Chapter Five.ppthhjhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

1
Elementary Probability
Chapter Five

Chapter Goals
After completing this chapter, you are expected to:
• Explain basic probability concepts and definitions
• Use a Venn diagram illustrate simple probabilities
• Determine the number of combinations or
permutations of n objects taking r at a time
• Apply common rules of probability
• Compute conditional probabilities
• Determine whether events are statistically independent
2

Introduction
 Probability is the language we use to model
uncertainty.
 Probability is the chance of an event occurring.
• One way to think about probability is the
proportion of individuals in a population that
have a particular characteristic.
• For example, suppose that in 2015 there were
18,500 students at AAU, and 6,400 of them were
females. If a single student was sampled at
random, the probability that s/he would be a
female is 6,400 / 18,500, or 0.35.
3

Introduction
 Some notions of probability:
– There is a 70% chance of rain today.
– The chances of winning a National Lottery game
are 1 in 2.5 million.
 Inferences are made about a population based on
a sample
 In probability, we use population information to
infer the probable nature of a sample.
 Probability is viewed as a Measure of Reliability
for an Inference.
 Probability is a number on a scale that runs from
zero to one inclusive.
4

5.1 Definition and Some Concepts
• Random Experiment – a process leading to an
uncertain outcome.
• Outcome is the result of a single trial of an experiment.
• A sample point is the most basic outcome of an
experiment.
5
A four
A Head

Definition . . .
• Sample space (S or Ω): the set of all possible outcomes
of a random experiment.
– The collection of all sample points.
– A sample space may have countable outcomes or
infinite outcomes.
– The sample space for rolling a die = {1, 2, 3, 4, 5, 6}
• Sample points and sample spaces are often represented
with Venn diagrams.
6

Definition . . .
Example: 1. Find the sample space for the sex of the
children if a family plans to have 3 children.
2. Find the sample space for rolling two dice.
7
1 2 3 4 5 6
1 (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
2 (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
3 (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
4 (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
5 (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
6 (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)
S = {FFF, FFM, FMM, FMF, MFF, MMF, MFM, MMM}

Definition . . .
• Event – any subset of basic outcomes from the sample
space
• Simple Event is the same as one outcome.
• Compound Event consists of two or more outcomes.
Example: Occurrence of odd number points
{1, 3, 5} in rolling a die.
• Sure (Certain) Event is same as sample space.
• Event “E” is called impossible event if E contains no
element belonging to a sample space S.
Example: Occurrence of a ‘7’ in single throw of a
die.
• Equally likely events are events with the same probability
of occurring.
8

Definition . . .
• Complement of an event A - the set of all basic outcomes in
the sample space that do not belong to A. The complement
of A is denoted by
• Intersection of Events – For two events A and B in S, the
intersection, A ∩ B, is the set of all outcomes in S that
belong to both A and B.
9
c
A
or
A
A A
AB
A B

Definition . . .
• Two or more events are mutually exclusive if they cannot
occur at the same time.
• Union of Events – For two events A and B in S, the union,
A U B, is the set of all outcomes in S that belong to either
A or B or both.
10
A B
A B

Example
• Let the Sample Space be the collection of all possible
outcomes of rolling one die:
• S = {1, 2, 3, 4, 5, 6}
• Let A be the event “Number rolled is even”
• Let B be the event “Number rolled is at least 4”
• Then A = {2, 4, 6} and B = {4, 5, 6}
a) Complements of A and B,
b) A intersection B, and
c) A union B
d) Are A and B mutually exclusive?
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Ac = {1, 3, 5}; Bc = {1, 2, 3}
{4, 6}
{2, 4, 5, 6}
No

5.2 Counting Rules
 The Addition Rule
 If A ∩ B = Ø, then n(A ∪ B) = n(A) + n(B)
 If A1, A2, . . . , Ak are k pair-wise mutually exclusive
events, then n(A1∪A2 ∪ · · · ∪Ak )=∑ n(Ai)
 For any events A & B, n (A ∪ B)=n (A) + n (B) – n(A∩ B).
• Example: In a survey, 100 people are asked whether they
drink or smoke or do both or neither. The results are: 60
drink, 30 smoke, 20 do both and 30 do neither. Are these
numbers compatible with each other?
12
N = 100; N(D) = 60; N(S) = 30; N(DnS) = 20; N(DcnSc) = 30
N(D u S) = N(S) + N(D) – N(DnS) = 30 + 60 – 20 = 70
N = N(DuS) + N(DcnSc)
= 70 + 30 They are compatible!

