types of window display for Christmas. There
are only 3 display windows. Will there be a sufficient number of display
arrangements?
7.
Among the 12 nominees for the board of directors of a farm cooperative, there are
7 men and 5 women. In how many ways can the members elect as directors
8.
a.
any 3 of the nominees?
b.
two of the male nominees and one female nominee?
c.
three female nominees?
In how many ways can a cook select 4 recipes from 6 equally tempting recipes?
15
9.
In how many ways can a chairman, a vice chairman and a secretary be selected
from a group of 14 members of a parish religious committee? Note that the order
of selection is important here.
10.
The hermana mayor of a town fiesta has to present a cultural show consisting of a
singing contest, dance contest, short plays and a band competition. In how many
ways will these happen if there are 5 singing contestants, 3 dance contestants, 3
plays and 5 competing bands?
Compare your answers with those in the Answer Key on pages 43 to 45. Did you get a
perfect score? If you did, that’s very good. If you did not, that’s okay. Just review the parts of
the lesson you did not understand very well. Afterward, you may move on to Lesson 2.
Let’s Remember
♦
The fundamental principle of counting states that if an activity n1 has different
outcomes, a second activity n2 has different outcomes and a third activity n3 has
different outcomes, then the three activities performed together have n1 × n2 × n3
different outcomes.
♦
Each possible outcome of an activity/experiment is called a sample point and the
total number of possible expected outcomes is called the sample space.
♦
Permutations are the total number of ways by which items may be arranged in
definite order.
♦
When there are n items are to be arranged all at a time, and where none of these
items are repeated, the number of permutations is computed by taking n! The
formula is
n
♦
Pn = n!
Where there are n items taken r at a time with no items repeated, the number of
permutations is computed through the formula
n
Pr =
n!
(n − r )!
♦
Combinations also refer to the number of ways a set of items or objects may be
arranged. However, the order is not taken into consideration.
♦
If we have n items and we take them r at a time and we don’t consider order, we
use the following formula:
n
Cr =
n!
(n − r )! r!
16
LESSON 2
Basic Concepts of Probability
So far you have learned the possible outcomes of any given event. In some instances,
you listed all possibilities and in others, you determined how many different possibilities
there are. Now you will go one step further, that is, you will determine what is probable and
what is improbable.
The most common way of measuring the uncertainties connected with events (say, the
results of an election, what the weather will be tomorrow or the number of crates of
mangoes that a mango tree yields) is to assign them probabilities. Hence, probability is the
ratio of the number of outcomes in a set of equally likely outcomes that produce a given
event to the total number of possible outcomes.
Probability is a mathematical concept whose applications cover nearly every area of
human activity. The probability theory is used in medical statistics, physics, astronomy,
meteorology, game theory, decision making in the business world and so on.
Probability has been studied in mathematics for a long period of time. The concept of
probability was first formally studied in the sixteenth century. At that time, it was the
outgrowth of a study on gambling and games of chance.
In the probability theory, one seeks answers to questions like:
“If a coin is flipped, what are the chances of getting heads?”
“If a die is rolled, what is the probability of getting a 3?”
“What is the probability of hitting the jackpot prize in a mega-lotto draw?”
“What is the probability of catching more tilapia under a half-moon?”
In this lesson you will learn the basic concepts of probability. You will learn how they
can be used to make choices and decisions.
After studying this lesson, you should be able to:
♦
know the concept of probability and discuss the classical approach to probability;
♦
calculate probabilities; and
♦
state and apply the laws of probability in everyday life situations.
17
Let’s Study and Analyze
EXAMPLE 1
Marlene is playing solitaire with a standard deck of cards. A standard deck consists
of 52 cards, with 13 cards in each of the 4 suits. The suits are clubs, diamonds, hearts and
spades. Clubs and spades are black while diamonds and hearts are red. The 13 cards in
each suit are ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, jack, queen and king.
Marlene mixes up the cards and picks a card at random from the deck. At random
means that each card is equally likely to have been chosen. What is the
probability of drawing a red card?
SOLUTION
There are 52 cards in a standard deck.
There are 26 red cards, 13 from the deck of diamonds and 13 from the deck of hearts.
So the probability of drawing a red card is
26
1
or .
52
2
EXAMPLE 2
If you toss a die, what is the probability of getting a 3?
SOLUTION
The set of all possible outcomes of tossing a die consists of the following:
So there are 6 possible outcomes and the probability of getting a 3 is
18
1
.
6
The set of all possible outcomes of an activity is called sample space, usually denoted
by S.
For instance, the sample space of the outcomes of tossing a die consists of
S = {1, 2, 3, 4, 5, 6}
While the sample space of flipping a coin is
S = {heads, tails}
An event is a subset of a sample space.
For example,
♦
When a pair of dice is tossed, the occurrence of the two dice totaling 7 is an event.
♦
When two coins are tossed, the outcome of having both tails is also an event.
♦
When you’re checking what the weather will be like, that it will be rainy is an
event.
EXAMPLE 3
A mailbox contains 6 bills and 4 letters. One piece of mail is chosen at random. What
is the probability that a bill is chosen?
SOLUTION
How many bills and letters are there in the mailbox?__________
How many bills are there?__________
If you answered there are 10 letters and 6 bills you are correct. Hence, the
probability that a bill is chosen is
6
3
or .
10
5
Let’s Learn
The probability that a certain event A will occur is equal to the ratio of the number of
outcomes in A to the number of outcomes in the sample space.
