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Probability AND Counting Rules
Statistic (Bulacan Polytechnic College)
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PROBABILITY AND COUNTING RULES
Lesson Outline
1.1 Introduction
1.2 Counting Rules
1.3 Sample space and Probability
1.4 The Addition Rules and Multiplication Rules for Probability
1.5 Marginal and Conditional Probabilities
1.6 Random Variables and Discrete Probability Distribution
1.7 Mean, Variance, Standard Deviation, and Mathematical Expectation
1.8 Binomial Probability Distribution
1.9 Poisson Probability Distribution
1
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1.1 Introduction
In this chapter we will deal with some counting techniques without direct
enumeration of the number of possible outcomes of a particular experiment or the
number of elements in particular set. Such techniques are sometimes called
combinatorial analysis.
Also we will deal with probability theory. These include topics such as probability
distribution, mathematical expectation, binomial distribution, and Poisson
distribution.
1.2 Counting Rules
Fundamental Counting Rule
Sum Rule. Suppose that an event can be performed by either of two different
procedures, with m possible outcomes for the first procedure and n possible outcomes
for the second. If the two sets of possible outcomes is disjoint, then the number of
possible outcomes for the event is
m+n
(Formula 1-1)
Product Rule. In a sequence of n events in which the first has m possibilities and the
second event has n2, and the third has n3, and so forth, the total number of
probabilities of the sequence will be
n1(n2)(n3) … (nk)
(Formula 1-2)
Example for Sum Rule: A scholarship is available, and the professor to receive this
scholarship must be chosen from the Accountancy, Business Administration or Tourism
Department. How many different choices are there for these scholarship if there are
15 qualified professors from the Accountancy Department, 50 qualified professors from
the Business Administration Department and 26 qualified professors from the Tourism
Department?
Solution:
The procedure of choosing a form the Accountancy Department has 15 possible
outcomes, the procedure of choosing professor from the Business Administration
Department has 50 possible outcomes, and a procedure of choosing a professor from
the Tourism Department has 26 possible outcomes.
Therefore, there are 15+50+26= 91 possible choices to award the scholarship.
Example for Product Rule: A student has a choice of 5 sandwiches and 6 juices. In
how many ways can he choose 1 sandwich and 1 juice?
Solution:
He can choose a sandwich in 5 ways, and with each of these choices there are 6 ways of
choosing a juice.
Hence, the required number of ways =5(6) =30 ways.
Factorial Notation.n! (which read “n factorial”) is the product of the first
inconsecutive natural numbers.0! is defined to be 1.
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n!= n(n-1)(n-2)(n-3)…(3)(2)(1)
(Formula 1-3)
Example 3: Evaluate the following:
a.
b.
c.
d.
e.
f.
1! = 1
3! = 3(2)(1) = 6
5! = 5(4)(3)(2)(1) = 120
6! – 4! = [6(5)(4)(3)(2)(1)
2!(3!) =2(1)[3(2)(1)] = 2(6) = 12
10!
5!
=
10(9)(8)(7)(6)(5)(4)(3)(2)(1) 3,628,800
= 120
=
5(4)(3)(2)(1)
30,240
Permutation. A permutation is an arrangement of all or part of a number of things (or
objects) in a definite order. The number of permutations of n objects taken r at a time
is given by
P (n, r) =n P r =
𝑛!
, 0 ≤ r ≤ n.
(− )!
(Formula 1-4)
Permutations with Repeated Elements. It is often happens that object which are
virtually identical get arranged. Our inability to distinguish between this items reduces
the number of possible permutations by the number of ways these identical items
themselves can be arranged.
𝑛!
Pn = (((
1!(2!)(𝑛3!) ...
where n1 + n2 + n3 ...= n
(Formula 1-5)
Circular Permutations. When things are arranged in places along a closed curve
or a circle, in which any place may be regarded as the first or last place, they form a
circular permutations. Thus with n distinguishable objects we have (n - 1)!
arrangements. In symbol,
Pc = (n - 1)
(Formula 1- 6)
Example 4: Evaluate the following:
4!
4!
4(3)(2)(1)!
24
a. P(4,0) =(4−0)! = 4! = 4(3)(2)(1)!= 24= 1
b. P(5,2) = 5!
5! 5(4)(3)(2)(1) 120
= 6 = 20
(5 − 2)! =3! = 3(2)(1)
P (7, 7) =
7!
7! 7(6)(5)(4)(3)(2)(1)
=
(7−7)! = 0!
1
=
5040
=
1
5,040
Example 5: How many different ways can a manager and a supervisor can be selected
for a company branch in Manila if there are 8 employees available?
Solution:
Since order is important for the managerial and supervisor position we would need to
apply permutation. Then,
8!
8! 8(7)(6)(5)(4)(3)(2)(1)
6(5)(4)(3)(2)
(1)
P (8, 2) =(8−2)! 6!
= =
40,320
=720
= 56
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Hence, there would be 56 ways to select a manager and a supervisor.
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Example 6: How many different ways can 5 televisions commercial can be played
during a 30-minute television program?
