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Nor Fashihah Mohd NoorInstitut Matematik Kejuruteraan
Universiti Malaysia Perlis
ІМќINSTITUT MATEMATIK K E J U R U T E R A A NU N I M A P
When you toss a single coin, you will see either a head (H) or a tail (T). If you toss the coin repeatedly, you will generate an infinitely large number of Hs and Ts – the entire population. What does the population look like? If the coin is fair, then the population should contain 50% Hs and 50% Ts. Now toss the coin one more time. What is the chance to getting a head? Most people would say that the “probability” or chance is ½.
3
Probability Probability is a measure of the likelihood of an event A
occurring in one experiment or trial and it is denoted by P (A).
Experiment An experiment is any process of making an observation
leading to outcomes for a sample space.
4
number of ways that the event can occur ( ) ( )
total number of outcomes( ) ( )A A n A
P AS n S
The mathematical basis of probability is the theory of sets.
Definition 1.2: Sample Spaces, Sets and Events
Sets
A set is a collection of elements or components
Sample Spaces, S
A sample space consists of points that correspond to all possible outcomes.
Events
An event is a set of outcomes of an experiment and a subset of the sample space.
5
Let A denote the events of obtaining a number which could divide by 3 in an experiment of tossing a dice, Hence,
A = {3, 6} is a subset of S = {1,2,3,4,5,6}
Basic Operations
Figure 1.1: Venn diagram representation of events
6
CB
A
S
1. The union of events A and B, which is denoted as , is the set of all elements that belong to A or B or both. Two or more events are called collective exhaustive events if the unions of these events result in the sample space.
2. The intersection of events A and B, which is denoted by ,
is the set of all elements that belong to both A and B. When A and B have no outcomes in common, they are said to be mutually exclusive or disjoint sets.
3. The event that contains all of the elements that do not belong to an event A is called the complement of A and is denoted by
7
A B
A B
A
Given the following sets;
A = {2, 4, 6, 8, 10}B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}C = {1, 3, 5, 11, ……} , the set of odd numbersD = {failure of a structure due to an earth quake}E = {failure of a structure due to strong winds}
8
9
State the set of;
i 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
ii 2, 4, 6, 8, 10
iii 2, 4, 6, 8, 10, , the set of even numbers
iv failure of a structure due to an earth quake or strong w
A B B
A B A
C
D E
inds
v non-failure of the structure due to an earth quakeD
Example 1.3
In a carom game, there are 3 white seeds and 2 black seeds. Three seeds have been successfully converted in the game. Assuming that the convert is being done one after another and the rules of carom game are disobeyed, calculate the probability.
SolutionLet, M = white seed and K = black seed
11
1
2 1
1 2
1 2 3 1 2 3 1 2 3 1 2 3
3a) ( )
51
b) ( | )2
3 2 3c) ( )
5 4 10d) ( ) ( ) ( ) ( )
3 2 1 3 2 2 2 3 2 2 1 3
5 4 3 5 4 3 5 4 3 5 4 3
3
5
P M
P K M
P M M
P M M M P M K M P K M M P K K M
3
5
2
5
1M
1K
4
2
4
2
3
4
1
4
2M
First
convert
Second
convert
Third
convert
2K
2K
3K
2M
3M
3M
3K
3K
3K
3M
3M
1
3
2
32
3
1
32
3
1
33
3
0
Results
1 2 3M M M
1 2 3M M K
1 2 3M K M
1 2 3M K K
1 2 3K M M
1 2 3K M K
1 2 3K K M
1 2 3K K K
This counting rule allows one to count the number of outcomes when the experiment involves selecting r objects from a set of n objects.
!
! !
n nnCr r r n r
8 8! 8 7 656
3 5!3! 3 2 1
Example 1.4Suppose that in the taste test, each participant samples eight products and is asked the three best products, butnot in any particular order.
SolutionThe number of possible answer test is then
This counting rule allows one to compute the number of outcomes when r objects are to be selected from a set of n objects where the order of selection is important.
!
!
nnPr n r
Example 1.5
Three lottery tickets are drawn from a total of 50. If the tickets will be distributed to each of the three employees in the order in which they are drawn, the order will be important. How many simple events are associated with the experiment?