P( A) =
number of successful/favorable outcomes
total number of possible outcomes
P( A) =
n( A)
n( S )
This is called classical probability.
The probability that an event will occur is assigned a value which ranges from 0 to 1.
An event that cannot happen has a probability of 0. An event that will certainly happen has a
probability of 1.
19
Consider a box containing 3 red balls, 5 black balls and 2 green balls. A ball is drawn
at random. What is the probability of
a.
drawing a red ball?
b.
drawing a white ball?
SOLUTION
How many balls are there in the box? __________
How many red balls are there in the box? __________
Using classical probability, what is the probability of drawing a red ball? __________
3
.
10
Likewise, the probability of drawing a white ball is 0 because there is no white ball in
the box.
Are your answers correct? The probability of drawing a red ball is
Two events, A and B, are said to be mutually exclusive or disjoint if they cannot occur
together.
Examples of mutually exclusive events are:
♦
The two events “heads come up” and “tails come up” in a single toss of a coin are
mutually exclusive events or disjoint because they cannot occur together at the
same time. In other words, it is impossible to have both heads and tails in a single
toss of a coin.
♦
Similarly, the two events “drawing a jack” and “drawing a queen” are also
mutually exclusive events because you cannot have a jack and a queen at the same
time in one draw from a deck of cards.
The probability of the occurrence of either A or B where both are mutually exclusive
events is the sum of their individual probabilities. Thus,
P(A or B) = P(A) + P(B)
This rule can be extended to the case of more than two mutually exclusive events.
Let’s Try This
1.
Two dice are tossed. What is the probability of getting a sum of 7 or 11?
SOLUTION
When a pair of dice is tossed, how many possible outcomes do you
expect to get? __________
Did you answer 36 possible outcomes?
Can you list these 36 possible outcomes in a table?
20
The following table shows two possible outcomes. You can work on the remaining 34
outcomes.
1st 2nd
1
1
(1, 1)
2
3
4
2
(3, 4)
3
4
5
6
The entry (1, 1) means 1 on the first die and 1 on the second. Similarly, the entry (3, 4)
means 3 on the first die and 4 on the second.
Compare your answers with those in the Answer Key on page 45.
Based on the table given above, what are the outcomes whose sum is 7?
_____________________________________________________________________
Likewise, what are the outcomes whose sum is 11?
_____________________________________________________________________
Is there an outcome whose sum is 7 and at the same time 11?
_____________________________________________________________________
Getting a sum of 7 and getting a sum of 11 in tossing a pair of dice are mutually
exclusive events.
Hence,
P (sum of 7 or 11) = P (sum of 7) + P (sum of 11)
But the event of getting a sum of 7 consists of
{(1,6), (6,1), (2,5), (5,2), (3,4),(4,3)}
while the event of getting a sum of 11 consists of
{(5, 6), (6, 5)}
So,
P (sum of 7 or 11) =
6
36
21
+
2
36
=
8
2
or
36
9
Let’s Review
1.
The probabilities that a certain television show will get a rating of excellent, very
satisfactory, satisfactory, poor and needs improvement are 0. 15, 0.35, 0.10, 0.25,
and 0.15, respectively. What is the probability that the television show will get
one of the ratings?
2.
There are 12 books on a shelf—6 math books, 4 science books and 1 dictionary. If
a book is selected at random, what is the probability that it is a math book, a
science book or a dictionary?
Compare your answers with those in the Answer Key on pages 45 and 46.
Let’s Learn
If two events A and B can occur together, they are said to be nonmutually exclusive
events.
For instance, the two events drawing a queen and drawing a heart from a deck of 52
playing cards are nonmutually exclusive events since both can occur together. In other
words, there is a card that is both a queen and a heart.
The probability of the occurrence of either A or B is equal to the sum of their individual
probabilities minus their joint probability. Thus,
P(A and B) = P (A) + P(B) – P(A and B)
Again this rule can be extended to the case of more than two nonmutually exclusive
events.
Consider the following examples.
EXAMPLE 1
What is the probability of drawing a queen or a heart from a
well-shuffled deck of 52 playing cards?
SOLUTION
From the deck of 52 playing cards, how many cards are queens?
4 cards are queens.
So the probability of drawing a queen is
P(queen) =
4
52
How many are hearts?
There are 13 cards that are hearts.
Hence, the probability of drawing a heart is P(heart) =
13
.
52
How many cards are a queen and a heart of the same time?
22
1
52
Therefore, the probability of drawing a queen or a heart is:
4
13
1
+
−
52 52 52
16 4
P(queen or heart) =
or
52 13
P(queen or heart) =
Let’s Try This
1.
Of the 25 employees of the municipal hall of Talisay, Bulacan, 20 are political
science graduates, 5 are law graduates and 3 are both political science and law
graduates. What is the probability of selecting an employee with a law or political
science degree?
Let’s identify first the given facts.
♦
There are 25 employees in the municipal hall of Talisay, Bulacan.
♦
20 are political science graduates.
♦
5 are law graduates.
♦
3 are both political graduates.
P(political science or law graduate) = P(political science graduate) + P(law
graduate) – P(political science and law
graduate)
=
+
–
=
Is your answer the same as this?
20 5
3
+
−
25 25 25
22
=
25
P(political science or law graduate) =
2.
A survey conducted in Barangay Mabini shows that out of 520 families, 460 own a
colored television set, 250 own a stereo set and 280 own both a colored TV set
and a stereo set. What is the probability that a family in Barangay Mabini will
own a colored TV set or a stereo set?
