The document discusses binomial probability distributions. It shows that if an event has two possible outcomes A and B, and is repeated n times, the probability of getting k successes can be calculated as:
P(X=k) = nCk * pk * (1-p)(n-k)
Where X is the random variable for the number of successes, p is the probability of outcome A, and 1-p is the probability of outcome B. It provides an example calculating the probability of drawing all black balls or 4 black balls from a bag with a given number of black and white balls.
12X1 T09 08 binomial probability (2010)Nigel Simmons
The document discusses binomial probability distributions. It explains that if an event has two possible outcomes and is repeated, the probability of each outcome follows a binomial distribution. It provides examples of calculating binomial probabilities for 1, 2, 3, and 4 events. The key points are:
- Binomial probabilities use the formula P(X=k) = nCk * pk * (1-p)(n-k)
- This calculates the probability of k successes in n trials with probability of success p
- It works through examples such as drawing balls from a bag to calculate various probabilities
This presentation discusses binomial probability distributions through the following key points:
- It defines basic terminology related to random experiments, events, and variables. The binomial distribution specifically describes discrete data from Bernoulli processes.
- It outlines the notation and assumptions for binomial distributions, including that there are two possible outcomes for each trial (success/failure), a fixed number of trials, and constant probabilities of success/failure.
- It presents three methods for calculating binomial probabilities: the binomial probability formula, table method, and using technology like Excel.
- It discusses measures of central tendency and dispersion for binomial distributions and how the shape of the distribution depends on the number of trials and probability of success.
- Real-world
Slideshare is discontinuing its Slidecast feature as of February 28, 2014. Existing Slidecasts will be converted to static presentations without audio by April 30, 2014. The document informs users that new slidecasts can be found on myPlick.com or the author's blog starting in 2014. However, myPlick proved unreliable, so future slidecasts will instead be hosted on the author's YouTube channel.
The document discusses different methods for factorising expressions:
1) Looking for a common factor and dividing it out of all terms
2) Using the difference of two squares formula (a2 - b2 = (a - b)(a + b))
3) Factorising quadratic trinomials into two binomial factors by identifying the values that multiply to give the constant term and sum to give the coefficient of the linear term.
12X1 T09 08 binomial probability (2010)Nigel Simmons
The document discusses binomial probability distributions. It explains that if an event has two possible outcomes and is repeated, the probability of each outcome follows a binomial distribution. It provides examples of calculating binomial probabilities for 1, 2, 3, and 4 events. The key points are:
- Binomial probabilities use the formula P(X=k) = nCk * pk * (1-p)(n-k)
- This calculates the probability of k successes in n trials with probability of success p
- It works through examples such as drawing balls from a bag to calculate various probabilities
This presentation discusses binomial probability distributions through the following key points:
- It defines basic terminology related to random experiments, events, and variables. The binomial distribution specifically describes discrete data from Bernoulli processes.
- It outlines the notation and assumptions for binomial distributions, including that there are two possible outcomes for each trial (success/failure), a fixed number of trials, and constant probabilities of success/failure.
- It presents three methods for calculating binomial probabilities: the binomial probability formula, table method, and using technology like Excel.
- It discusses measures of central tendency and dispersion for binomial distributions and how the shape of the distribution depends on the number of trials and probability of success.
- Real-world
Slideshare is discontinuing its Slidecast feature as of February 28, 2014. Existing Slidecasts will be converted to static presentations without audio by April 30, 2014. The document informs users that new slidecasts can be found on myPlick.com or the author's blog starting in 2014. However, myPlick proved unreliable, so future slidecasts will instead be hosted on the author's YouTube channel.
The document discusses different methods for factorising expressions:
1) Looking for a common factor and dividing it out of all terms
2) Using the difference of two squares formula (a2 - b2 = (a - b)(a + b))
3) Factorising quadratic trinomials into two binomial factors by identifying the values that multiply to give the constant term and sum to give the coefficient of the linear term.
The document provides information on index laws and the meaning of indices in algebra:
- Index laws state that am × an = am+n, am ÷ an = am-n, and (am)n = amn. Exponents can be added or subtracted when multiplying or dividing terms with the same base.
- Positive exponents indicate a term is raised to a power. Negative exponents indicate a root is being taken. Terms with exponents are evaluated from left to right.
- Examples demonstrate how to simplify expressions using index laws and interpret different types of indices.