SolutionThe total number of simple events is
503
50!50 49 48 117600
47!P
1) 0 ( ) 1
2) ( ) ( ) 1
3) ( ) ( ) ( )
4) ( ) ( ) ( )
5) ( ) 1 ( )
6) (( ) ) ( )
7) (( ) ) ( )
8) ( ( )) ( )
9) ( ) [( ) ( )]
P A
P A P A
P A B P A P A B
P A B P B P A B
P A B P A B
P A B P A B
P A B P A B
P A A B P A B
P B P A B A B
A B A BA B
A B
S
Probability Axioms:
Axiom 1 : For any event , A, P(A) ≥ 0
Axiom 2 : P(S) = 1
Axiom 3 : For any countable collection A1 , A2, ……… of mutually exclusive
events
Theorem 1.1 : Laws of Probability
1 2
a) ( ) 1 –
b) ( ) – ( )
c) ( ) – ( ) – ( ) – ( ) ( )
d) If and are mutually exclusive events, then ( ) 0
e) If and are the subset of where
P A P A
P A B P A P B P A B
P A B C P A P B P C P A B P A C P B C P A B C
A B P A B
A A S A
1 2 , 1 2 then A P A P A
1 2 1 2( ..) ( ) ( ) P A A P A P A
Example 1.6
Two fair dice are thrown. Determine a) the sample space of the experimentb) the elements of event A if the outcomes of both dice thrown
are showing the same digit. c) the elements of event B if the first thrown giving a greater
digit than the second thrown.d) probability of event A, P(A) and event B, P(B)
1 2 3 4 5 6
1 (1, 1) (1, 2) (1, 3) (1, 2) (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)
b) A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
c) B = {(2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3), (5, 4), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}
( ) 6 1d)
( ) 36 6
( ) 15 5
( ) 36 12
n AP A
n S
n BP B
n S
Example 1.7Consider randomly selecting a KUKUM Master Degreeinternational student, and let A denote the event that the selected individual has a Visa Card and B has a Master Card. Suppose that P(A) = 0.5 and P(B) = 0.4 and = 0.25.
a) Compute the probability that the selected individual has at least one of the two types of cards ?
b) What is the probability that the selected individual has neither type of card?
( )P A B
a) ( ) – ( )
= 0.5 0.4 – 0.25 0.65
b) 1 ( ) 1– 0.65 0.35
P A B P A P B P A B
P A B
Example 1.8 In an assessment for students who tookEngineering Mathematics I course, it is known that the percentage of students who passed in their monthly test is 80% while 85% passed their quiz and 75% passed in both monthly test and quiz. A students is selected at random, calculate the probability that a) passed the monthly test or quizb) passed the monthly test but failed quizc) failed both monthly test and quiz
SolutionsLet,
U = student passed the monthly testK = student passed the quiz
a) ( ) ( ) ( ) ( )
= 0.8 + 0.85 0.75 0.9
b) ( ) ( ) ( )
0.8 0.75 0.05
c) ( ) 1 ( )
1 0.9 0.1
P U K P U P K P U K
P U K P U P U K
P U K P U K
Example 1.9 In a certain residential area, 60% of all households subscribe to Berita Harian newspaper, 80% subscribe to The Star paper, and 50% of all households subscribe to both papers. If a household is selected at random, what is the probability that it subscribes a) at least one of the two newspaperb) exactly one of the two newspaper
Solutions Let , A = subscribes to Berita Harian paper B = subscribes to The Star paper
a) ( ) – ( )
0.6 0.8 – 0.5 0.9
b) ( ) ( ) ( ) ( )
0.1 0.3 0.4
P A B P A P B P A B
P A B P A B P A B P A B
Example 1.10 A box in a certain supply room contains four 40-W lightbulbs, two 60-W bulbs and eight 75-W bulbs. One lightbulb is randomly selected, calculate the probability that the selected lightbulb is rated 40-W or 60-W ?
SolutionSample space, S = {four 40-W lightbulbs, two 60-W bulbs, eight 75-W bulbs }Then, n(S) = 14.