Let us identify the given facts in the problem.
♦
♦
There are 520 families in Barangay Mabini.
460 own a colored TV set.
23
♦
250 own a stereo set.
♦
280 own both a colored TV set and a stereo set.
P(a family owns a colored TV set or a stereo set) = P(a family owns a colored TV)
+ P(a family owns a stereo) –
P(a family owns a colored TV
and a stereo set)
=
+
–
=
Did you get 43 as your answer? Very good! You’re doing well.
52
Let’s Learn
The complement of an event E is the set of outcomes that are not included in the
outcomes of event E. The complement of E is denoted by E‘.
For example, the complement of the event of getting heads in a single toss of a coin is
that of getting tails. Hence,
E is the occurrence of heads.
E‘ (read as “E complement”) is the occurrence of tails.
Another example of complementary events are
E = it will rain tomorrow
E‘ = it will not rain tomorrow
Also
E = getting a double when a pair of dice is rolled
E‘ = not getting a double
When E and E‘ are complementary events, the probability of occurrence of their
separate probabilities is always equal to one, that is,
P(E) + P(E‘) = 1
Look at the following examples:
EXAMPLE 1
What is the probability of not getting a double in rolling a pair of dice?
SOLUTION
There are six possible outcomes for getting a double in rolling a pair of
dice:
{(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}
24
Not getting a double is the complement of having a double.
But what is the probability of not getting a double? __________
So
P(double) + P(not double) = 1
6
+ P(not double) = 1
36
6
36
36 − 6
=
36
30 5
=
or
36 6
P (not double) = 1 –
Therefore, the probability of not getting a double in rolling a pair of
dice is 5 .
6
EXAMPLE 2
The probability that Eddie will get the first prize in the contest is
80%. What is the probability that he will not get the first prize in the
contest?
SOLUTION
Determine the given facts.
The probability that Eddie will get the first prize in the contest is
80%.
Now the probability that he will not get the first prize is the
complement.
Therefore,
P(Eddie will get the first prize) + P(Eddie will not get the first
prize) = 1
0.8 + P(Eddie will not get the first prize) = 1
P(Eddie will not get the first prize) =
Did you get 0.2? If you did, that’s very good!
25
Let’s Review
A box contains 15 mixed cookies, 6 of which are chocolate, 7 are vanilla and 2 are
plain butter. If someone draws a cookie from the box, what is the probability of his/her not
getting a plain butter cookie?
Compare you answer with the one in the Answer Key on pages 46 and 47.
Let’s See What You Have Learned
Read the problems carefully and encircle the correct answers.
1.
Mang Ben has 4 screwdrivers, each a different size, in his workshop. He sent his
son to the shop to bring back any screwdriver. What is the probability that his son
brought back the smallest one?
a.
2.
d.
1/4
3
b.
4
c.
6
d.
8
0
b.
1
c.
3.75
d.
7/8
1/2
b.
11/15
c.
1/4
d.
1
0.35
b.
0.45
c.
0.65
d.
0.25
What is the probability that in a throw of a die, an even number will appear?
a.
7.
1/2
If the probability that one housewife will name Clean as her favorite detergent
soap is 0.65, what is the probability that the said housewife will not name Clean
as her favorite detergent soap?
a.
6.
c.
Among 600 married women interviewed, 440 said that they prefer to be called
“Mrs.” rather than “Ms.”. Estimate the probability that a married woman prefers
to be called “Mrs.” rather than “Ms.”.
a.
5.
1
Which of the following numbers could not possibly be a probability?
a.
4.
b.
Three coins are tossed at the same time. The sample space consists of how many
possible outcomes?
a.
3.
3
1/6
b.
1/2
c.
1/4
d.
1/3
An envelope contains 8 pink, 6 blue and 10 black slips of paper. What is the
probability of drawing a pink or a black slip?
a.
3/4
b. 1/3
c. 5/12
26
d.
1/4
8.
9.
Which of the following pairs of events are mutually exclusive?
a.
drawing a black card and a jack card from a deck of 52 playing cards
b.
singing and dancing
c.
a guy wearing a blue shirt and a red tie
d.
being 18 years old and being the president of the Philippines
The probability that a traveler stopping at a bus station in Laguna will eat
merienda is 0.45. The probability that he will buy souvenir items is 0.18. The
probability that he will both eat merienda and buy souvenir items is 0.25. What is
the probability that a traveler stopping at the said bus station will neither eat
merienda nor buy souvenir items?
a.
10.
0.5
b.
0.82
c.
0.75
d.
0.62
The probability that a person visiting his/her dentist will have his/her teeth
cleaned is 0.42, will have a cavity filled is 0.24 and will have his/her teeth
cleaned and a cavity filled is 0.13. What is the probability that a person visiting
his/her dentist will have his/her teeth cleaned or a have a cavity filled?
a.
0.87
b.
0.58
c.
0.76
d.
0.53
Compare your answers with those in the Answer Key on page 47. Did you get a perfect
score? If you did, that’s very good! If you did not, don’t worry. Just review the parts of the
lesson you did not understand very well before you move on to Lesson 3.
Let’s Remember
♦
Probability is the chance that something will happen.
♦
Probability is computed as
P=
number of successful/favorable outcomes
total number of possible outcomes
This is called classical probability.
♦
The probability of an event is always between zero and one.
♦
A probability of zero means that the event cannot happen.
♦
A probability of one means that the event will surely happen.