The Multiplication Rule
 Rule1: If each event in a sequence of n events has K
possibilities, then the total number of possibilities will be.
K.… K = Kn
Examples: 1. Seven dice are rolled. How many different
outcomes are there?
2. An instructor gives a five question multiple-choice
examination. There are four possible responses to each
question. How many different answer keys can be made?
13
K = 6; n = 7 Thus, Total = 67
K = 4; n = 5 Thus, Total = 45

The Multiplication Rule …
 Rule-2: In a sequence of n events, if there are m1 ways a
first event can occur and m2 ways a second event can
occur, … mk ways a kth event can occur, then the total
number of ways the k events can occur is given by m1 x
m2 x . . . xmk.
Example. There are 8 different statistics, 6 different calculus
and 3 different physics books. A student must select one
book of each type and make arrangement. How many
different ways can this be done?
14
m1 = 8; m2 = 6; m3 = 3
Total = 8 x 6 x 3 = 144

Permutation
 An arrangement of n objects in a specific order.
• Factorial: n! = n x (n – 1) x (n – 2) x ... x 1
Note that 1! = 0! = 1 by definition.
 The number of permutation of n objects taken all together is
given by nPn (read as n permutation n) = n!
 An arrangement of n objects in a specific order using r
objects at a time is given by:
15
r)!
(n
n!
r
nP



Permutation . . .
 Example
1. A class in probability theory consists of 10 men and 4
women. An exam is given and the students are ranked
according to their performance. Assuming that no two
students obtain the same score, how many different
rankings are possible?
2. How many different four – letter permutations can be
formed from the letters in the word DECAGON?
16
n(F) = 4; n(M) = 10; Total number of students = 4 + 10 = 14
Total number of ranking = 14P14 = 14! = 87, 178,291,200
n = 7; r = 4
Total number of ways = 7P4 = 7!/ (7-4)! = 7x6x5x4 = 840

Permutation . . .
• The number of permutation of n objects in which K1 are alike,
K2 are alike… and Kr are alike is given by
Example
 A house has 12 rooms. We want to paint 4 yellow, 3 purple and
5 red. In how many ways can this be done?
17
)
!
r
!.....K
2
K
!
1
(K
n!
n = 12; K1 = 4; K2 = 3; K3 = 5; Total = 12!/(4!x3!x5!) =
27720

Combination
 A selection of objects without regard to order.
 Example: Given the letters A, B, C and D. List the
permutations and combinations for selecting two letters.
 The number of combination of r objects selected from n
objects is denoted by nCr or
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r!
nPr
r)!r!
(n
n! 


nCr
Permutation AB; AC; AD; BA;
BC; BD; CA; CB;
CD; DA; DB; DC
Combination AB; AC; AD; BC; BD;
CD








r
n

Combination . . .
Example
1. Suppose you plan to invest equal amounts of money in each
of five business ventures. If you have 20 ventures from which
to make the selection, how many different samples of five
ventures can be selected from the 20?
2. How many ways can a jury of 6 men and 4 women be
selected from 10 men and 8 women?
19
n = 20; r = 5; Total = 20C5 = 20!/ (20-5)!x5! = 20!/ 15!x5! = 15, 504
n(W) = 8; n(M) = 10; r(W) = 4; r(M) = 6
Total(M) = 10C6 = 210; Total(W) = 8C4 = 70
Grand Total = 210 x 70 = 14, 700

5.3 Approaches in Probability Definition
 Classical Probability
 Note: all outcomes in the sample space should be equally likely
to occur.