Let the events A and B be,
A = 40-W lightbulbs is selected
B = 60-W lightbulbs is selected
Hence,
Events A and B are said to be mutually exclusive because of the
two events couldn’t occur at the same time. Thus,
( ) 4 2( )
( ) 14 7
( ) 2 1( )
( ) 14 7
n AP A
n S
n BP B
n S
( ) ( ) ( )
2 1 3
7 7 7
P A B P A P B
Definition 1.4 : For any two events A and B with P(B) > 0, the conditional probability of A given that B has occurred is defined by
( )( | )
( )
P A BP A B
P B
Area/Gender Male Female Total
Urban 35 10 45
Rural 25 30 55
Total 60 40 100
If a student is selected at random and have been told that the individual is a male student, what is the probability of he is from urban area?
Probability of a person is from urban area and it is known that the individual is a male student ,
( ) 35( ) 35( ) 100( | )
( ) 60( ) 60( ) 100
n A CP A C n S
P A Cn CP Cn S
Transmission type/Colour Grey Blue Black Red
Automatic 0.15 0.10 0.10 0.10
Manual 0.15 0.05 0.15 0.20
a) ( ), ( ) and ( )
b) ( | ) and ( | )
c) ( | ) and ( | )
P A P B P A B
P A B P B A
P A C P A C
a)P(A) 0.15 0.10 0.10 0.10 0.45
P(B) 0.10 0.15 0.25
P(A B) 0.10
( ) 0.10b) | 0.4 ,
( ) 0.25
( ) 0.10 | 0.222
( ) 0.45
P A BP A B
P B
P B AP B A
P A
( ) 0.15c) P A | C 0.5 ,
( ) 0.30
( ) 0.30 P(A|C ) 0.428
( ) 0.70
P A C
P C
P A C
P C
1 2 If , ,..., is a partition of a sample space, then the
of events conditional on an event can be obtained
from the probabilities and | using the formula,
|
n
i
i i
i
A A A
A B
P A P B A
P A
posterior
probabilities
1
| |
|
i i i i in
j jj
P A B P A P B A P A P B AB
P B P B P A P B A
Solution
Let denote the label of the selected box and let denote the indicator random variable
for choosing a red ball. Then 1 2 3 1/ 3. Also,
1| 1 1/ 4, 1| 2 2 / 4, and 1| 3 3 / 4.
Thus, the
X Y
P X P X P X
P Y X P Y X P Y X
probability of choosing a red ball is given by,
1 1 1| 1 2 1| 2 3 1| 3
1 1 1 2 1 3
3 4 3 4 3 41
2
Therefore, by Bayes’ theorem, the conditional probability t
P Y P X P Y X P X P Y X P X P Y X
hat box 1 was selected,
given that a red ball is chosen, is given by
1 11 1| 1 13 4
1| 111 62
P X P Y XP X Y
P Y
Definition 1.5 : Two events A and B are said to be independent if and only if either
Otherwise, the events are said to be dependent.
( | ) ( )
or
( | ) ( )
P A B P A
P B A P B
The probability that both two events and , occur is
( ) |
|
If and are independent,
( )
A B
P A B P A P B A
P B P A B
A B
P A B P A P B
3 1Suppose that ( ) and ( ) . Are events and independent or
5 3mutually exclusive if ,
1a) ( )
514
b) ( )15
P A P B A B
P A B
P A B
a) ( ) ( ) ( ) ( )
3 1 1 11
5 3 5 15
1Notice that ( ) 0, then and are mutually non exclusive events.
5
3 1 1( ). ( ) ,
5 3 5
Notice that, ( ) ( ). ( ), then and are i
P A B P A P B P A B
P A B A B
P A P B
P A B P A P B A B
ndependent events.
b) ( ) ( ) ( ) ( )
3 1 14 0
5 3 15
( ) 0, and are mutually exclusive events.
3 1 1( ). ( )
5 3 5
Notice that ( ) ( ). ( ), then and are dependent events.
P A B P A P B P A B
P A B A B
P A P B
P A B P A P B A B