Solution:
Since order is important, the solution is
5!
5!
5(4)(3)(2)(1)
120
=1
= 120
P (5, 5) =(5−5)! =0! = 1
Hence, there would be 120 ways to set up to commercials.
Example 7: In how many ways can 4 students be seated at a round table?
Solution:
Let one of them be seated anywhere. Then the 3 remaining can be seated in 3! ways.
Thus there are Pc= (n-1)! = (4-1)! =3! =6 ways of arranging for 4 persons in a circle
Example 8: There are 4 copies of Statistics book, 5 copies of Probability book, and 3
copies of Forecasting book. In how many ways can they be arranged on a shelf?
Solution:
There are 4 + 5 + 3 = 12 books of which are 4 copies of Statistics book, 5 copies of
Probability book, and 3 copies of Forecasting book.
The number of arrangements is
=27,720
Pn=
=
𝑛!
12!
4!(5!)(3!)
!1!2!𝑛3!
!!
Combination. A combination is a grouping or selection of all or part of a number
of things (or objects) without reference to the arrangement of the things selected.
The number of combinations of n objects taken r at a time is given by
C (n, r)=nCr =(𝑛)=
𝑛𝑛𝑛
𝑛
𝑛!
=
𝑛!
, 0≤r≤n
(Formula 1-7)
(−)!!
(−!
Combinations of different things taken any number at a time. The total number of
combinations Cn of n different things taken 1, 2, 3,.., n at a time is
Cn=2n -1
(Formula 1-8)
Example 9: Evaluate the following:
a. C(5,0)=(5)=
0
5!
=
5!
(5−0)!0! 5!
(0!)
120
=
=
120(1) 120
b. C(4,4)=( 4) =(4−4)!4!=
4!
4(3)(2)(1)
24
24
= (1)[4(3)(2)(1)]
= 1(24)
= 24
=1
0!(4!)
c. C(6,2)= (6)=
6!
4
4!
(4)(3)(2)(1)
120
=
5(4)(3)(2)(1)
[1]
6!
=
=
6(5)(4)(3)(2)(1)
=
720
=2
=1
720
=15
(6−2)!2!
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4(3)(2)(1)[2(1)] 24(2)
48
Example 10: In how many ways can board members be selected out of 15 board
members of a company to represent the body of the stockholders meeting?
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Solution:
Since this is combination problem, the answer is
15!
C (15, 4) =(15−4)!4!=
15
15(14)(13)(12)(11!)
= 11!(4)(3)(2)(1) =1,365
11!(4!)
Hence, there would be 1,365 ways to be select a committee to represent the body.
Example 11: How many different sums of money can be drawn from a wallet containing
one bill of each of 20, 50, 100, 200, and 500 pesos?
Solution:
Each bill may be dealt with two ways, as it can be chosen or not. Since each of the
two ways of dealing with a bill is associated with two ways of dealing with the 5
bills=2(2)
(2) (2) (2) =25 ways. But 25 ways includes the case which no bill is chosen.
Hence, the required number of ways in Cn=2n - 1=25- 1=32 - 1=31 ways.
Or we can solved the problem in different way such as
Cn = C(5,1)+C(5,2)+C(5,3)+C(5,4)+C(5,5)=5+10+10+5+1=31
Take note that we did not consider C (5, 0) because it refers to not taking any bill among
the peso bills.
1.3 Sample Spaces and Probability
This section of the chapter discussed some basic concepts of probability, types,
and rules. The process of flipping a coin, rolling die, or drawing a card from an
ordinary deck of cards refers to probability experiments, and also defined as a chance
process that leads to well defined results called outcome.
A. Sample Space
An outcome is a result of a single trial of a probability experiment, while a
sample space is a set of all possible outcomes of a probability experiment.
We can also represent the sample space using a Venn diagram or tree diagram.
A Venn diagram a picture that deficits all possible outcomes for an experiment, while
tree diagram is a device consisting of line segments emanating from a starting point
and in its outcome point and it determines all possible outcomes of a probability
experiment.
Table 1.1: Example of Experiments, Outcomes, and Sample Spaces
Experiment
Toss a coin once
Toss a coin twice
Roll a die
Exam Result
Game Result
Outcomes
Head, Tail
HH, HT, TH, TT
1,2,3,4,5,6
Pass, Fail
Win, Lose
Sample Space
S={Head, Tail}
S={HH, HT, TH, TT}
S={1,2,3,4,5,6}
S={Pass, Fail}
S={Win, Lose}
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Example 1: Determine the sample space for rolling two dice.
Solution:
Given that each die can land in 6 different ways, two dice (or pair of dice) are rolled,
the same space can be presented by an array. Table 1.2 shows the list of pairs of
sample.
Die 1
1
2
3
4
5
6
1
(1,1)
(2,1)
(3,1)
(4,1)
(5,1)
(6,1)
Table 1.2: Sample Space Rolling Two Dice
Die 2
2
3
4
5
6
(1,2)
(1,3)
(1,4)
(1,5) (1,6)
(2,2)
(2,3)
(2,4)
(2,5) (2,6)
(3,2)
(3,3)
(3,4)
(3,5) (3,6)
(4,2)
(4,3)
(4,4)
(4,5) (4,6)
(5,2)
(5,3)
(5,4)
(5,5) (5,6)
(6,2)
(6,3)
(6,4)
(6,5) (6,6)
Example 2: Determine the sample space for drawing one card from ordinary deck of
cards.