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Approaches in Definitions
Classical Subjective Axiomatic
Frequency
space
sample
the
in
outcomes
of
number
total
event
the
satisfy
that
outcomes
of
number
N
A
N
A
event
of
y
probabilit 

)
A
(
P
1
)
C
A
(
P
)
C
A
(
P
1
)
A
(
P
1
)
C
A
(
P
)
A
(
P







Approaches …
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S = {1, 2, 3, 4, 5, 6}
E = {1, 3, 5}
P(E) = 3/6


1
i
)
i
A
(
P

Properties of the Probability of an Event
• Probability – the chance that an
uncertain event will occur (always
between 0 and 1)
• The probabilities of all the sample
points must sum to 1
0 ≤ P(A) ≤ 1 For any event A
Certain
Impossible
.5
1
0
P(Impossible Event) = 0
P(Sure Event) =1
1
1
i
n
i
p



P(A)
1
)
A
P( 

22

5.4 Some Probability Rules
• The Addition Rule for Probability
• It is concerned with the probability of a union of outcomes.
 i.e., for two events A and B: simple probability that A
occurs added to the simple probability that B occurs.
 If events A and B are mutually exclusive, the probability that
A or B occurs is given by P (A U B) = P (A) + P (B)
Example
A day of the week is selected at random. Find the probability
that it is weekend day {Saturday or Sunday}.
23
S = { Mo, Tu, We, Th, Fr, Sa, Su}
A = {Sa}; B = {Su}
P(A) = 1/7; P(B) = 1/7
P(A u B) = P(A) + P(B) = 1/7 + 1/7

Probability Rules . . .
• If two events A and B are not mutually exclusive, then
Pr (A U B) = Pr (A) + Pr (B) – P r(A and B)
Example
• Consider the experiment of having two children and let
A = {event that 1st child is a girl}
B = {event that 2nd child is a girl}
• Assume Pr(A) = Pr(B) = ½
• Then find the probability of having at least one girl.
24

Exercise. There are 80 nurses and 40 physicians in a hospital. Of
these, 70 nurses and 15 physicians are females. If a staff person
is selected at random, find the probability that the subject is a
nurse or male. (Answer: 105/120)
Note
• If A1, A2, …, An are pair –wise mutually exclusive
25
B)
U
P(A
1
)
B
A
P( 



B)
P(A
1
c
B)
P(A 
 

B)
P(A
1
)
B
U
A
P( 




B)
P(A
1
c
B)
P(A 
 

C)
B
P(A
C)
P(B
C)
P(A
B)
P(A
P(C)
P(B)
P(A)
P(AUBUC) 



 













n
1
i
)
i
P(A
)
n
P(A
....
)
2
P(A
)
1
P(A
)
n
....UA
U
3
A
U
2
A
U
1
P(A

• Example: In a freshman dorm there are 90 students, who watch
different TV programs. 60 students watch African Journal, 40
watch Ethiopian Idols and 30 watch Feature Film. Moreover, 20
watch African Journal and Ethiopian Idols, 15 watch Ethiopian
Idols and Feature Film, 10 watch African Journal and Feature
Film. In addition, 5 of them watch all the three programs. What
is the probability that a randomly selected student watches at
least one TV program?
26
Let the following events are defined
A = Watching African Journal
B = Watching Ethiopian Idol
C = Watching Feature Film
n(A) = 60 ; n(B) = 40 ; n(C) = 30 ; n(A n B) = 20 ; n(B n C) = 15
n(A n C) = 10 ; n(A n B n C) = 5 ; n(S) = 90
P(A u B u C) = P(A) + P(B) + P(C) – P(AnB) – P(AnC) – P(BnC) +
P(AnBnC)
= 60/90 + 40/90 + 30/90 – 20/90 – 10/90 - 15/90 + 5/90
= 1.00

The Multiplication Rules for Probability
• Concerned with the probability of intersection of outcomes.
• When two events A and B are independent the probability of
both occurring is =
• Example
• An urn contains 3 red balls, 2 blue balls and 5 white balls. A
ball is selected, its color noted, and then replaced back. A
second ball is selected and its color noted. Find the
probability of each of the following.
• Selecting 2 blue balls
• Selecting a red ball and then a white ball
27
P(B)
x
P(A)
B)
P(A 

n(R) = 3; n(B) = 2; n(W) = 5
P(R) = 3/10; P(B) = 2/ 10; P(W) = 5/10
P(B1 n B2) = P(B1) x P(B2) = 2/10 x 2/10 = 4/100
P(R1 n W2) = P(R1) x P(W2) = 3/10 x 5/10 = 15/100