12 x1 t01 03 integrating derivative on function (2013)Nigel Simmons
The document discusses integrating derivatives of functions. It states that the integral of the derivative of a function f(x) is equal to the natural log of f(x) plus a constant. It then provides examples of integrating several derivatives: (i) ∫(1/(7-3x)) dx = -1/3 log(7-3x) + c, (ii) ∫(1/(8x+5)) dx = 1/8 log(8x+5) + c, and (iii) ∫(x5/(x-2)) dx = 1/6 log(x6-2) + c. It also discusses techniques for integrating fractions by polynomial long division and finds
The document discusses logarithms and their properties. Logarithms are defined as the inverse of exponentials. If y = ax, then x = loga y. The natural logarithm is log base e, written as ln. Properties of logarithms include: loga m + loga n = loga mn; loga m - loga n = loga(m/n); loga mn = n loga m; loga 1 = 0; loga a = 1. Examples of evaluating logarithmic expressions are provided.
The document discusses relationships between the coefficients and roots of polynomials. It states that for a polynomial P(x) = axn + bxn-1 + cxn-2 + ..., the sum of the roots equals -b/a, the sum of the roots taken two at a time equals c/a, and so on for higher order terms. It also provides examples of using these relationships to find the sums of roots for a given polynomial.
P
4
3
2
The document discusses properties of polynomials with multiple roots. It first proves that if a polynomial P(x) has a root x = a of multiplicity m, then the derivative of P(x), P'(x), will have a root x = a of multiplicity m-1. It then provides an example of solving a cubic equation given it has a double root. Finally, it examines a quartic polynomial and shows that its root α cannot be 0, 1, or -1, and that 1/
The document discusses factorizing complex expressions. The main points are:
- If a polynomial's coefficients are real, its roots will appear in complex conjugate pairs.
- Any polynomial of degree n can be factorized into a mixture of quadratic and linear factors over real numbers, or into n linear factors over complex numbers.
- Odd degree polynomials must have at least one real root.
- Examples of factorizing polynomials over both real and complex numbers are provided.
The document describes the Trapezoidal Rule for approximating the area under a curve between two points. It shows that the area A is estimated by dividing the region into trapezoids with height equal to the function values at the interval endpoints and bases equal to the intervals. In general, the area is approximated as the sum of the areas of each trapezoid, which is equal to the average of the endpoint function values multiplied by the interval length.
The document discusses methods for calculating the volumes of solids of revolution. It provides formulas for finding volumes when an area is revolved around either the x-axis or y-axis. Examples are given for finding volumes of common solids like cones, spheres, and others. Steps are shown for using the formulas to calculate volumes based on given functions and limits of revolution.
The document discusses different methods for calculating the area under a curve or between curves.
(1) The area below the x-axis is given by the integral of the function between the bounds, which can be positive or negative depending on whether the area is above or below the x-axis.
(2) To calculate the area on the y-axis, the function is solved for x in terms of y, then the bounds are substituted into the integral of this new function with respect to y.
(3) The area between two curves is calculated by taking the integral of the upper curve minus the integral of the lower curve, both between the same bounds on the x-axis.
The document discusses 8 properties of definite integrals:
1) Integrating polynomials results in a fraction.
2) Constants can be factored out of integrals.
3) Integrals of sums are equal to the sum of integrals.
4) Splitting an integral range results in the sum of the integrals.
5) Integrals of positive functions over a range are positive, and negative if the function is negative.
6) Integrals can be compared based on the relative values of the integrands.
7) Changing the limits of integration flips the sign of the integral.
8) Integrals of odd functions over a symmetric range are zero, and integrals of even functions are twice the integral over
0
The document discusses the relationship between integration and calculating the area under a curve. It shows that the area under a curve from x=a to x=b can be calculated as the integral from a to b of the function f(x) dx. This is equal to the antiderivative F(x) evaluated from b to a. The area under a curve can also be estimated using rectangles, and as the width of the rectangles approaches 0, the estimate becomes the exact area. The derivative of the area function A(x) gives the equation of the curve f(x). Examples are given to calculate the exact area under curves.
The document discusses nth roots of unity. It states that the solutions to equations of the form zn = ±1 are the nth roots of unity. These solutions form a regular n-sided polygon with vertices on the unit circle when placed on an Argand diagram. As an example, it shows that the solutions to z5 = 1 are the fifth roots of unity located at angles that are integer multiples of 2π/5 around the unit circle. It then proves that if ω is a root of z5 - 1 = 0, then ω2, ω3, ω4 and ω5 are also roots. Finally, it proves that 1 + ω + ω2 + ω3 + ω4 = 0.
Executive Directors Chat Leveraging AI for Diversity, Equity, and InclusionTechSoup
Let’s explore the intersection of technology and equity in the final session of our DEI series. Discover how AI tools, like ChatGPT, can be used to support and enhance your nonprofit's DEI initiatives. Participants will gain insights into practical AI applications and get tips for leveraging technology to advance their DEI goals.
The document provides information on index laws and the meaning of indices in algebra:
- Index laws state that am × an = am+n, am ÷ an = am-n, and (am)n = amn. Exponents can be added or subtracted when multiplying or dividing terms with the same base.
