An application of the hyperfunction theory to numerical integrationHidenoriOgata
The slide of a speech in the conference "ECMI2016" (The 19th European Conference on Mathematics for Industry) held at Santiago de Compostela, Spain in June 2016.
X std maths - Relations and functions (ex 1.1), Maths, IX std Maths, Samacheerkalvi maths, II year B.Ed., Pedagogy, Mathematics, Relation, Functions, Cartesian product, ordered pair, definition of cartesian product, standard infinite set, cartesian product of three sets,
An application of the hyperfunction theory to numerical integrationHidenoriOgata
The slide of a speech in the conference "ECMI2016" (The 19th European Conference on Mathematics for Industry) held at Santiago de Compostela, Spain in June 2016.
X std maths - Relations and functions (ex 1.1), Maths, IX std Maths, Samacheerkalvi maths, II year B.Ed., Pedagogy, Mathematics, Relation, Functions, Cartesian product, ordered pair, definition of cartesian product, standard infinite set, cartesian product of three sets,
On homogeneous biquadratic diophantineequation x4 y4=17(z2-w2)r2eSAT Journals
Abstract
Five different methods of the non-zero non-negative solutions of non- homogeneous cubic Diophantine equation x4 – y4 = 17( z2 –
w2) R2 are obtained. Some interesting relations among the special numbers and the solutions are exposed.
Keywords: The Method of Factorization, Integer Solutions, Linear Transformation, Relations and Special Numbers
On cubic diophantine equation x2+y2 xy=39 z3eSAT Journals
Abstract
Four different methods of the non-zero non-negative solutions of non- homogeneous cubic Diophantine equation x2 + y2 – xy =
39z2 are obtained. Some interesting relations among the special numbers and the solutions are exposed.
Keywords: The Method of Factorization, Integer Solutions, Linear Transformation, Relations and Special Numbers
Fortran is a general-purpose programming language, mainly intended for mathematical computations in
science applications
this chapter is the third chapter
Accurate Numerical Method for Singular Initial-Value Problemsaciijournal
In this paper, an accurate numerical method is presented to find the numerical solution of the singular initial value problems. The second-order singular initial value problem under consideration is transferred into a first-order system of initial value problems, and then it can be solved by using the fifth-order Runge Kutta method. The stability and convergence analysis is studied. The effectiveness of the proposed methods is confirmed by solving three model examples, and the obtained approximate solutions are compared with the existing methods in the literature. Thus, the fifth-order Runge-Kutta method is an accurate numerical method for solving the singular initial value problems.
ملزمة الرياضيات للصف السادس التطبيقي الفصل الاول الاعداد المركبة 2022anasKhalaf4
طبعة جديدة ومنقحة
حل تمارين الكتاب
شرح المواضيع الرياضية بالتفصيل وبأسلوب واضح ومفهوم لجميع المستويات
حلول الاسألة الوزارية
اعداد الدكتور أنس ذياب خلف
email: anasdhyiab@gmail.com
On Optimization of Manufacturing of a Sense-amplifier Based Flip-flopBRNSS Publication Hub
The paper describes an approach for increasing of density of field-effect heterotransistors in a sense-amplifier based flip-flop. To illustrate the approach, we consider manufacturing of an amplifier of power in a heterostructure with specific configuration. One shall dope some specific areas of the heterostructure by diffusion or ion implantation. After that, it should be done optimized annealing of radiation defects and/or dopant. We introduce an approach for decreasing of stress between layers of heterostructure. Furthermore, it has been considered an analytical approach for prognosis of heat and mass transport in heterostructures, which can be take into account mismatch-induced stress.
On homogeneous biquadratic diophantineequation x4 y4=17(z2-w2)r2eSAT Journals
Abstract
Five different methods of the non-zero non-negative solutions of non- homogeneous cubic Diophantine equation x4 – y4 = 17( z2 –
w2) R2 are obtained. Some interesting relations among the special numbers and the solutions are exposed.
Keywords: The Method of Factorization, Integer Solutions, Linear Transformation, Relations and Special Numbers
On cubic diophantine equation x2+y2 xy=39 z3eSAT Journals
Abstract
Four different methods of the non-zero non-negative solutions of non- homogeneous cubic Diophantine equation x2 + y2 – xy =
39z2 are obtained. Some interesting relations among the special numbers and the solutions are exposed.