Solution:
Given that there are 4 suits (diamonds, clubs, heart, and spades) and 13 for each suits
(ace through king), there are 52 outcomes in the sample space for an ordinary deck of
cards. Figure 1.2 shows the sample space for ordinary deck of cards.
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Figure 1.2: Sample Space for Ordinary Deck of Cards
Example 3: Determine the sample space for the three True or False quiz. Use tree
diagram to represent the sample space.
Solution:
There are two possible outcomes, True or False, and each question could either be
True or False.
Hence there are eight possibilities, as shown in Figure 1.3.
Figure 5.3: Tree Diagram for Three True or False Quiz
Start
True
True
True
TTT
False
False
TTF
1st
Quarter
False
True
TFT
True
True
False
False
True
2nd Quarter
False
False
TFF
FTT
Outcomes
FTF
FFT
FFF
B. Simple and Compound Events
An event is a collection of one or more outcomes of an experiment, it maybe a
simple event or a compound event. A simple event is an event that includes one and
only one of the outcomes for an experiment and is denoted by E, it is also called the
elementary event. On the other hand, a compound event is a collection of more than
one outcome for an experiment; it is called a composite event.
Example: In a group of Catholic Christian, some are in favor of death penalty as major
punishment for heinous crimes and others are against it. Three persons are selected in
random and asked whether they are in favor of or against death penalty. How many
distinct outcomes are possible? List all the outcomes included in each of following
events and mention whether they are simple or compound events?
a. All three persons are in favor of death penalty.
b. At least one person is in favor of death penalty.
c. Exactly one person is against death penalty.
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Solution:
Let F=a person is in favor of death penalty as major punishment for heinous crimes.
A=a person is against death penalty as major punishment for heinous crimes.
This experiment has the following 8 outcomes: FFF, FFA, FAF, AFF, FAA, AFA, AAF, AAA.
a. The event "all three persons are in favor of death penalty" will occur if FFF is
obtained. Thus, all three persons are in favor of death penalty = {FFF}.
Because this event includes only one of the eight outcomes, it is simple event.
b.
The event "At least one person is in favor of death penalty" will occur if one of
the person is in favor, or two of the persons are in favor, or three of the
persons are in favor. Thus, at least one person is in favor of death penalty
={FAA, AFA, AAF, FFA, FAF, AFF, FFF}. Because this event includes more than
one outcome, it is a compound event.
c.
The event "Exactly one person is against death penalty" will occur if one of
the three persons selected are against death penalty. Thus, it includes the
three possible outcomes:
Exactly one person is against death penalty = {FFA, FAF, AFF}. Because this
event includes more than one outcome, it is compound event.
C. Probability
Probability is a numerical measure of the likelihood that a specific event will
occur. An event that cannot occur has zero probability which is called an impossible
event and if an event that is certain to occur has a probability equal to 1 which is
called sure event. There are four basic probability rules that will be helpful in solving
probability problems.
a. The probability of an event is with in the range 0 to 1.
0≤P(E)≤1.
b. The sum of the probabilities of all simple events for an experiment is
always 1.
For an experiment: ∑ P(Ei) = P(E1)+P(E2)+P(E3)+…=1
c. If an event cannot occur , its probability is 0.
d. If an event is certain, then the probability is 1.
Complimentary Events. The complement of an event E is a set of outcomes in the
sample space that are not included in the outcome of event E. The complement of E is
denoted by E'(read as E prime). The rule for complementary events are denoted by
P(E') = 1 - P(E) or P(E) = 1 - P(E') or P(E)+P(E') = 1
(Formula 1-9)
D. Three Conceptual Approaches to Probability
a. Classical Probability. Classical probability assumes that all outcomes in
the sample space are equally likely to occur.
P(E) = 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 𝑛𝑛 𝑛𝑛𝑛𝑛𝑛𝑛𝑛 𝑛𝑛 ℎ 𝑛𝑛𝑛 𝑛𝑛𝑛
𝑛𝑛𝑛𝑛𝑛
=
()
()
(Formula 1-10)
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b. Empirical or Relative Frequency Probability. Empirical probability is tje
type of probability that uses frequency distribution based on observations
to determine numerical probabilities of events.
P(E) =
𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 𝑛𝑛𝑛 ℎ 𝑛𝑛𝑛𝑛𝑛
𝑛𝑛𝑛𝑛𝑛 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 ℎ
𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛
𝑛
=𝑛
c. Subjective Probability. Subjective probability is the probability assigned
to an event based on subjective judgment, experience, information, and
belief. For example, a sportwriter may say that there is 90% probability
that San Sebastian Stags will win the NCAA championships. A physician may
say that, on the basis of his diagnosis, there is a 60% chance that the
patient will recover. A financial analysis may say that there is 80%
probability that peso dollar exchange rate will decrease by 3 pesos.