- Positive exponents indicate a term is raised to a power. Negative exponents indicate a root is being taken. Terms with exponents are evaluated from left to right.
- Examples demonstrate how to simplify expressions using index laws and interpret different types of indices.
12 x1 t01 03 integrating derivative on function (2013)Nigel Simmons
The document discusses integrating derivatives of functions. It states that the integral of the derivative of a function f(x) is equal to the natural log of f(x) plus a constant. It then provides examples of integrating several derivatives: (i) ∫(1/(7-3x)) dx = -1/3 log(7-3x) + c, (ii) ∫(1/(8x+5)) dx = 1/8 log(8x+5) + c, and (iii) ∫(x5/(x-2)) dx = 1/6 log(x6-2) + c. It also discusses techniques for integrating fractions by polynomial long division and finds
The document discusses logarithms and their properties. Logarithms are defined as the inverse of exponentials. If y = ax, then x = loga y. The natural logarithm is log base e, written as ln. Properties of logarithms include: loga m + loga n = loga mn; loga m - loga n = loga(m/n); loga mn = n loga m; loga 1 = 0; loga a = 1. Examples of evaluating logarithmic expressions are provided.
The document discusses relationships between the coefficients and roots of polynomials. It states that for a polynomial P(x) = axn + bxn-1 + cxn-2 + ..., the sum of the roots equals -b/a, the sum of the roots taken two at a time equals c/a, and so on for higher order terms. It also provides examples of using these relationships to find the sums of roots for a given polynomial.
P
4
3
2
The document discusses properties of polynomials with multiple roots. It first proves that if a polynomial P(x) has a root x = a of multiplicity m, then the derivative of P(x), P'(x), will have a root x = a of multiplicity m-1. It then provides an example of solving a cubic equation given it has a double root. Finally, it examines a quartic polynomial and shows that its root α cannot be 0, 1, or -1, and that 1/
The document discusses factorizing complex expressions. The main points are:
- If a polynomial's coefficients are real, its roots will appear in complex conjugate pairs.
- Any polynomial of degree n can be factorized into a mixture of quadratic and linear factors over real numbers, or into n linear factors over complex numbers.
- Odd degree polynomials must have at least one real root.
- Examples of factorizing polynomials over both real and complex numbers are provided.
The document describes the Trapezoidal Rule for approximating the area under a curve between two points. It shows that the area A is estimated by dividing the region into trapezoids with height equal to the function values at the interval endpoints and bases equal to the intervals. In general, the area is approximated as the sum of the areas of each trapezoid, which is equal to the average of the endpoint function values multiplied by the interval length.
The document discusses methods for calculating the volumes of solids of revolution. It provides formulas for finding volumes when an area is revolved around either the x-axis or y-axis. Examples are given for finding volumes of common solids like cones, spheres, and others. Steps are shown for using the formulas to calculate volumes based on given functions and limits of revolution.
The document discusses different methods for calculating the area under a curve or between curves.
(1) The area below the x-axis is given by the integral of the function between the bounds, which can be positive or negative depending on whether the area is above or below the x-axis.
(2) To calculate the area on the y-axis, the function is solved for x in terms of y, then the bounds are substituted into the integral of this new function with respect to y.
(3) The area between two curves is calculated by taking the integral of the upper curve minus the integral of the lower curve, both between the same bounds on the x-axis.
The document discusses 8 properties of definite integrals:
1) Integrating polynomials results in a fraction.
2) Constants can be factored out of integrals.
3) Integrals of sums are equal to the sum of integrals.
4) Splitting an integral range results in the sum of the integrals.
5) Integrals of positive functions over a range are positive, and negative if the function is negative.
6) Integrals can be compared based on the relative values of the integrands.
7) Changing the limits of integration flips the sign of the integral.
8) Integrals of odd functions over a symmetric range are zero, and integrals of even functions are twice the integral over
0
The document discusses the relationship between integration and calculating the area under a curve. It shows that the area under a curve from x=a to x=b can be calculated as the integral from a to b of the function f(x) dx. This is equal to the antiderivative F(x) evaluated from b to a. The area under a curve can also be estimated using rectangles, and as the width of the rectangles approaches 0, the estimate becomes the exact area. The derivative of the area function A(x) gives the equation of the curve f(x). Examples are given to calculate the exact area under curves.
The document discusses nth roots of unity. It states that the solutions to equations of the form zn = ±1 are the nth roots of unity. These solutions form a regular n-sided polygon with vertices on the unit circle when placed on an Argand diagram. As an example, it shows that the solutions to z5 = 1 are the fifth roots of unity located at angles that are integer multiples of 2π/5 around the unit circle. It then proves that if ω is a root of z5 - 1 = 0, then ω2, ω3, ω4 and ω5 are also roots. Finally, it proves that 1 + ω + ω2 + ω3 + ω4 = 0.