Keywords: The Method of Factorization, Integer Solutions, Linear Transformation, Relations and Special Numbers
Fortran is a general-purpose programming language, mainly intended for mathematical computations in
science applications
this chapter is the third chapter
Accurate Numerical Method for Singular Initial-Value Problemsaciijournal
In this paper, an accurate numerical method is presented to find the numerical solution of the singular initial value problems. The second-order singular initial value problem under consideration is transferred into a first-order system of initial value problems, and then it can be solved by using the fifth-order Runge Kutta method. The stability and convergence analysis is studied. The effectiveness of the proposed methods is confirmed by solving three model examples, and the obtained approximate solutions are compared with the existing methods in the literature. Thus, the fifth-order Runge-Kutta method is an accurate numerical method for solving the singular initial value problems.
ملزمة الرياضيات للصف السادس التطبيقي الفصل الاول الاعداد المركبة 2022anasKhalaf4
طبعة جديدة ومنقحة
حل تمارين الكتاب
شرح المواضيع الرياضية بالتفصيل وبأسلوب واضح ومفهوم لجميع المستويات
حلول الاسألة الوزارية
اعداد الدكتور أنس ذياب خلف
email: anasdhyiab@gmail.com
On Optimization of Manufacturing of a Sense-amplifier Based Flip-flopBRNSS Publication Hub
The paper describes an approach for increasing of density of field-effect heterotransistors in a sense-amplifier based flip-flop. To illustrate the approach, we consider manufacturing of an amplifier of power in a heterostructure with specific configuration. One shall dope some specific areas of the heterostructure by diffusion or ion implantation. After that, it should be done optimized annealing of radiation defects and/or dopant. We introduce an approach for decreasing of stress between layers of heterostructure. Furthermore, it has been considered an analytical approach for prognosis of heat and mass transport in heterostructures, which can be take into account mismatch-induced stress.
Manual Solution Probability and Statistic Hayter 4th EditionRahman Hakim
All of material inside is un-licence, kindly use it for educational only but please do not to commercialize it.
Based on 'ilman nafi'an, hopefully this file beneficially for you.
Thank you.
Estimate the hidden States of a Non-linear Dynamic Stochastic System from Noisy Measurements. Estimation is a prerequisite. The Probability Theory summary is included.
The presentation is at graduate level in math and engineering.
For comments please connect me at solo.hermelin@gmail.com.
For more presentations on different subjects visit my website at http://www.solohermelin.com.
Introduction to Mathematical ProbabilitySolo Hermelin
This is a lecture I've put together summarizing the topics of mathematical probability.
The presentation is at a Undergraduate in Science (Math, Physics, Engineering) level..
In the Upload Process a part of Figures and Equations are missing. For a better version of this presentation please visit my website at http://solohermelin.com at Math Folder and open Probability presentation.
Please feel free to comment and suggest improvements to solo.hermelin@gmail.com.Thanks!
1 Probability Please read sections 3.1 – 3.3 in your .docxaryan532920
1
Probability
Please read sections 3.1 – 3.3 in your textbook
Def: An experiment is a process by which observations are generated.
Def: A variable is a quantity that is observed in the experiment.
Def: The sample space (S) for an experiment is the set of all possible outcomes.
Def: An event E is a subset of a sample space. It provides the collection of outcomes
that correspond to some classification.
Example:
Note: A sample space does not have to be finite.
Example: Pick any positive integer. The sample space is countably infinite.
A discrete sample space is one with a finite number of elements, { }1,2,3,4,5,6 or one that
has a countably infinite number of elements { }1,3,5,7,... .
A continuous sample space consists of elements forming a continuum. { }x / 2 x 5< <
2
A Venn diagram is used to show relationships between events.
A intersection B = (A ∩ B) = A and B
The outcomes in (A intersection B) belong to set A as well as to set B.
A union B = (A U B) = A alone or B alone or both
Union Formula
For any events A, B, P (A or B) = P (A) + P (B) – P (A intersection B) i.e.
P (A U B) = P (A) + P (B) – P (A ∩ B)
3
cA complement not A A ' A A = = = =
A complement consists of all outcomes outside of A.
Note: P (not A) = 1 – P (A)
Def: Two events are mutually exclusive (disjoint, incompatible) if they do not intersect,
i.e. if they do not occur at the same time. They have no outcomes in common.
When A and B are mutually exclusive, (A ∩ B) = null set = Ø, and P (A and B) = 0.