Example for Classical Probability: A card is drawn from a ordinary deck of card. Find
these probabilities (a) of getting king of hearts,(b) of getting spade, (c) of getting a 5
or a clubs(d) of getting a 5 or a 7, (e) of getting a card which is not a spade, (f) of
getting 11 of clubs.
Solution:
a. Since there is only one king pf hearts in an event E and 52 possible outcomes in
the sample
space.The probability of getting king of heart is
P(king of hearts) = () = 1
)(
52
b. There are 13 spades so there are 13 outcomes in an event E, the probability of
a getting a spade is
P(spade)= () =
)(
13
= 1/4
52
c. There are four 5s and 13 spades in an event E, but the 5 of spades are counted
twice in this listing. Thus, there 16 possibilities of drawing a 5 or a spade, so
P(5 or spade) = () = 4+13−1= 16= 4
)(
52
52
13
This is an example of the inclusive or.
d. Given that there are four 5s and four 7s,
P(5 or 7) = () = 4+4= 8 = 2
)(
52
52
13
This is an example of the exclusive or.
e. Recall that P(spade) = ¼, we simply deduct this to 1 to obtained the probability
of getting a non-spade card. We will use P(E') = 1 - P(E), where P(E') is the
probability of getting a non-spade card.
P(E') = 1 - P(E)= 1 - () = 1 - 1 = 4
1
3
)(
4
4 - 4 =4
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f. It is impossible to to get an 11of clubs in the sample space of an ordinary deck
of card. Hence, the probability is
P(11) )( ) = 0 = 0
)(
52
This is an example of impossible event.
Example for Empirical Probability: In a sample of 50 college students, 18 are
freshmen, 23 are sophomore, 2 are junior, and 7 are senior. Set up a frequency
distribution and find the following probabilities:
a. a student is a freshman.
b. a student is a freshman or sophomore.
c. a student is neither a freshman nor a junior.
d. a student is not a senior.
Solution:
Year Levels
Freshmen
Sophomore
Junior
Senior
Total
Frequency
18
23
2
7
50
a. To obtain the probability of selecting a freshmen we simply divide the number
of freshmen by the sample space.
P(freshmen) = () = 18= 9
)(
50
25
b. We need to add the frequency of the two level (or classes).
P(freshman or sophomore) = () = 18+23= 41
)(
50
50
c. Note that neither a freshman nor a junior means that the student is either a
sophomore or a senior.
P(neither a freshman nor a junior)= P(sophomore or senior)
P(sophomore or senior) = () = 23+7 = 30= 3
)(
50
50
5
d. In order to find the probability of not a senior, we need to subtract the
probability of senior from 1.
P(not a senior) = 1 - P(senior) = 1- () = 1- 7 = 50- 7 = 43
)(
50
50
50
50
1.4 The Addition Rules and Multiplication Rules for Probability
A lot of problems involve determining ng probability of two or more events. This is
when independent, dependent, and mutually exclusive comes into the picture in
dealing with probability.
There are important notes to think about mutually exclusive, independent, and
dependent events. The first is of which is mutually exclusive are always dependent,
secondly is independent events are never mutually exclusive, and lastly is dependent
events may or may not be mutually exclusive. The exception of first and second is that
when at least one of the two events has a zero probability.
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A. Independent, Dependent, Mutually Exclusive Events
Two events A and B are independent events if the fact that A occurs does not
affect the probability of B occurring . In other words, A and B are independent events
if
P(A|B) = P(A)
or
P(B|A) = P(B)
Two events A and B are dependent events for which the outcome or occurrence
of event A affects the outcome or occurrence of event B in such a way that the
probability is changed. In other words, A and B are dependent events if
P(A|B) ≠ P(A)
or
P(B|A) ≠ P(B)
Two events A and B are mutually exclusive events if they cannot occur at the
same time.
B. Addition Rules for Probability
Rule 1: When two events A and B are mutually exclusive,
the probability that A or B will occur is
P(A or B) = P(A)+P(B)
P(S) = 1
P(A)
(Formula 5-12)
Rule 2: If A and B are not mutually exclusive, then
P(A or B) = P(A) + P(B) - P(A and B)
(Formula 5-13)
P(S) = 1
P(A)P(B)
Note: Addition Rule 1 can also be used when the events are mutually exclusive,
since P(A and B) will always equal to 0
Example 1: A box contains 4 red marbles, 8 blue marbles, and 7 green marbles. If a
person selects a marble at random, find the probability that is either a red or a green
marble.
Solution:
Since the box contains 4 red marbles, 7 green marbles, and a total of 19 marbles, P(red
or green) = P(red)+ P(green) = 4 + 7 = 11
9
19
19
The events are mutually exlusive.
Example 2: A single card in drawn from an ordinary deck of card. Find the probability
that it is a queen and once as a diamond; thus, one of the outcomes must be
deducted, as shown.