Executive Directors Chat Leveraging AI for Diversity, Equity, and InclusionTechSoup
Let’s explore the intersection of technology and equity in the final session of our DEI series. Discover how AI tools, like ChatGPT, can be used to support and enhance your nonprofit's DEI initiatives. Participants will gain insights into practical AI applications and get tips for leveraging technology to advance their DEI goals.
Strategies for Effective Upskilling is a presentation by Chinwendu Peace in a Your Skill Boost Masterclass organisation by the Excellence Foundation for South Sudan on 08th and 09th June 2024 from 1 PM to 3 PM on each day.
The simplified electron and muon model, Oscillating Spacetime: The Foundation...RitikBhardwaj56
Discover the Simplified Electron and Muon Model: A New Wave-Based Approach to Understanding Particles delves into a groundbreaking theory that presents electrons and muons as rotating soliton waves within oscillating spacetime. Geared towards students, researchers, and science buffs, this book breaks down complex ideas into simple explanations. It covers topics such as electron waves, temporal dynamics, and the implications of this model on particle physics. With clear illustrations and easy-to-follow explanations, readers will gain a new outlook on the universe's fundamental nature.
Main Java[All of the Base Concepts}.docxadhitya5119
This is part 1 of my Java Learning Journey. This Contains Custom methods, classes, constructors, packages, multithreading , try- catch block, finally block and more.
Exploiting Artificial Intelligence for Empowering Researchers and Faculty, In...Dr. Vinod Kumar Kanvaria
Exploiting Artificial Intelligence for Empowering Researchers and Faculty,
International FDP on Fundamentals of Research in Social Sciences
at Integral University, Lucknow, 06.06.2024
By Dr. Vinod Kumar Kanvaria
A workshop hosted by the South African Journal of Science aimed at postgraduate students and early career researchers with little or no experience in writing and publishing journal articles.
it describes the bony anatomy including the femoral head , acetabulum, labrum . also discusses the capsule , ligaments . muscle that act on the hip joint and the range of motion are outlined. factors affecting hip joint stability and weight transmission through the joint are summarized.
ISO/IEC 27001, ISO/IEC 42001, and GDPR: Best Practices for Implementation and...PECB
Denis is a dynamic and results-driven Chief Information Officer (CIO) with a distinguished career spanning information systems analysis and technical project management. With a proven track record of spearheading the design and delivery of cutting-edge Information Management solutions, he has consistently elevated business operations, streamlined reporting functions, and maximized process efficiency.
Certified as an ISO/IEC 27001: Information Security Management Systems (ISMS) Lead Implementer, Data Protection Officer, and Cyber Risks Analyst, Denis brings a heightened focus on data security, privacy, and cyber resilience to every endeavor.
His expertise extends across a diverse spectrum of reporting, database, and web development applications, underpinned by an exceptional grasp of data storage and virtualization technologies. His proficiency in application testing, database administration, and data cleansing ensures seamless execution of complex projects.
What sets Denis apart is his comprehensive understanding of Business and Systems Analysis technologies, honed through involvement in all phases of the Software Development Lifecycle (SDLC). From meticulous requirements gathering to precise analysis, innovative design, rigorous development, thorough testing, and successful implementation, he has consistently delivered exceptional results.
Throughout his career, he has taken on multifaceted roles, from leading technical project management teams to owning solutions that drive operational excellence. His conscientious and proactive approach is unwavering, whether he is working independently or collaboratively within a team. His ability to connect with colleagues on a personal level underscores his commitment to fostering a harmonious and productive workplace environment.
Date: May 29, 2024
Tags: Information Security, ISO/IEC 27001, ISO/IEC 42001, Artificial Intelligence, GDPR
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How to Fix the Import Error in the Odoo 17Celine George
An import error occurs when a program fails to import a module or library, disrupting its execution. In languages like Python, this issue arises when the specified module cannot be found or accessed, hindering the program's functionality. Resolving import errors is crucial for maintaining smooth software operation and uninterrupted development processes.
How to Add Chatter in the odoo 17 ERP ModuleCeline George
In Odoo, the chatter is like a chat tool that helps you work together on records. You can leave notes and track things, making it easier to talk with your team and partners. Inside chatter, all communication history, activity, and changes will be displayed.