Thus, when A and B are mutually exclusive, P (A or B) = P (A) + P (B)
(This is exactly the same statement as rule 3 below)
Axioms of Probability
Def: A probability function p is a rule for calculating the probability of an event. The
function p satisfies 3 conditions:
1) 0 ≤ P (A) ≤1, for all events A in the sample space S
2) P (Sample Space S) = 1
3) If A, B, C are mutually exclusive events in the sample space S, then
P(A B C) P(A) P(B) P(C)∪ ∪ = + +
4
The Classical Probability Concept: If there are n equally likely possibilities, of which one
must occur and s are regarded as successes, then the probability of success is s
n
.
Example:
Frequency interpretation of Probability: The probability of an event E is the proportion of
times the event occurs during a long run of repeated experiments.
Example:
Def: A set function assigns a non-negative value to a set.
Ex: N (A) is a set function whose value is the number of elements in A.
Def: An additive set function f is a function for which f (A U B) = f (A) + f (B) when A and
B are mutually exclusive.
N (A) is an additive set function.
Ex: Toss 2 fair dice. Let A be the event that the sum on the two dice is 5. Let B be the
event that the sum on ...
It is a consolidation of basic probability concepts worth understanding before attempting to apply probability concepts for predictions. The material is formed from different sources. ll the sources are acknowledged.
JEE Mathematics / Lakshmikanta Satapathy / Probability / Multiplication theorem of probability, for two or more events and independent events and Addition theorem of probability
This Presentation was prepared to tell the students how to write a letter to the teacher and business type letters were discussed in this presentation.
June 3, 2024 Anti-Semitism Letter Sent to MIT President Kornbluth and MIT Cor...Levi Shapiro
Letter from the Congress of the United States regarding Anti-Semitism sent June 3rd to MIT President Sally Kornbluth, MIT Corp Chair, Mark Gorenberg
Dear Dr. Kornbluth and Mr. Gorenberg,
The US House of Representatives is deeply concerned by ongoing and pervasive acts of antisemitic
harassment and intimidation at the Massachusetts Institute of Technology (MIT). Failing to act decisively to ensure a safe learning environment for all students would be a grave dereliction of your responsibilities as President of MIT and Chair of the MIT Corporation.
This Congress will not stand idly by and allow an environment hostile to Jewish students to persist. The House believes that your institution is in violation of Title VI of the Civil Rights Act, and the inability or
unwillingness to rectify this violation through action requires accountability.
Postsecondary education is a unique opportunity for students to learn and have their ideas and beliefs challenged. However, universities receiving hundreds of millions of federal funds annually have denied
students that opportunity and have been hijacked to become venues for the promotion of terrorism, antisemitic harassment and intimidation, unlawful encampments, and in some cases, assaults and riots.
The House of Representatives will not countenance the use of federal funds to indoctrinate students into hateful, antisemitic, anti-American supporters of terrorism. Investigations into campus antisemitism by the Committee on Education and the Workforce and the Committee on Ways and Means have been expanded into a Congress-wide probe across all relevant jurisdictions to address this national crisis. The undersigned Committees will conduct oversight into the use of federal funds at MIT and its learning environment under authorities granted to each Committee.
• The Committee on Education and the Workforce has been investigating your institution since December 7, 2023. The Committee has broad jurisdiction over postsecondary education, including its compliance with Title VI of the Civil Rights Act, campus safety concerns over disruptions to the learning environment, and the awarding of federal student aid under the Higher Education Act.
• The Committee on Oversight and Accountability is investigating the sources of funding and other support flowing to groups espousing pro-Hamas propaganda and engaged in antisemitic harassment and intimidation of students. The Committee on Oversight and Accountability is the principal oversight committee of the US House of Representatives and has broad authority to investigate “any matter” at “any time” under House Rule X.
• The Committee on Ways and Means has been investigating several universities since November 15, 2023, when the Committee held a hearing entitled From Ivory Towers to Dark Corners: Investigating the Nexus Between Antisemitism, Tax-Exempt Universities, and Terror Financing. The Committee followed the hearing with letters to those institutions on January 10, 202
How to Make a Field invisible in Odoo 17Celine George
It is possible to hide or invisible some fields in odoo. Commonly using “invisible” attribute in the field definition to invisible the fields. This slide will show how to make a field invisible in odoo 17.