P(queen or diamond) = P(queen)+P(diamond)-P(queen and diamond)
4
13
=52 + 52 -
1
16
52= 52
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Example 3: In a certain insurance company there are 20 senior and 20 junior
salespersons; 8 senior and 14 junior salespersons are males. If a salesperson is
selected, find the probability that the salesperson is a senior or a male.
Solution:
The sample space is shown here.
Salesperson
Male
8
Senior
14
Junior
Total
22
Female
12
16
Total
20
30
28
50
The probability is P(senior or female) = P(senior)+P(female) - P(female and senior)
=
20 28 12
50 + 50 - 50
36
18
=50 =25
C. Multiplication Rules for Probability
The multiplication rules can be applied to determine the probability of two or
more events that occur in sequence. The probability of the intersection of two events
is called their joint probability. It is written as P(A and B). These rules are for
independent or dependent events.
Rule 1: When two events are dependent, the probability of both occuring is
P(A and B)= P(A)•P(B)
(Formula 1-14)
Rule 2: When two events are dependent, the probability of both occuring is
P(A and B)= P(A)•P(A|B)
(Formula 1-15)
Rule 3: When two events are mutually exclusive their joint probability is always zero.
If A and B are two
mutually exclusive events, then
P(A and B)=0
(Formula 1-16)
Note: Multiplication Rule 1 can be extended to three or more independent events by applying
the formula
P(N1 and N2 and N3 ...and Nk) = P(N1)•P(N2)•P(N3)•••P(NK)
Example 4: A die is rolled and a coin is flipped. Find the probability of getting a 5 on
the die and tail on the coin.
Solution:
P(5 and tail)= P(5)•P(tail) = 1 ( 1)= 1
6
Recall that the sample space
Head,Tail.
2
12
for the die is 1,2,3,4,5,6; and for the coin it is
Example 5: A box contains 3 red balls, 8 blue balls, and 9 green balls. A first ball is
selected, and then it is replaced. A second ball is selected. Find the probability of
selecting: (a) 2 red balls,(b) 1 blue ball and then 1 green ball.
Solution:
a. P(red and red)= P(red)•P(red)= 3
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(
3
9
20
)=
b. P(blue and green)= P(blue)•P(green)=
20
8
9
20 ( 20)
72
=400 =
50
9
Example 6: A SJS survey found that one out of 5 Filipinos say they are in favor of the
death penalty for heinous crimes. If the people are selected at random, find the
probability that all three will say that they are in favor death penalty.
Solution:
Let D denote that a person is in favor of death penalty. Then
P(D and D and D)= P(D)•P(D)•P(D)= (1 )(1 1 = 1
5 5
5 20
)
Example 7: Reina owns a collection of 25 bags, of which 6 are made by Guess. If the
two bags are selected at random, find the probability that both are made by Guess.
Solution:
Since the events are dependent
P(G1 and G2)= P(G1)•P(G2|G2)=
6
25
=
30
600
=
20
1
Example 8: The RSS Financing Inc. found that 50% of the members had salary loan(S)
with the financing company. Of these members 8% also had a calamity loan(C). If a
member is selected at random find the probability tha the member has both loans with
the company.
Solution:
Note that the events are dependent,
P(S and C)= P(S)•P(C|S)= 0.50(0.08)=0.04 or 4%
1.5 Marginal and Conditional Probabilities
A. Marginal Probability.Marginal probability is a probability of a single event without
consideration of any other event; it is also called single probability. It can be
computed using the formula
P(A)= P(A and B1)+P(A and B2)+...+P(A and Bk)
(Formula 1-17)
where B1,B2, ... Bk, are k mutually exclusive and collectively exhaustive events.
Recall that two events are mutually exclusive if both the events cannot occur
simultaneous, while collectively exhaustive if one of the events must occur.
B. Conditonal Probability. Conditional probability is probability that an event will
occur given that another event has already occurred. If A and B are two events, then
the conditional probability is given as P(A|B) and reads as "the probability of A given
that B has already occurred." In symbol,
P(A|B)= P(B and A)/P(B) and
P(B|A)= P(A and B)/P(A)
given that P(B)≠ 0 and P(A) ≠ 0.
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(Formula 1-18)
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Example 1: A box contains blue and red balls. A person selects two balls without
replacement. If the probability of selecting a blue ball and a red ball is 12/30, and a
probability of selecting a blue ball on the first draw is 3/5, find the probability of
selecting a red ball on the second draw, given that the first ball selected was a blue
ball.
Solution:
Let B = selecting a blue ball.
R = selecting a red ball.
Then
P(R|B)=
(
𝑛𝑛𝑛 �)
12
30
5
3
()
5
=
12
30
.
60
3
=
=
2
90 3
Thus, the probability of selecting a red ball on the second draw given that the first
ball selected was blue is ⅔.
Example 2: In a fast-food chain, 75% of the customers’ orders chicken meal. If 40% of
the customers’ orders chicken meal and sundae, find the probability that the customer
orders chicken meal will also order a sundae.
Solution:
Let C = the customer orders chicken meal.
S = the customer orders sundae.