♦
Event denotes the outcome of an activity.
♦
Sample space is the set of all possible outcomes of an activity.
♦
Sample points are the members of a sample space.
♦
Mutually exclusive events are two or more events that cannot occur at the same
time.
27
♦
The law of addition states that if two events A and B are mutually exclusive, then
P(A or B) = P(A) + P(B)
♦
The general law of addition states that for two events A and B,
P(A or B) = P(A) + P(B) – P(A and B)
♦
When an activity has exactly two possible outcomes, then each one is the
complement of the other. The probability of the occurrence of their separate
probabilities is equal to one, that is
P(E) + P(E‘) = 1
28
LESSON 3
Probability of Simple and Compound Events
The basic concepts of probability were discussed in the previous lesson. In this lesson,
we will continue discussing the concept of probability. You will first learn about compound
events—events that are combinations of other events. You will also know the concept of
conditional probability—the probability of an event occurrence given another event that has
occurred. Finally, you will also learn about independence of two or more events.
The knowledge and understanding you will gain from this lesson will help you solve
practical problems involving the calculation of specific probabilities.
After studying this lesson, you should be able to:
♦
calculate the probabilities of simple and compound events;
♦
define and apply conditional probability to solve practical problems; and
♦
define and solve real-life problems involving the concept of independence.
Let’s Learn
In many activities, it is sometimes convenient or necessary to combine events to answer
probability questions. The compounding of events may lead to a solution without the need to
list all points in the sample space.
The compounding of events can be manifested in two ways:
♦
union
♦
intersection
To define the union of events:
Suppose A and B are two events defined in a sample space S. The union of A and B,
denoted by A ∪ B, is the set of all sample points in the event A or in the event B or
in both events A and B.
For example, suppose a die is rolled. The sample space S consists of
S = {1, 2, 3, 4, 5, 6}
Let event A consist of all odd numbers. Thus,
A = {1, 3, 5}
Let event B consist of all even numbers. Thus,
B = {2, 4, 6}
29
Then the union of events A and B is the set of all the elements of A and B. Hence,
A ∪ B = {1, 2, 3, 4, 5, 6}
Therefore, the compound event is
A ∪ B = {1, 2, 3, 4, 5, 6}
To define intersection:
Suppose A and B are two events defined in the sample space S. The intersection of A
and B, denoted by A ∩ B, is the set of sample points common to both A and B.
For instance, suppose a pair of dice is tossed. Then the sample space S consists of 36
outcomes.
Let the following events be:
A:
total of the dots on the top two faces of the dice is less than or equal to 4
B:
the dots on the top two faces of the dice is a double
So, A = {(1,1), (1,2), (2,1), (1,3), (3,1), (2,2)}
B = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}
Then, the compound event A ∩ B is the set of all sample points common to both events
A and B. That is,
A ∩ B = {(1,1), (2,2)}
Let’s Study and Analyze
To determine the probability of compound events, we use the following formula:
number of sample points in the comp
number of sample points in the sam
What is the probability of getting a sum of less than or equal to 4 and a
double when a pair of dice is tossed?
P(compound event) =
EXAMPLE 1
SOLUTION
What is the probability of getting a sum of less than or equal to 4? __________
What is the probability of getting a double? __________
How many sample points are double and at the same time have a sum less than or equal
to 4? __________
Did you get the correct answers?
My answers:
The probability of getting a sum less than or equal to 4 is
P(getting a sum less than or equal to 4) =
6
.
36
30
The probability of getting a double is P(getting a double) =
6
.
36
The probability getting a sum of less than or equal to 4 and a double is
P (sum of less than or equal to 4 and a double) =
2
1
or
36 18 .
Now, applying the formula for finding the probability of compound events, you get
P(getting a sample point whose sum is less than or equal to 4 and a double) =
number of sample points whose sum is less than or equal to
number of sample points for when a pair of
Therefore, the probability of getting a sum of less than or equal to 4 and a double when
a pair of dice is tossed is
1
.
18
EXAMPLE 2
If one card is drawn from a deck of 52 playing cards, what is the
probability that it will be either a diamond or a face card (king, queen,
or jack)?
SOLUTION
Let
D—denote the event of drawing a diamond
F—denote the event of drawing a face card
But there are cards that are both diamond and face cards.
What are they?
How many are they ?
Did you answer 3? If you did, you are correct!
Thus,
P(D ∩ F) =
3
52
Therefore, P(D ∪ F) = P(D) + P(F) – P(D ∩ F)
=
=
13 12
3
+
–
52 52 52
22
11
or
52
26
Hence, the probability of drawing either a diamond or a face card from a deck of 52
playing cards is 11 .
26
31
Let’s Try This
A survey was conducted in a certain neighborhood to determine the size of toothpaste
bought by the households in the neighborhood and the number of users in each household.
Size of
Toothpaste
Bought
Number of Persons in the Household
1–2
3–4
5 or mor
Large
22
75
78
Small
54
25
16
Sachet
30
65
35
Total
106
165
129
Suppose that a person is randomly selected from among these 400 surveyed buyers.
What is the probability that the buyer surveyed
a.
uses a large tube?
b.
uses a sachet?
c.
belongs to a household with 1 to 2 members?
d.
belongs a to a household with 5 or more members?
e.
uses a sachet and belongs to a household with 1 to 2 members?
f.
uses a small tube or belongs to a household with 5 or more members?
Let’s answer all the questions together.
a.
How many use the large toothpaste?
P(large tube) =
b.