Multiplication Rules . . .
 If the probability of occurrence of one event affects the
other, the two events are said to be dependent.
 For dependent events (A&B), the probability of both
occurring is: P (A n B) = P (A) x P (B|A) = P(B) x P(A|B)
Example:
• A flashlight has six batteries, two of which are defective. If
two are selected at random without replacement, find the
probability that both are defective.
28
n = 6; n(D) = 2
P(D) = 2/ 6 = 1/3
P(D1D2) = P(D1) x P(D2|D1) = 1/3 x 1/5 = 1/15

A Probability Table
B
A
A
B
)
B
P(A 
)
B
A
P( 
B)
A
P( 
P(A)
B)
P(A 
)
A
P(
)
B
P(
P(B) 1.0
P(S) 
Probabilities and joint probabilities for two events A and
B are summarized in this table:
29

5.5 Conditional probability and independence
• Additional information may have an impact on the
probability of an event.
– P(Rolling a 4 in a die) is one-sixth (unconditionally).
– If we know an even number was rolled, the probability
of a 4 goes up to one-third.
• A conditional probability is the probability of one event
given that another event has occurred.
• The sample space is reduced to only the conditioning event.
• To find P(A), once we know B has occurred (i.e., given B),
we ignore BC (including the A region within BC).
30
B
A
BC
A

Conditional probability . . .
Example
1. Consider the experiment of tossing a coin 3 times
S= {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}
Let A: The outcome of the first toss is tail
B: The outcome of the second toss is tail.
• Find P (B|A)
31
0
)
B
(
P
;
P(B)
B)
P(A
B)
|
P(A 


0
)
A
(
P
;
P(A)
B)
P(A
A)
|
P(B 


The conditional
probability of A given
that B has occurred
The conditional
probability of B given
that A has occurred

• A = {TTT, TTH, THT, THH}; B = {TTT, TTH, HTT, HTH}
• A and B = {TTT, TTH}; P(A) = 4/8; P(A and B) = 2/8
• P(B|A) = P(A and B)/ P(A)= (2/8)/ 4/8= 1/2
2. In a class of 100 students 30% of them read Statistics, 20%
of them read Mathematics and 15% of them read both
Mathematics and Statistics. One student is selected at
random. Find the probability
a) that a student reads Statistics, if it is known that s/he reads
Mathematics;
b) that s/he reads Mathematics, if it is known that the student
reads Statistics.
32
n = 100; P(S) = 0.30; P(M) = 0.20; P(M n S) = 0.15
a) P(S|M) = P(M n S)/ P(M) = 0.15/ 0.20 =0.75
b) P(M|S) = P(M n S) / P(S) = 0.15/ 0.30 = 0.50

• Exercise. Consider again the experiment of having two
children. Suppose that the first child is known to be a
boy, and the second child’s gender is unknown. What is
the conditional probability of at least one girl given that
the first child is a boy? (Answer: ½)
33

Statistical Independence
• Events A and B are independent when the probability of one
event is not affected by the other event
• Two events are statistically independent if and only if:
• If A and B are independent, then
• If the events A and B are independent, then so are all three
sets of events: A and Bc; Ac and B; Ac and Bc.
• i.e., P(A ∩ Bc) = P(A)P(Bc); P(Ac ∩ B) = P(Ac)P(B)
• P(Ac ∩ Bc) = P(Ac)P(Bc)
34
P(B)
P(A)
B)
P(A 

0
P(B)
P(A)
B)
|
P(A 
 where
;
0
P(A)
where
P(B);
A)
|
P(B 


Statistical Independence Example
• Suppose that we roll a pair of fair dice, so each of the 36
possible outcomes is equally likely. Let A denote the event
that the first die lands on 3, and let B be the event that the
sum of the dice is 8.
• Are A and B independent?
A = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)}
B = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}
Since A ∩ B is the event that the first die lands on 3 and the
second on 5, A ∩ B = {(3,5)}
P(A) = 6/36
P(B) = 5/36
P(A ∩ B) = 1/36
Since 1/36  (6/36)(5/36) events A and B are not independent.
35

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