2. Binomial Probability
If an event has only two possibilities and this event is repeated, then the
probability distribution is as follows;
3. Binomial Probability
If an event has only two possibilities and this event is repeated, then the
probability distribution is as follows;
A
B
4. Binomial Probability
If an event has only two possibilities and this event is repeated, then the
probability distribution is as follows;
A p
Bq
5. Binomial Probability
If an event has only two possibilities and this event is repeated, then the
probability distribution is as follows;
AA
A p
AB
Bq
6. Binomial Probability
If an event has only two possibilities and this event is repeated, then the
probability distribution is as follows;
AA
A p
AB
BA
Bq
BB
7. Binomial Probability
If an event has only two possibilities and this event is repeated, then the
probability distribution is as follows;
AA p 2
A p
AB pq
BA pq
Bq
BB q 2
8. Binomial Probability
If an event has only two possibilities and this event is repeated, then the
probability distribution is as follows; AAA
AA p 2
AAB
A p ABA
AB pq
ABB
BAA
BA pq
BAB
Bq BBA
BB q 2
BBB
9. Binomial Probability
If an event has only two possibilities and this event is repeated, then the
probability distribution is as follows; AAA p 3
AA p 2
AAB p q
2
A p ABA p 2 q
AB pq
ABB pq 2
BAA p 2 q
BA pq
BAB pq
2
Bq BBA pq 2
BB q 2
BBB p
3
10. Binomial Probability
If an event has only two possibilities and this event is repeated, then the
probability distribution is as follows; AAA p 3 AAAA
AAAB
AA p
2
AABA
AAB p q
2
AABB
A p ABA p q
2
ABAA
AB pq ABAB
ABBA
ABB pq 2
ABBB
BAAA
BAA p 2 q BAAB
BA pq BABA
BAB pq
2
BABB
Bq BBA pq 2 BBAA
BB q 2 BBAB
BBB p
3 BBBA
BBBB
11. Binomial Probability
If an event has only two possibilities and this event is repeated, then the
probability distribution is as follows; AAA p 3 AAAA p 4
AAAB p 3 q
AA p 2 AABA p 3 q
AAB p q AABB p 2 q 2
2
A p ABA p 2 q ABAA p 3 q
AB pq ABAB p 2 q 2
ABB pq 2 p q
ABBA 2 2
ABBB pq 3
BAA p q
BAAA 3
p q
BAAB p 2 q 2
2
BA pq BABA p 2 q 2
BAB pq BABB pq
2 3
Bq BBA pq 2 BBAA p q
2 2
BB q 2
BBAB pq 3
BBB p pqq
3 BBBA 3
4
BBBB
15. 1 Event 2 Events
P A p P AA p 2
P B q P AandB 2 pq
P BB q 2
16. 1 Event 2 Events 3 Events
P A p P AA p 2
P B q P AandB 2 pq
P BB q 2
17. 1 Event 2 Events 3 Events
P A p P AA p 2 P AAA p 3
P B q P AandB 2 pq P2 AandB 3 p 2 q
P BB q 2 P Aand 2 B 3 pq 2
P BBB q 3
18. 1 Event 2 Events 3 Events
P A p P AA p 2 P AAA p 3
P B q P AandB 2 pq P2 AandB 3 p 2 q
P BB q 2 P Aand 2 B 3 pq 2
P BBB q 3
4 Events
19. 1 Event 2 Events 3 Events
P A p P AA p 2 P AAA p 3
P B q P AandB 2 pq P2 AandB 3 p 2 q
P BB q 2 P Aand 2 B 3 pq 2
P BBB q 3
4 Events
P AAAA p 4
P3 AandB 4 p 3 q
P2 Aand 2 B 6 p 2 q 2
P Aand 3B 4 pq 3
P BBBB q 4
20. 1 Event 2 Events 3 Events
P A p P AA p 2 P AAA p 3
P B q P AandB 2 pq P2 AandB 3 p 2 q
P BB q 2 P Aand 2 B 3 pq 2
P BBB q 3
4 Events If an event is repeated n times and P X p
P AAAA p 4 and P X q then the probability that X will
P3 AandB 4 p 3 q occur exactly k times is;
P2 Aand 2 B 6 p 2 q 2
P Aand 3B 4 pq 3
P BBBB q 4
21. 1 Event 2 Events 3 Events
P A p P AA p 2 P AAA p 3
P B q P AandB 2 pq P2 AandB 3 p 2 q
P BB q 2 P Aand 2 B 3 pq 2
P BBB q 3
4 Events If an event is repeated n times and P X p
P AAAA p 4 and P X q then the probability that X will
P3 AandB 4 p 3 q occur exactly k times is;
P2 Aand 2 B 6 p 2 q 2
P X k nCk q nk p k
P Aand 3B 4 pq 3
P BBBB q 4 Note: X = k, means X will occur exactly k times
22. e.g.(i) A bag contains 30 black balls and 20 white balls.
Seven drawings are made (with replacement), what is the
probability of drawing;
a) All black balls?