Then
P(S|C) =
(
𝑛𝑛𝑛 𝑛)
)(
40%
=75
%
0.40
=
0.75
=15
8
Thus, the costumer has a 8/15 probability of ordering sundae, given that he ordered
chicken meal.
Example 3: A survey conducted by WSS asked 250 whether or not they have shopped
on the new Shopping Mall. The following table gives the two-way classification of the
responses.
Gender
Male
Female
Total
Have shopped
20
130
150
Have never shopped
70
30
100
Total
90
160
250
Suppose one person is selected at random from these 250 persons. Find the following
probabilities.
a. The respondent answered has shopped, given that the respondent is a male.
b. The respondent is a female, given that the respondent answered has never shopped.
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Solution:
Let M = respondent is a male.
F = respondent is a female.
S = have shopped.
N = have never shopped.
a.
The problem to find P(S|M). The rule
states
P(S|M)
=
)(
(
𝑛𝑛𝑛 𝑛)
The probability P(S and M) is the number of males who shopped, divided by the
number of respondents.
P(M and S) =
20
250
The probability P(M) is the probability of selecting a male:
P(M) =
90
250
Then
P(S|M) =
b.
( 𝑛𝑛𝑛 𝑛)
)(
=
20/250
20
250
90/25= 250∙ 90
0
=9
2
The problem is to find P(F|N)
30
)(
100
250
25
0
100
10
1.6 Random Variables and Discrete Probability Distribution1
Suppose Table 5.3 shows the frequency and relative frequency distribution of the
number of TV owned by 500 families residing in the City of Manila.
Table 1.3: Frequency and Reltive Frequency Distributions of the number of TVs
owned by City of Manila Families
Numbers of TVs owned
Frequency
Relative frequency
0
34
34/500 = 0.068
1
316
316/500 = 0.632
2
132
132/500 = 0.264
3
18
18/500 = 0.036
N=500
Sum = 1.000
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Suppose one family is randomly selected from this population. The process of randomly
selecting a family is called a random or chance experiment. Let X denotes the
number of TVs owned by the selected family. Tjen X can assume any of the 4 possible
values (0, 1, 2, and 3) recorded in the leftmost column of the table. The value
assumed by X depends on the family is selected. Hence, this value depends on the
outcome of an random experiment. Therefore, X is referred to random variable. A
random variable is a function or rule that assigns a number to each outcome of an
experiment, it is called chance variable. In general, a random variable is denoted by
X. A random variable can be discrete or continuous. A discrete random variable
assumes values that can be counted, while discrete random variable that can assume
all values between any two specific values; a variable obtained by measuring or are
contained one or more intervals.
A discrete probability distribution consists of the values a random variable can
assume and the corresponding probabilities of the values. The probabilities are
determined theoritically or by observation. There are several requirements for a
distribution of a discrete random variable. For a discrete random variable X that can
assume values Xi,
1.
2.
3.
4.
0≤P(X)≤1, for all Xi(The probability outcome is between 0 and 1).
(The sum of all possible outcomes is 1.0)
The listing is exhaustive(all possible outcomes are included).
The outcomes are mutually exclusive(The outcomes cannot occur at the same
time).
Example 1: Construct a probability distribution for rolling a die.
Solution:
Since the sample spaces of a die is 1, 2, 3, 4, 5, 6 and each outcome has a probability
of 1/6, the number of tails.
Outcome X
Probability P(X)
1
2
1⁄
6
3
1⁄
6
4
1⁄
6
5
1⁄
6
1⁄
6
6
1⁄
6
Example 2: Construct a probability distribution for tossing three coins. Let X
represents the number of tails.
Solution:
When three coins are tossed, the sample space represented as HHH, HHT, HTH, THH,
HTT, THT, TTH, TTT; and if X is the random variable for the number of tails, thenX
assumes the value 0,1,2, or 3.Probabilities for the values X can be determined as
follows:
No tails
HHH
1⁄
8
1⁄
8
One tail
HHT
1⁄
8
HTH
1⁄
8
3⁄
8
THH
1⁄
8
HTT
1⁄
8
Two
tails
THT
1⁄
3⁄
8
8
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TTH
1⁄
8
Three
tails
TTT
1⁄
8
1⁄
8
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Hence, the probability of getting no tails is ⅛, one tail is ⅜, two tails ⅜, and three
tails is ⅛. From these results, a probability distribution can be constructed by listing
the outcomes and assigning the probability of each outcome, as presented here.
Number of tails
Probability(X)
0
1⁄
8
1
2
3
3⁄
3⁄
1⁄
8
8
8
Example 3: A wallet containing four ₱100 bills, two ₱200 bills, three ₱500 bills, and
one ₱1,000 bill. Construct a probability distribution for the data.
Solution:
The probability P(X) can be calculated for each X by dividing the number of particular
bills by the total number of bills.
For ₱100 bills:
For ₱200 bills:
For ₱500 bills:
For ₱1,000 bills:
4
= 0.40
10
2
= 0.20
10
3
= 0.30
10
1
= 0.10
10
The probability is shown here:
Number of bills X
Probability (X)
₱100 bills
0.40
₱200 bills
0.20
₱500 bills
0.30
₱1,000bills
0.10
1.7 Means Variance Standard Deviation Mathematical Expectation
The mean, variance, and standard deviation for a probability distribution are
calculated differently from the mean, variance, and standard deviation for samples.