7
175
or
400
16
How many use the sachet?
P(sachet) =
130
13
or
400
40
106
53
or
400
200
c.
Likewise, P(user belongs to household with 1 to 2 members) =
d.
How many belong to households with 5 or more members? __________
P (users belongs to a household with 5 or more members ) =
e.
129
400
How many of those surveyed use the sachet and belong to households with 1 to 2
members?
P(user uses the sachet and belongs to households with 1 to 2 members) =
32
30
3
=
400 40
To find the probability that the buyer surveyed uses a small tube or belongs to a
household with 5 or more members, you have to determine first if there is a buyer who uses
a small tube and belongs to a household with 5 or more members.
There are 16 buyers who use the small tube and belong to households with 5 or more
members.
Thus, the two events are nonmutually exclusive events. Hence,
P(user uses a small tube or belongs to a household with 5 or more members) = P(user
uses a small tube) + P(user belongs to a household with 5 or more members) – P(user uses a
small tube and belongs to a household with 5 or more members)
=
95
129
16
+
–
400
400 400
=
208
13
or
400
25
Hence, the probability that the buyer surveyed uses a small toothpaste or belongs to a
household with 5 or more members is
13
.
25
Let’s Learn
In some cases, the probability of an event can be determined only when another event
has already occurred. In other words, the probability of an event A is affected by the
occurrence of a previous event B. The resulting probability is called the conditional
probability of A given that B has already occurred. We write this as
P(A|B) (read as “the conditional probability of A given B”)
The conditional probability of event A, given that event B has occurred, is defined as:
P( A and B )
where P( B ) > 0
P( B)
A die is tossed. Given that the die comes up even, what is the
probability that it is a 2?
P( A B ) =
EXAMPLE 1
SOLUTION
Let A be the event of getting a 2
B be the event of getting an even number
The formula for conditional probability is:
P( A and B)
, where P( B) > 0
P( B )
But what is the probability of getting a 2?
P( A B ) =
A = {2}
33
Hence, the probability of getting a 2 is
P(A) = P(getting a 2) = 1
6
What is the probability of getting an even number?
B = {2, 4, 6}
Hence, the probability of getting an even number is
P(B) = P(getting an even number) = 3
6
Therefore,
1
1 6 1
P( A B ) = 6 = × =
3 6 3 3
6
Hence, the probability of getting a 2 when the die comes up even is
1/3.
Let’s Try This
The following table gives the distribution of 200 customers who purchased soft drinks
in the past 3 days from a certain sari-sari store.
Type of Soft
Drink
12
Ounces
1 Liter
1.5
Liters
Total
Pop Soda
25
15
10
50
Big Taste
10
40
10
60
Real Thing
65
15
10
90
100
70
30
200
Total
Let us use these symbols to describe the following events:
A—Pop Soda is selected.
B—Big Taste is selected.
C—Real Thing is selected.
Also
S : A 12-ounce bottle is selected.
1
S : A 1-liter bottle is selected.
2
S : A 1.5-liter bottle is selected.
3
34
If one purchase is selected at random,
a.
What is the probability a 12-ounce bottle was purchased, given that it was Big
Taste?
b.
What is the probability that Pop-Soda was purchased, given that it came in a
1-liter bottle?
SOLUTION
The different probabilities of the different events are:
P( A) =
50
1
or
200 4
P( S1 ) =
100 1
or
200 2
P( B ) =
60
3
or
200 10
P ( S2 ) =
70
7
or
200 20
P (C ) =
90
9
or
200 20
P ( S3 ) =
30
3
or
200 20
Compare your answers with these:
P(a 12 ounce Big Taste was purchased) = P( S1 ∩ B ) =
P(a1 liter Pop Soda was purchased) = P( S 2 ∩ A) =
P ( S1 B ) =
1
P ( S1 B )
1 10 1
= 20 =
× =
3
P( B )
20 3 6
10
P( A S2 ) =
3
3 20 3
P ( A/S2 )
= 40 =
×
=
7
40 7 14
P ( S2 )
20
10
200
15
or
200
The following table shows a set of 300 randomly selected voters of a barangay grouped
by age and party choice in an election.
Age
Below 30
30 or above
Total
Party
Makabayan
Kasama
94
33
70
103
164
136
Total
127
173
300
What is the probability of randomly selecting a voter in this barangay who is
a.
below 30 years old ?
b.
30 years old or above and chose Makabayan?
c.
below 30 years old and chose Kasama?
How many voters are below 30 years old?
P(voter is below 30 years old) =
35
127
300
How many voters are below 30 years old and chose Makabayan as their political
party?
How many voters are 30 years old or above?
P (voter
voter
P(voter chose Makabayan /voter is 30 or above) =
P
=
70 300 70
70 300
=
×
=
173 300 300 173 173
How many are below 30 years old and chose Kasama?
How many chose Kasama?
P(voter is below 30 voter and chose Kasama) =
=
P(voter is below 30 /voter chose Kasama)
P(voter chose Kasama)
33
300 = 33 × 300 = 33
136
300 136 136
300
Let’s Learn
Two events are said to be independent if the occurrence of one event does not affect
the probability of occurrence of the other.
If two events A and B are independent, the probability of the occurrence of both events
is equal to the product of their individual probabilities. That is,
P(A and B) = P(A) × P(B)
For instance, getting heads on the first toss of a coin will not affect the result of getting
1
2
regardless of how many tosses you have. In other words, the probability of getting heads on
heads on the second toss. This means that the probability of getting heads remains
the tenth toss or on the hundredth toss is still
1
.