23. e.g.(i) A bag contains 30 black balls and 20 white balls.
Seven drawings are made (with replacement), what is the
probability of drawing;
a) All black balls?
0 7
P X 7 7C7
2 3
5 5
24. e.g.(i) A bag contains 30 black balls and 20 white balls.
Seven drawings are made (with replacement), what is the
probability of drawing; Let X be the number of black balls drawn
a) All black balls?
0 7
P X 7 7C7
2 3
5 5
25. e.g.(i) A bag contains 30 black balls and 20 white balls.
Seven drawings are made (with replacement), what is the
probability of drawing; Let X be the number of black balls drawn
a) All black balls?
0 7
P X 7 7C7
2 3
5 5
7
C7 37
57
26. e.g.(i) A bag contains 30 black balls and 20 white balls.
Seven drawings are made (with replacement), what is the
probability of drawing; Let X be the number of black balls drawn
a) All black balls?
0 7
P X 7 7C7
2 3
5 5
7
C7 37
57
2187
78125
27. e.g.(i) A bag contains 30 black balls and 20 white balls.
Seven drawings are made (with replacement), what is the
probability of drawing; Let X be the number of black balls drawn
a) All black balls? b) 4 black balls?
0 7
P X 7 7C7
2 3
5 5
7
C7 37
57
2187
78125
28. e.g.(i) A bag contains 30 black balls and 20 white balls.
Seven drawings are made (with replacement), what is the
probability of drawing; Let X be the number of black balls drawn
a) All black balls? b) 4 black balls?
0 7 3 4
P X 7 7C7
2 3
P X 4 7C4
2 3
5 5 5 5
7
C7 37
57
2187
78125
29. e.g.(i) A bag contains 30 black balls and 20 white balls.
Seven drawings are made (with replacement), what is the
probability of drawing; Let X be the number of black balls drawn
a) All black balls? b) 4 black balls?
0 7 3 4
P X 7 7C7
2 3
P X 4 7C4
2 3
5 5 5 5
7
C7 37 7
C 4 2 33 4
7
5 57
2187
78125
30. e.g.(i) A bag contains 30 black balls and 20 white balls.
Seven drawings are made (with replacement), what is the
probability of drawing; Let X be the number of black balls drawn
a) All black balls? b) 4 black balls?
0 7 3 4
P X 7 7C7
2 3
P X 4 7C4
2 3
5 5 5 5
7
C7 37 7
C 4 2 33 4
7
5 57
2187 4536
78125 15625
31. e.g.(i) A bag contains 30 black balls and 20 white balls.
Seven drawings are made (with replacement), what is the
probability of drawing; Let X be the number of black balls drawn
a) All black balls? b) 4 black balls?
0 7 3 4
P X 7 7C7
2 3
P X 4 7C4
2 3
5 5 5 5
7
C7 37 7
C 4 2 33 4
7
5 57
2187 4536
78125 15625
(ii) At an election 30% of voters favoured Party A.
If at random an interviewer selects 5 voters, what is the
probability that;
a) 3 favoured Party A?
32. e.g.(i) A bag contains 30 black balls and 20 white balls.
Seven drawings are made (with replacement), what is the
probability of drawing; Let X be the number of black balls drawn
a) All black balls? b) 4 black balls?
0 7 3 4
P X 7 7C7
2 3
P X 4 7C4
2 3
5 5 5 5
7
C7 37 7
C 4 2 33 4
7
5 57
2187 4536
78125 15625
(ii) At an election 30% of voters favoured Party A.
If at random an interviewer selects 5 voters, what is the
probability that;
a) 3 favoured Party A?
Let X be the number
favouring Party A
33. e.g.(i) A bag contains 30 black balls and 20 white balls.
Seven drawings are made (with replacement), what is the
probability of drawing; Let X be the number of black balls drawn
a) All black balls? b) 4 black balls?
0 7 3 4
P X 7 7C7
2 3
P X 4 7C4
2 3
5 5 5 5
7
C7 37 7
C 4 2 33 4
7
5 57
2187 4536
78125 15625
(ii) At an election 30% of voters favoured Party A.
If at random an interviewer selects 5 voters, what is the
probability that; 2 3
P3 A 5C3
7 3
a) 3 favoured Party A?
10 10
Let X be the number
favouring Party A
34. e.g.(i) A bag contains 30 black balls and 20 white balls.
Seven drawings are made (with replacement), what is the
probability of drawing; Let X be the number of black balls drawn
a) All black balls? b) 4 black balls?
0 7 3 4
P X 7 7C7
2 3
P X 4 7C4
2 3
5 5 5 5
7
C7 37 7
C 4 2 33 4
7
5 57
2187 4536
78125 15625
(ii) At an election 30% of voters favoured Party A.