This section enlightens how this measures-and a new measure called the expectationsare computed for probability distributions.
A. The Mean of a Probability Distribution
The mean of a random variable with a discrete probability distribution is
M= X1•P(X1)+X2•P(X2)+X3•P(X3)+ ...+Xn•P(Xn)= E[X•P(X)]
(Formula 1-19)
where X1, X2, X3,...Xn = the outcomes.
P(X1), P(X2), P(X3),...P(Xn) = the corresponding probabilities.
E[X•P(X)] = means to sum the products.
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B. Variance and Standard Deviation of a Probability Distribution
The variance of a random variable with a discrete probability distribution is
𝑛 2 = ∑[𝑛2 ∙ 𝑛(𝑛 )] − 𝑛 2
(Formula 1-20)
The standard deviation of a random variable with a discrete probability distribution
is
𝑛 = √
2
or𝑛 = √∑[𝑛2 ∙ 𝑛(𝑛)] − 𝑛2
(Formula 1-21)
C. Expected Value
The expected value value of a discrete random variable of a probability
distribution is the theoretical
theoretical average of the variable. The formula is
M= ∑(X)= ∑[X•P(X)]
(Formula 1-22)
The symbol E(X) is used for the expected value.
Example 1: Five balls numbered 1,2,3,4 and 5/are placed in a box. One is selected, its
number is noted, and then it is replaced. If this experiment is repeated many times,
find the mean, variance, and standard deviation of the numbers on the balls.
Solution:
Let X = the number of each ball.
The probability distribution is
Number on ball X
1
Probability P(X)
1⁄
5
2
3
1⁄
1⁄
5
5
The mean is
µ = ∑[X•P(X)]
=1(1⁄5)+2(1⁄5)+3(1⁄5)+4(1⁄5)+5(1⁄5)
=0.20+0.40+0.60+0.80+1.00
=3
The variance is
𝑛2 = ∑[𝑛2 ∙ 𝑛(𝑛)] −
𝑛2
=[12(0.20)+22(0.20)+32(0.20)+42(0.20)+52(0.20)] - 32
= (0.20+0.80+1.80+3.20+5.00) - 9
=11 - 9
=2
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4
1⁄
5
5
1⁄
5
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The standard deviation is
𝑛 = √𝑛2 = √2 =
1.41
Example 2: Five hundred raffle tickets are sold at ₱25 each for three prizes of ₱4,000,
₱2,500 and ₱1,000. After each prize drawing, the winner ticket is then returned to the
collection of tickets. What is the expected value if a person purchases four tickets?
136
Gain X
Probability P(X)
₱3,900
₱2,400
₱900
-₱100
4
50
0
4
50
0
4
50
0
496
500
Solution:
E(X)= ∑[X∙P(X)]
E(X)= (3,900)(
4
500
)+(2,400)(
4
500
)+(900)(
4
500
)+(-100)(496)
500
=31.20+19.20+7.20 - 99.2
=-₱41.60
Thus, the expected value is -₱41.60.
Example 3: A financial analyst suggest that his client select one of two types of
investments in which to invest ₱100,000. Investment A pays a return of 8% and has a
failure to pay rate of 5%. Investment B has a 6% return and a failure to pay rate of 3%.
Find the expected rate of return and decide which investment would be better. (When
the investment fails to pay, the investor loses all the investment.)
Solution:
The return on investment A is ₱100,000(8%) = ₱8,000. The expected return then is
E(X) = ₱8,000(0.95) - ₱100,000(.05) = 7,600 - 5,000 = ₱2,600
The return on investment B is ₱100,000(6%)= ₱6,000. The expected return then is
E(X) = ₱6,000(0.97) - ₱100,000(.03) = 5,820 - 3,000= ₱2,820
Thus, investment B would be a better investment since the expected return is higher
compared to investment A
1.8 Binomial Probability Distribution
If the probability has only two outcomes or can be reduced to two outcomes(the
outcomes are considered as either success or failure these are called binomial
experiment. A binomial experiment is a probability experiment with the following
requirements:
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1.
2.
3.
4.
There must be a fix number of trials.
Each trial can have only two outcomes and are mutually exclusive outcomes.
The outcomes of each trial are independent.
The probability of each success is the same for each trial.
Binomial Formula: The probability of X success in a binomial experiment with n trials
and probability of success P is given by the formula
𝑛!
P(X) =
(−!) !
where
pXqn-X
(Formula 1-23)
P(X)= Binomial probability distribution.
n = the number of trials
X = the number of observed successes
p = the probability of success on each tail
q = the probability of failure, found by 1 – p
Note that 0 ≤ X ≤ n and X = 0, 1, 2, 3, …, n
Example 1: In a survey, 25% of the people interviewed said they bought their
Refrigerator during the last six months. If eleven people are selected at random, find
the probability that exactly six these people bought their refrigerator during last six
months.