2
Take a look of the following examples.
EXAMPLE 1
A pencil case contains 4 blue pens, a red pen and 5 pencils. Suppose
you get one from the pencil case at random, replace it then get another
one. What is the probability that you will get a red pen and a pencil?
SOLUTION
Since the first choice is replaced the second choice is not affected by the
first choice. These events are independent. So,
36
P (getting a red pen and a pencil) = P(getting a red pen) ×
 1  5 
=   
 20   10 
 1 
= 
 20 
Hence, the probability of getting a red pen and a pencil is
EXAMPLE 2
1
.
20
The manager of a chain of fast-food restaurants found out that 55% of the
customers order a dessert with their meals, 40% order a certain brand
of cola and 70% order a soup. What is the probability that a customer
orders a dessert and a soup?
SOLUTION
What is the probability that a customer will order a dessert?
What is the probability that a customer will order a soup?
Τhe event that a customer will order a dessert and the event that a customer will order
a soup are independent.
Thus P(customer will order a dessert and a soup) = P(dessert) × P(soup)
= 0.55 × 0.7
= 0.385
Therefore, the probability that a customer orders a dessert and a soup is .385.
Let’s See What You Have Learned
Read each of the following problems carefully and encircle the correct answers.
1.
Two coins are tossed. What is the probability of getting heads for both coins?
a. 1
2.
c. 1/3
d. 1/4
An envelope contains 12 P100 bills, 5 P500 bills and 3 P1000 bills. What is the
probability of getting a P100 and a P1000 bill?
a.
3.
b. 1/3
4/5
b.
3/4
c. 3/20
d.
2/5
From a box containing 3 black balls and 7 green balls, 2 balls are drawn in
succession, each ball being replaced in the box before the next one is drawn.
What is the probability that each color is represented?
a. 1
b.
7/10
c.
37
3/10
d.
21/100
4.
The probability that a patient recovers from a delicate heart operation is 0.78.
What is the probability that all of the next three patients who will have this
operation survive?
a.
5.
0.05
0.22
d.
0.61
b. 0.4
c.
0.7
d. 1
b. 0.49
c. 0.36
d.
0.31
A small town has a fire truck and an ambulance for emergencies. The probability
that the fire truck is available when needed is 0.93. The probability that the
ambulance is available when called is 0.96. In the event of a fire, what is the
probability that both the ambulance and the fire truck will be available?
a.
8.
c.
The probability that a married man watches a certain noontime TV show is 0.48.
The probability that a married woman watches the said TV show is 0.65. The
probability that a man watches the show, given that his wife does, is 0.75. What is
the probability that a married couple watches the show?
a. 0.62
7.
0.47
The probability that a housewife is at home when a cosmetic company
representative calls is 0.6. Given the housewife is at home, the probability that she
makes a purchase is 0.3. What is the probability that the housewife is at home and
makes a purchase when the cosmetic company representative calls?
a. 0.18
6.
b.
0.28
b. 0.04
c.
0.8928
d. 0.07
Consider the table below:
Civil Status
Sex
Single
Married
Male
18
40
Female
20
22
Total
38
62
What is the probability that a person chosen at random from among the 150 will be
single?
a.
9.
b.
78/150
c.
18/150
d.
38/150
Referring to the data in problem 8, what is the probability that a person chosen at
random is female and married?
a.
10.
18/38
11/75
b.
11/31
c. 11/36
d.
12/25
In tossing 3 coins at a time, what is the probability of getting all heads or all tails?
a. 1/2
b. 1/4
c. 1/8
d.
3/4
Compare your answers with those in the Answer Key on page 47. Did you get
everything right? If you did, that’s very good! If you did not, that’s okay. Just review the parts
of the lesson you did not understand before you move on to the next part of the module.
38
Let’s Remember
♦
A compound event is a combination of two or more events.
♦
A compound event can either be a union or an intersection of two or more events.
♦
The probability of the union of two events is
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
♦
The probability of the intersection of two events is
set of common sample points
set of points in the sample space
Conditional probability measures the chance that an event A will happen, given
that another event B has happened. That is,
P( A ∩ B ) =
♦
P(A/B) =
♦
P( A ∩ B)
, where P(B) > 0
P( B )
Two events are independent if the outcome of one event A is not affected by the
outcome of the other event B. Thus,
P(A and B) = P(A) × P(B)
You have now reached the end of the module. Congratulations! Did you enjoy studying
this module? Did you learn a lot from it? The following is a summary of its main points to
help you remember them better.
Let’s Sum Up
♦
Permutations are the total number of ways by which items may be arranged in
definite order.
♦
Combinations also refer to the ways a set of items or objects may be arranged.
However, the order is not taken into consideration.
♦
The probability of the occurrence of either A or B where both are mutually
exclusive events is the sum of their individual probabilities. Thus,
P(A or B) = P(A) + P(B)
♦
Probability is computed as
P=
number of successful /favorable outcomes
total number of possible outcomes
This is called classical probability.
39
♦
The probability of an event is always between zero and one.
♦
The sample space is the set of all possible outcomes of an activity.
♦
Sample points are the members of a sample space.
♦
Mutually exclusive events are two or more events that cannot occur at the same
time.