Palestine last event orientationfvgnh .pptxRaedMohamed3
An EFL lesson about the current events in Palestine. It is intended to be for intermediate students who wish to increase their listening skills through a short lesson in power point.
Acetabularia Information For Class 9 .docxvaibhavrinwa19
Acetabularia acetabulum is a single-celled green alga that in its vegetative state is morphologically differentiated into a basal rhizoid and an axially elongated stalk, which bears whorls of branching hairs. The single diploid nucleus resides in the rhizoid.
Macroeconomics- Movie Location
This will be used as part of your Personal Professional Portfolio once graded.
Objective:
Prepare a presentation or a paper using research, basic comparative analysis, data organization and application of economic information. You will make an informed assessment of an economic climate outside of the United States to accomplish an entertainment industry objective.
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Synthetic Fiber Construction in lab .pptxPavel ( NSTU)
Synthetic fiber production is a fascinating and complex field that blends chemistry, engineering, and environmental science. By understanding these aspects, students can gain a comprehensive view of synthetic fiber production, its impact on society and the environment, and the potential for future innovations. Synthetic fibers play a crucial role in modern society, impacting various aspects of daily life, industry, and the environment. ynthetic fibers are integral to modern life, offering a range of benefits from cost-effectiveness and versatility to innovative applications and performance characteristics. While they pose environmental challenges, ongoing research and development aim to create more sustainable and eco-friendly alternatives. Understanding the importance of synthetic fibers helps in appreciating their role in the economy, industry, and daily life, while also emphasizing the need for sustainable practices and innovation.
Biological screening of herbal drugs: Introduction and Need for
Phyto-Pharmacological Screening, New Strategies for evaluating
Natural Products, In vitro evaluation techniques for Antioxidants, Antimicrobial and Anticancer drugs. In vivo evaluation techniques
for Anti-inflammatory, Antiulcer, Anticancer, Wound healing, Antidiabetic, Hepatoprotective, Cardio protective, Diuretics and
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1.4 modern child centered education - mahatma gandhi-2.pptx
Probability And Random Variable Lecture(3)
1. Last Lecture
• We studied behavior of multiple events.
• Any Question??
2. Lecture 3
1.4 Conditional Probability
1.5 Probabilities of Event Intersections
1.6 Posterior Probabilities
3. 1.4 Conditional Probability
1.4.1 Definition of Conditional Probability(1/2)
• Conditional Probability
The conditional probability of event A conditional on event B is
for P(B)>0. It measures the probability that event A occurs when it is
known that event B occurs.
( )
( | )
( )
P A B
P A B
P B
∩
=
( ) 0
( | ) 0
( ) ( )
( ) ( )
( | ) 1
( ) ( )
A B
P A B
P A B
P B P B
B A
A B B
P A B P B
P A B
P B P B
∩ = ∅
∩
= = =
⊂
∩ =
∩
= = =
g
g
4. 1.4.1 Definition of Conditional Probability(2/2)
( ) 0.26
( | ) 0.464
( ) 0.56
( ) ( ) ( ) 0.27 0.26
( | ) 0.023
( ) 1 ( ) 1 0.56
( | ) ( | ) 1
Formally,
( ) ( )
( | ) ( | )
( ) ( )
1 1
( ( ) ( )) ( ) 1
( ) ( )
( |
P A B
P A B
P B
P A B P A P A B
P A B
P B P B
P A B P A B
P A B P A B
P A B P A B
P B P B
P A B P A B P B
P B P B
P A
∩
= = =
′∩ − ∩ −
′ = = = =
′ − −
′⇒ + =
′∩ ∩
′+ = +
′= ∩ + ∩ = =
g
g
( ( ))
)
( )
P A B C
B C
P B C
∩ ∪
∪ =
∪
5. 1.4.2 Examples of Conditional Probabilities(1/3)
• Example 4 Power Plant Operation
A = { plant X is idle } and P(A) = 0.32
Suppose it is known that at least two
out of the three plants are generating
electricity ( event B ).
B = { at least two out of the three
plants generating electricity }
P(A) = 0.32
=> P(A|B) = 0.257
( ) 0.18
( | ) 0.257
( ) 0.70
P A B
P A B
P B
∩
= = =
6. 1.4.2 Examples of Conditional Probabilities(2/3)
• GAMES OF CHANCE
- A fair die is rolled.
- A red die and a blue die are thrown.