Solution;
In this case, n = 11, X = 6, p = 25%, and q = 1- p = 1 – 0.25 = 0.75. Hence,
P(6) =
11!
(11−6)!6!(0.25)6(0.75)11−6
= 0.0268
The probability that exactly six of people bought a refrigerator during the last six
months is 0.0268.
Example 2: Department of Labor and Employment (DOLE) found that 83% of Filipinos
think that having a college Education is important to succeed in life if a random
sample of seven Filipinos is selected, find these probabilities.
a. Exactly four people will agree with that statement
b. At most two people will agree with that statement
c. At least Five two people will agree with that statement.
Solution:
a. In this case, n = 7, X = 4, p = 83% = 0.83, and q = 0.17. Hence,
P(4) =
7!
(7−4)!4! (0.83)4(0.17)7−4
= 0.0816
b. Recall that n = 7, p = 0.83, and q = 0.17. “At most two people” means 0, 1, or
2. Hence, the solution i
P(0) = 71 (0.83)0(0.17)7−0 = 0.0000
(7−0)!0!
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P(1) =
P(2) =
7!
(7−1)!
1!
7!
(7−2)!
2!
(0.83)1(0.17)7−1 = 0.0001
(0.83)2(0.17)7−2 = 0.0020
P(0) + P(1) + P(2) = 0.0000 + 0.0001 + 0.0020 = 0.0023
c. Note that n = 7, p = 0.83, and q = 0.17. To find the probability that at least five
people think that having a college education is important to succeed in life, it is
necessary to find the individual probabilities for 5, 6, 7 and then add them to get
the probability.
P(5) =
P(6) =
P(7) =
Thus,
7!
(7−5)!
5!
7!
(7−6)!
6!
(0.83)5(0.17)7−5 = 0.2390
(0.83)6(0.17)7−6 = 0.3890
(0.83)7(0.17)7−7 = 0.2713
71
(7−7)!
7!
P(at least five people think that having a college education is important)
= P(5) + P(6) + P(7) = 0.2390 + 0.3890 + 0.2713 = 0.8993
1.9 Poisson Probability Distribution
The Poisson probability distribution is another important probability distribution
on a discrete random variable that has a large number of applications. It is applied to
experiment with random and independent occurrences. The occurrences (successes)
are random which does not follow any pattern and it is unpredictable. Independence
of occurrences (successes) means that the event does not influence the successive
occurrences or non-occurrences of an event. The occurrences are always measured
with respect to interval and interval maybe in terms of time, a space or a volume. The
actual number of occurrences within an interval is random and independent. If the
average number of occurrences for a given interval is known, then by using the Poison
Probability distribution, the probability of a certain number of occurrences in an
interval is represented by X. The Poisson Probability distribution was named after the
French Mathematician Samuel D. Poisson, who describes it in 1837. The Poisson
probability distributions can be described using Formula 1-24.
𝑛
P(X = x) = 𝑛� −For
x = 0, 1, 2, … (Formula 1-24)
𝑛!
where:
𝑛 = the mean number of occurrence in a particular interval of time (𝑛 =
np). e = is approximately 2.17828 (base on the Naperian logarithmic
system).
X = the number of occurrence (success).
P(X) = the probability to be computed for a specified value of X.
Example: A vintage car breaks down an average for times per month. Using the
Poisson probability distribution formula, find the probability that during the next
month this vintage car will have (a) exactly three breakdowns, (b) at most two
breakdowns, and
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(c) more than one breakdown.
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Solution:
Let 𝑛= the number of breakdowns per month.
X = the actual number of breakdowns observed during the next month for this
vintage car.
a. The probability that exactly three breakdowns observed during the next month is
P(X = 3) =
b.
𝑛 𝑛𝑛−
𝑛!
=
43𝑛−4
3!
=
64(0.183156)
6
= 0.1954
The probability that at most two breakdown will be observed during the next
month is given by the sum of the probabilities of zero, one and two breakdowns.
Hence,
P(at most 2 breakdowns) = P(0 or 1 or 2 breakdowns)
= P(X = 0) + P(X = 1) + P(X = 2)
=
40𝑛−4
0!
=
1(0.0183156)
1
+
41𝑛−4
1!
+
+
42𝑛−4
2!
4(0.0183156)
1
+
16(0.0183156)
1
= 0.0183 + 0.0733 + 0.1465
= 0.2381
c. The probability that at least two breakdown will be observed during the next
month is given by P(X > 1) = P(X ≥ 2). We could try
P(𝑛 ≥ 2) = P(X = 2) + P(X = 3) + P(X = 4) + …
= P(2) + P(3) + P(4) + …
There are infinite number of terms. A much better way is to use the fact that these
probabilities sum to 1. Consequently,
P(X = 2) = 1 – P(X < 2)
= 1 – P(X ≤ 1)
= 1 – [P(X = 0) + P(X = 1)]
=1–
[4
41𝑛−4
+
1!
0𝑛−4
]
0!
= 1- [0.0183 + 0.0733]
= 1- 0.0916
= 0.9084
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