♦
The law of addition states that if two events A and B are mutually exclusive, then
P(A or B) = P(A) + P(B)
♦
The general law of addition states that for two events A and B,
P(A or B) = P(A) + P(B) – P(A and B)
♦
When an activity has exactly two possible outcomes, then each outcome is the
complement of the other. The probability of the occurrence of their separate
probabilities is equal to one, that is
P(E) + P(E‘) = 1
♦
A compound event is a combination of two or more events.
♦
The probability of the union of two events is
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
♦
The probability of the intersection of two events is
set of common sample points
set of points in the sample space
Conditional probability measures the chance that an event A will happen, if
another event B has happened, that is
P( A ∩ B ) =
♦
P(A|B) =
♦
P( A ∩ B)
, where P(B) > 0
P( B )
Two events are independent if the outcome of one event A is not affected by the
outcome of the other event B. Thus
P(A and B) = P (A) × P(B)
40
What Have You Learned?
Solve the following problems.
1.
A fast-food restaurant owner decided to offer “combo” meals for P 65.00 each,
consisting of any one of 7 different meat dishes, any one of 4 different vegetable
dishes and any one of 3 different desserts. In how many ways can a person choose
one of these combo meals?
2.
In how many different ways can an NFE Instructional Manager assign 20 learning
modules to his/her 5 learners?
3.
In how many ways can the names of 7 candidates be listed in a sample ballot?
4.
Given a pair of dice, what is the probability of getting a sum of 5 or 7? List all
possible outcomes in a table.
5.
A survey in Barangay Masipag shows that out of 600 families, 360 own a colored
television set, 100 own a stereo set and 140 own both a colored television set and
a stereo set. What is the probability that a family in Barangay Masipag owns a
colored TV set or a stereo set?
41
6.
If the probability of winning in a raffle game is 0.15, what is the probability of not
winning, given that you have purchased a raffle ticket?
7.
Two coins are tossed. What is the probability of getting tails on both coins?
8.
The following table shows a list of 300 voters of a barangay grouped according to
age and party choice in an election.
Party
Age
PCC
PMC
Total
Below 30
90
60
150
30 or above
60
140
200
Total
150
200
350
a.
What is the probability of randomly selecting a voter who is below 30?
b.
What is the probability of selecting a voter in this barangay who chose PCC
given that he/she is 30 years old or above?
c.
What is the probability of selecting a voter in this barangay who is below 30
given that he/she chose PMC?
Compare your answers in the Answer Key on pages 48 to 50. If you got a score of:
9–10
Very good! That means you learned a lot from this module. You are ready to
move on to the next module.
6–8
Good! Just review the parts of the module you did not understand very well
before you move on to the next module.
0–5
You should study the whole module again.
42
Answer Key
A.
B.
Let’s See What You Already Know (pages 1–3 )
1.
a
2.
a
3.
a
4.
c
5.
c
6.
b
7.
b
8.
b
9.
a
10.
b
11.
b
12.
c
13.
a
14.
c
15.
a
Lesson 1
Let’s Try This (page 9)
Using FPC, we have:
Blood Type
4
Sex
×
2
Blood Pressure
×
3
Possibilities
=
Let’s See What You Have Learned (pages 15–16)
1.
outcomes of tossing a coin = (heads, tails) = 2
outcomes of tossing a die = (1, 2, 3, 4, 5, 6) = 6
possible outcomes of tossing a coin and a die = 2 × 6 = 12
2.
5! = 5 × 4 × 3 × 2 × 1 = 120
43
24
3.
Method of Placing Order Method of Confirmation
s
s
s
s
by telephone
by sending somebody
s
s
by herself
s
by telephone
by sending somebody
s s
s
by sending someone
through tricycle
s
s
by telephone
by sending somebody
by telephone
P
(
(
(s
(s
(b
(b
4.
number of choices = 6 heads × 3 colors × 5 sizes = 90 choices
5.
number of choice for the promo package = 6 × 4 × 5 = 120 choices
6.
7.
5!
5! 5 × 4 × 3 × 2 × 1
= =
= 5 × 4 × 3 = 60 = number of display
(5 − 3)! 2!
2 ×1
arrangements
a.
n = 12; r = 3
12
b.
12!
12! 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2
=
=
(12 − 3)! 9!
9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 ×1
n =12; x1 = 7; x2= 3
12
c.
P3
12!
7! 3!
12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
=
(7 × 6 × 5 × 4 × 3 × 2 × 1)(3 × 2 × 1)
95040
=
6
= 15840
P(7,3) =
n = 12; r = 3
n!
(n − r )! r!
12!
=
(12 − 3)! 3!
12!
=
9! 3!
12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
=
(9 × 8 × 7 × 6 × 5 × 4)(3 × 2 × 1)
1320
=
6
= 220
C=
44
8.
n=6
r=4
n!
(n − r )! r!
6!
=
(6 − 4)! 4!
6 × 5 × 4 × 3 × 2 ×1
=
( 2 × 1)( 4 × 3 × 2 × 1)
38
=
2
= 15
C=
9.
n = 14
r=3
n
10.
C.
n!
14!
14!
=
=
(n − r )! (14 − 3) 11!
14 × 13 × 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
=
11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
= 2184
Pr =
5 × 3 × 3 × 5 = 225
Lesson 2
Let’s Try This (page 21)
The list of all the possible outcomers when a pair of dice is tossed.
2nd
1st
1
2
3
4
5
6
1
(1, 1)
(2, 1)
(3, 1)
(4, 1)
(5, 1)
(6, 1)
2
(1, 2)
(2, 2)
(3, 2)
(4, 2)
(5, 2)
(6, 2)
3
(1, 3)
(2, 3)
(3, 3)
(4, 3)
(5, 3)
(6, 3)
4
(1, 4
(2, 4
(3, 4
(4, 4
(5, 4
(6, 4
Let’s Review (page 22)
1.