A = { the red die scores a 6 }
B = { at least one 6 is obtained on the two dice }
1
(6)
6
(6 ) (6)
(6 | )
( ) ( )
(6) 1/ 6 1
(2) (4) (6) 1/ 6 1/ 6 1/ 6 3
P
P even P
P even
P even P even
P
P P P
=
∩
= =
= = =
+ + + +
6 1 11
( ) and ( )
36 6 36
( )
( | )
( )
( )
( )
1/ 6 6
11/36 11
P A P B
P A B
P A B
P B
P A
P B
= = =
∩
=
=
= =
7. 1.4.2 Examples of Conditional Probabilities(3/3)
C = { exactly one 6 has been scored }
( )
( | )
( )
5/36 1
10/36 2
P A C
P A C
P C
∩
=
= =
8. 1.5 Probabilities of Event Intersections
1.5.1 General Multiplication Law
• Probabilities of Event Intersections
The probability of the intersection of a series of events
can be calculated from the expression
( )
( | )
( )
( )
( | )
( )
( ) ( ) ( | ) ( ) ( | )
( )
( | )
( )
( ) ( ) ( | )) ( ) ( | ) ( | )
P A B
P A B
P B
P A B
P B A
P A
P A B P B P A B P A P B A
P A B C
P C A B
P A B
P A B C P A B P C A B P A P B A P C A B
∩
• =
∩
=
⇒ ∩ = =
∩ ∩
• ∩ =
∩
⇒ ∩ ∩ = ∩ ∩ = ∩
1 2, ,..., nA A A
1 1 2 1 1 1( ) ( ) ( | ) ( | )n n nP A A P A P A A P A A A −∩ ∩ = ∩ ∩L L L
9. 1.5.2 Independent Events
• Independent Events
Two events A and B are said to be independent
events if
one of the following holds:
Any one of these conditions implies the other
two.
( | ) ( )
( ) ( ) ( | ) ( ) ( )
( ) ( ) ( )
and ( | ) ( )
( ) ( )
P B A P B
P A B P A P B A P A P B
P A B P A P B
P A B P A
P B P B
=
⇒ ∩ = =
∩
= = =
g
( | ) ( ), ( | ) ( ), and ( ) ( ) ( )P A B P A P B A P B P A B P A P B= = ∩ =
10. • The interpretation of two events being
independent is that knowledge about one
event does not affect the probability of the
other event.
• Intersections of Independent Events
The probability of the intersection of a series of
independent events is simply given
by
1 2, ,..., nA A A
1 1 2( ) ( ) ( ) ( )n nP A A P A P A P A∩ ∩ =L L
11.
12. 1.5.3 Examples and Probability Trees(1/3)
• Example 7: Car Warranties
A company sells a certain type of car, which it assembles in one of
four possible locations. Plant I supplies 20%; plant II, 24%; plant III,
25%; and plant IV, 31%. A customer buying a car does not know
where the car has been assembled, and so the probabilities of a
purchased car being from each of the four plants can be thought of
as being 0.20, 0.24, 0.25, and 0.31.
Each new car sold carries a 1-year bumper-to-bumper warranty.
P( claim | plant I ) = 0.05, P( claim | plant II ) = 0.11
P( claim | plant III ) = 0.03, P( claim | plant IV ) = 0.08
For example, a car assembled in plant I has a probability of 0.05 of
receiving a claim on its warranty.
Notice that claims are clearly not independent of assembly
location because these four conditional probabilities are unequal.
13. 1.5.3 Examples and Probability Trees(2/3)
P( claim ) = P( plant I, claim ) + P( plant II, claim )
+ P( plant III, claim) + P( plant IV, claim )
= 0.0687
14. 1.5.3 Examples and Probability Trees(3/3)
• GAMES OF CHANCE
- A fair die
even = { 2,4,6 } and high score = { 4,5,6 }
Intuitively, these two events are not independent.
- A red die and a blue die are rolled.