If at random an interviewer selects 5 voters, what is the
probability that; 2 3
P3 A 5C3
7 3
a) 3 favoured Party A?
10 10
Let X be the number 5
C3 7 233
favouring Party A
105
35. e.g.(i) A bag contains 30 black balls and 20 white balls.
Seven drawings are made (with replacement), what is the
probability of drawing; Let X be the number of black balls drawn
a) All black balls? b) 4 black balls?
0 7 3 4
P X 7 7C7
2 3
P X 4 7C4
2 3
5 5 5 5
7
C7 37 7
C 4 2 33 4
7
5 57
2187 4536
78125 15625
(ii) At an election 30% of voters favoured Party A.
If at random an interviewer selects 5 voters, what is the
probability that; 2 3
P3 A 5C3
7 3
a) 3 favoured Party A?
10 10
Let X be the number 5
C3 7 233 1323
favouring Party A
105 10000
44. 2005 Extension 1 HSC Q6a)
There are five matches on each weekend of a football season.
Megan takes part in a competition in which she earns 1 point if
she picks more than half of the winning teams for a weekend, and
zero points otherwise.
The probability that Megan correctly picks the team that wins in
any given match is 2
3
(i) Show that the probability that Megan earns one point for a
given weekend is 0 7901, correct to four decimal places.
45. 2005 Extension 1 HSC Q6a)
There are five matches on each weekend of a football season.
Megan takes part in a competition in which she earns 1 point if
she picks more than half of the winning teams for a weekend, and
zero points otherwise.
The probability that Megan correctly picks the team that wins in
any given match is 2
3
(i) Show that the probability that Megan earns one point for a
given weekend is 0 7901, correct to four decimal places.
Let X be the number of matches picked correctly
46. 2005 Extension 1 HSC Q6a)
There are five matches on each weekend of a football season.
Megan takes part in a competition in which she earns 1 point if
she picks more than half of the winning teams for a weekend, and
zero points otherwise.
The probability that Megan correctly picks the team that wins in
any given match is 2
3
(i) Show that the probability that Megan earns one point for a
given weekend is 0 7901, correct to four decimal places.
Let X be the number of matches picked correctly
2 3 1 4 0 5
P X 3 5C3 5C4 5C5
1 2 1 2 1 2
3 3 3 3 3 3
47. 2005 Extension 1 HSC Q6a)
There are five matches on each weekend of a football season.
Megan takes part in a competition in which she earns 1 point if
she picks more than half of the winning teams for a weekend, and
zero points otherwise.
The probability that Megan correctly picks the team that wins in
any given match is 2
3
(i) Show that the probability that Megan earns one point for a
given weekend is 0 7901, correct to four decimal places.
Let X be the number of matches picked correctly
2 3 1 4 0 5
P X 3 5C3 5C4 5C5
1 2 1 2 1 2
3 3 3 3 3 3
5
C3 23 5C4 2 4 5C5 25
35
48. 2005 Extension 1 HSC Q6a)
There are five matches on each weekend of a football season.
Megan takes part in a competition in which she earns 1 point if
she picks more than half of the winning teams for a weekend, and
zero points otherwise.
The probability that Megan correctly picks the team that wins in
any given match is 2
3
(i) Show that the probability that Megan earns one point for a
given weekend is 0 7901, correct to four decimal places.
Let X be the number of matches picked correctly
2 3 1 4 0 5
P X 3 5C3 5C4 5C5
1 2 1 2 1 2
3 3 3 3 3 3
5
C3 23 5C4 2 4 5C5 25
35
0 7901
49. (ii) Hence find the probability that Megan earns one point every
week of the eighteen week season. Give your answer correct
to two decimal places.
50. (ii) Hence find the probability that Megan earns one point every
week of the eighteen week season. Give your answer correct
to two decimal places.
Let Y be the number of weeks Megan earns a point
51. (ii) Hence find the probability that Megan earns one point every
week of the eighteen week season. Give your answer correct
to two decimal places.
Let Y be the number of weeks Megan earns a point
PY 1818C18 0 2099 0 7901
0 18
52. (ii) Hence find the probability that Megan earns one point every
week of the eighteen week season. Give your answer correct
to two decimal places.
Let Y be the number of weeks Megan earns a point
PY 1818C18 0 2099 0 7901
0 18
0 01 (to 2 dp)
53. (ii) Hence find the probability that Megan earns one point every
week of the eighteen week season. Give your answer correct
to two decimal places.
Let Y be the number of weeks Megan earns a point
PY 1818C18 0 2099 0 7901
0 18
0 01 (to 2 dp)
(iii) Find the probability that Megan earns at most 16 points
during the eighteen week season. Give your answer correct
to two decimal places.