The 5 events are mutually exclusive, so the probability that such a TV
show will get one of the ratings is
P(the TV show will get one of the ratings) = 0.15 + 0.35 + 0.10 + 0.25 + 0.15
= 1.0
2.
The three events are:
Getting a math book
Getting a science book
Getting a dictionary
45
Since no book is a math book and a science book at the same time, nor a
book is both a science book and a dictionary, we can say that the three events
are mutually exclusive.
So,
P(getting a math book, a science book or a dictionary)
= P(getting a math book) + P(getting a science book) + P(getting a
dictionary)
=
6
4
1
+
+
12 12
12
=
11
12
Let’s Review (page 26)
The events of getting a plain butter cookie and not getting a plain butter cookie are
complementary events.
But what is the probability of getting a plain butter cookie?
P(getting a plain butter cookie) + P(not getting a plain butter cookie) = 1
2 + P (getting not the plain butter cookies) = 1
15
Therefore,
P(not getting a plain butter cookie) = 1 –
=
2
15
15 − 2
15
=
Therefore, the probability of not getting a plain butter cookie is 13 .
15
An alternative solution:
The probability of not getting a plain butter cookie is equivalent to the probability
of getting a chocolate or a vanilla cookie.
How many chocolate cookies are there?
How many vanilla cookies are there?
46
Therefore, the probability of getting a chocolate or a vanilla cookie is
P(getting a chocolate or a vanilla cookie)
= P(getting a chocolate cookie) + P(getting a vanilla cookie)
= 6 + 7
15
15
= 13 ,
15
which is equivalent to the probability of not getting a plain butter cookie.
Let’s See What Have You Learned (pages 26–27)
C.
1.
d
2.
d
3.
c
4.
b
5.
a
6.
b
7.
a
8.
d
9.
d
10.
d
Lesson 3
Let’s See What Have You Learned (pages 37–38)
1.
d
2.
b
3.
d
4.
b
5.
a
6.
b
7.
c
8.
d
9.
a
10.
b
47
D.
What Have You Learned? (pages 41–42)
1.
Solution: 7 meats × 4 vegetables × 3 desserts = 84 combination of combo
meals
2.
Solution: n = 20
r=5
20
20!
(20 − 5)!
20 × 19 × 18 × 17 × 16 × 15 × 14 × 13 × 12 × 11 × 10 × 9 ×
=
15 × 14 × 13 × 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5
= 20 × 19 × 18 × 17 × 16
= 1860480 ways
P5 =
3.
7! = 7 × 6 × 5 × 4 × 3 × 2 ×1 = 5040
4.
Possible outcomes of tossing a pair of dice:
2nd die
1st die
1
2
3
4
5
6
1
(1 ,
(2 ,
(3 ,
(4 ,
(5 ,
(6 ,
2
1)
1)
1)
1)
1)
1)
(1 ,
(2 ,
(3 ,
(4 ,
(5 ,
(6 ,
3
2)
2)
2)
2)
2)
2)
(1 ,
(2 ,
(3 ,
(4 ,
(5 ,
(6 ,
4
3)
3)
3)
3)
3)
3)
(1 ,
(2 ,
(3 ,
(4 ,
(5 ,
(6 ,
4
4
4
4
4
4
(1) There are a total of 36 possible outcomes.
(2) The outcomes whose sum is 5 are: {(2,3), (1,4), (3,2), (4,1)}
(3) The outcomes whose sum is 7 are: {(1,6), (3,4), (4,3) (5,2) (2,5)
(6,1)}
(4) There is no outcome that has a sum of 5 and 7 at the same time,
therefore the two events are mutually exclusive.
P (sum of 5 or 7 ) = P (sum of 5) + P (sum of 7 )
4
6
=
+
36 36
10
=
36
5
=
18
48
5.
(1) There are 600 families in Barangay Masipag.
(2) 360 own a colored television set.
(3) 100 own a stereo set.
(4) 140 own both a colored tv set or a stereo set.
The events of a family having a colored tv set and having a stereo set are
not mutually exclusive, hence
P(of a family owning a colored tv set or stereo set)
= P (a family owns a colored tv set ) + P (a family owns a ster
P (a family owns a colored tv set and a stereo set )
360 100 140
=
+
−
600 600 600
460 140
=
−
600 600
320
=
600
8
=
15
6.
probability of winning = 0.15
probability of not winning = 1 – 0.15
= .85
7.
Outcomes of tossing 2 coins:
2nd
H
T
H, H
T, H
H, T
T, T
1st
H
T
Probability of getting 2 tails =
8.
a.
1
4
number of voters who are below 30 years old = 90
P(selecting a voter below 30 years old) =
49
90
350
b.
P(selecting a voter who chose PCC and is 30 years old or ab
 selecting a voter who chose PCC/selecting 
P

a voter who is 30 years old or above

= 
P (selecting a voter who is 30 years old or above )
90 350
=
200 350
60 350
=
×
350 200
90
=
200
9
=
200
c.
P (selecting a voter who chose PMC and is below 30 years o
 selecting a voter who is below 30 years 
P

old/selecting a voter who chose PMC 

=
P (selecting a voter who chose PMC )
60 350
=
200 350
60 350
=
×
350 200
60
=
200
3
=
10
50