A = { the red die has an even score }
B = { the blue die has an even score }
1 2
( ) and ( | )
2 3
P even P even high score= =
1 1 1
( ) ( ) ( )
2 2 4
P A B P A P B∩ = = × =
15. 1.6 Posterior Probabilities
1.6.1 Law of Total Probability(1/3)
1
1
1
1 1
{1,2,3,4,5,6}
and : mutually exclusive
( ) ( ) and ( ) : mutually exclusive
( ) ( ) ( )
( ) ( | ) ( ) ( | )
n i
n i
n
n n
S
S A A A
B A B A B A B
P B P A B P A B
P A P B A P A P B A
=
= ∪ ∪
⇒ = ∩ ∪ ∪ ∩ ∩
⇒ = ∩ + + ∩
= + +
Lg
L
L
L
16. 1.6.1 Law of Total Probability(2/3)
• Law of Total Probability
If is a partition of a sample space, then the probability of an
event B can be obtained from the probabilities
and using the formula
1 1( ) ( ) ( | ) ( ) ( | )n nP B P A P B A P A P B A= + +L
1 2, ,..., nA A A
( )iP A ( | )iP B A
17. 1.6.1 Law of Total Probability(3/3)
• Example 7 Car Warranties
If are, respectively, the events that a car is assembled in
plants I, II, III, and IV, then they provide a partition of the sample
space, and the probabilities are the supply proportions of the
four plants.
B = { a claim is made }
= the claim rates for the four individual plants
1 2 3 4, , , andA A A A
( )iP A
1 1 2 2 3 3 4 4( ) ( ) ( | ) ( ) ( | ) ( ) ( | ) ( ) ( | )
(0.20 0.05) (0.24 0.11) (0.25 0.03) (0.31 0.08)
0.0687
P B P A P B A P A P B A P A P B A P A P B A= + + +
= × + × + × + ×
=
18. 1.6.2 Calculation of Posterior Probabilities
• Bayes’ Theorem
If is a partition of a sample space, then the posterior
probabilities of the event conditional on an event B can be
obtained from the probabilities and using the formula
1
1
1
( ) and ( | ) ( | ) ?
( ), , ( ) : the prior probabilities
( | ), , ( | ) : the posterior probabilities
( ) ( ) ( | ) ( ) ( | )
( | )
( ) ( )
( ) ( | )
i i i
n
n
i i i i i
i n
j j
j
P A P B A P A B
P A P A
P A B P A B
P A B P A P B A P A P B A
P A B
P B P B
P A P B A
=
⇒ =
∩
⇒ = = =
∑
g
Lg
Lg
1 2, ,..., nA A A
( )iP A ( | )iP B A
1
( ) ( | )
( | )
( ) ( | )
i i
i n
j j
j
P A P B A
P A B
P A P B A
=
=
∑
iA
19. 1.6.3 Examples of Posterior Probabilities(1/2)
• Example 7 Car Warranties
- The prior probabilities
- If a claim is made on the warranty of the car, how does this change
these probabilities?
( ) 0.20, ( ) 0.24
( ) 0.25, ( ) 0.31
P plant I P plant II
P plant III P plant IV
= =
= =
( ) ( | ) 0.20 0.05
( | ) 0.146
( ) 0.0687
( ) ( | ) 0.24 0.11
( | ) 0.384
( ) 0.0687
( ) ( | ) 0.
( | )
( )
P plant I P claim plant I
P plant I claim
P claim
P plant II P claim plant II
P plant II claim
P claim
P plant III P claim plant III
P plant III claim
P claim
×
= = =
×
= = =
= =
25 0.03
0.109
0.0687
( ) ( | ) 0.31 0.08
( | ) 0.361
( ) 0.0687
P plant IV P claim plant IV
P plant IV claim
P claim
×
=
×
= = =
20. 1.6.3 Examples of Posterior Probabilities(2/2)
- No claim is made on the warranty
( ) ( | )
( | )
( )
0.20 0.95
0.204
0.9313
( ) ( | )
( | )
( )
0.24 0.89
0.229
0.9313
( ) ( |
( | )
P plant I P no claim plant I
P plant I no claim
P no claim
P plant II P no claim plant II
P plant II no claim
P no claim
P plant III P no claim plan
P plant III no claim
=
×
= =
=
×
= =
=
)
( )
0.25 0.97
0.261
0.9313
( ) ( | )
( | )
( )
0.31 0.92
0.306
0.9313
t III
P no claim
P plant IV P no claim plant IV
P plant IV no claim
P no claim
×
= =
=
×
= =
21.
22.
23.
24.
25. Home take assignment
• Solve the relevant examples and end
exercise questions of text book.
Study before next lecture
• Permutations
• Combinations
26. Assignment #1
• Will be e-mailed
• Due date same day same class.
Assignments will be accepted only in
class.