54. (ii) Hence find the probability that Megan earns one point every
week of the eighteen week season. Give your answer correct
to two decimal places.
Let Y be the number of weeks Megan earns a point
PY 1818C18 0 2099 0 7901
0 18
0 01 (to 2 dp)
(iii) Find the probability that Megan earns at most 16 points
during the eighteen week season. Give your answer correct
to two decimal places.
PY 16 1 PY 17
55. (ii) Hence find the probability that Megan earns one point every
week of the eighteen week season. Give your answer correct
to two decimal places.
Let Y be the number of weeks Megan earns a point
PY 1818C18 0 2099 0 7901
0 18
0 01 (to 2 dp)
(iii) Find the probability that Megan earns at most 16 points
during the eighteen week season. Give your answer correct
to two decimal places.
PY 16 1 PY 17
118C17 0 2099 0 7901 18C18 0 2099 0 7901
1 17 0 18
56. (ii) Hence find the probability that Megan earns one point every
week of the eighteen week season. Give your answer correct
to two decimal places.
Let Y be the number of weeks Megan earns a point
PY 1818C18 0 2099 0 7901
0 18
0 01 (to 2 dp)
(iii) Find the probability that Megan earns at most 16 points
during the eighteen week season. Give your answer correct
to two decimal places.
PY 16 1 PY 17
118C17 0 2099 0 7901 18C18 0 2099 0 7901
1 17 0 18
0 92 (to 2 dp)
57. 2007 Extension 1 HSC Q4a)
In a large city, 10% of the population has green eyes.
(i) What is the probability that two randomly chosen people have
green eyes?
58. 2007 Extension 1 HSC Q4a)
In a large city, 10% of the population has green eyes.
(i) What is the probability that two randomly chosen people have
green eyes?
P 2 green 0.1 0.1
0.01
59. 2007 Extension 1 HSC Q4a)
In a large city, 10% of the population has green eyes.
(i) What is the probability that two randomly chosen people have
green eyes?
P 2 green 0.1 0.1
0.01
(ii) What is the probability that exactly two of a group of 20 randomly
chosen people have green eyes? Give your answer correct to three
decimal eyes.
60. 2007 Extension 1 HSC Q4a)
In a large city, 10% of the population has green eyes.
(i) What is the probability that two randomly chosen people have
green eyes?
P 2 green 0.1 0.1
0.01
(ii) What is the probability that exactly two of a group of 20 randomly
chosen people have green eyes? Give your answer correct to three
decimal eyes.
Let X be the number of people with green eyes
61. 2007 Extension 1 HSC Q4a)
In a large city, 10% of the population has green eyes.
(i) What is the probability that two randomly chosen people have
green eyes?
P 2 green 0.1 0.1
0.01
(ii) What is the probability that exactly two of a group of 20 randomly
chosen people have green eyes? Give your answer correct to three
decimal eyes.
Let X be the number of people with green eyes
P X 2 C2 0.9 0.1
20 18 2
62. 2007 Extension 1 HSC Q4a)
In a large city, 10% of the population has green eyes.
(i) What is the probability that two randomly chosen people have
green eyes?
P 2 green 0.1 0.1
0.01
(ii) What is the probability that exactly two of a group of 20 randomly
chosen people have green eyes? Give your answer correct to three
decimal eyes.
Let X be the number of people with green eyes
P X 2 C2 0.9 0.1
20 18 2
0.2851
0.285 (to 3 dp)
63. (iii) What is the probability that more than two of a group of 20
randomly chosen people have green eyes? Give your answer
correct to two decimal places.
64. (iii) What is the probability that more than two of a group of 20
randomly chosen people have green eyes? Give your answer
correct to two decimal places.
P X 2 1 P X 2
65. (iii) What is the probability that more than two of a group of 20
randomly chosen people have green eyes? Give your answer
correct to two decimal places.
P X 2 1 P X 2
1 C2 0 9 0 1 C1 0 9 0 1 C0 0 9 0 1
20 18 2 20 19 1 20 20 0
66. (iii) What is the probability that more than two of a group of 20
randomly chosen people have green eyes? Give your answer
correct to two decimal places.
P X 2 1 P X 2
1 C2 0 9 0 1 C1 0 9 0 1 C0 0 9 0 1
20 18 2 20 19 1 20 20 0
0.3230
0 32 (to 2 dp)
67. (iii) What is the probability that more than two of a group of 20
randomly chosen people have green eyes? Give your answer
correct to two decimal places.
P X 2 1 P X 2
1 C2 0 9 0 1 C1 0 9 0 1 C0 0 9 0 1
20 18 2 20 19 1 20 20 0
0.3230
0 32 (to 2 dp)
Exercise 10J;
1 to 24 even