1. Probability is a measure of the likelihood of the occurrence of an event. It is represented by a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty.
2. The classical definition of probability of an event E is the ratio of the number of favorable cases to the total number of possible cases, when the possible cases are equally likely.
3. For independent events, the occurrence of one event does not affect the probability of the other. Mutually exclusive events cannot occur simultaneously. The sum of probabilities of exhaustive events covering the sample space is 1.
This document contains 34 quantitative probability problems involving concepts like binomial, normal, and Poisson distributions. The problems cover topics like finding the probability of certain outcomes occurring based on given probabilities, estimating values from sample data, and calculating confidence intervals.
The document is a sample physics test with multiple choice and numerical questions covering various physics topics. It contains 8 objective questions, 4 reasoning questions, and 2 passages with 4 questions each related to the passages. It also includes 2 matrix matching questions and 2 numerical/subjective questions requiring numeric answers. The test examines concepts in mechanics, thermodynamics, electromagnetism, and other core areas of physics.
This Mathematics Learner's module discusses about the basic concepts of Probability and its strategies. It also teaches includes some examples about Probability.
This document discusses various concepts in probability, including empirical and theoretical probability, compound events, addition and multiplication rules, dependent and independent events, and counting principles. It provides examples to illustrate key probability concepts such as determining the probability of drawing certain cards or dice outcomes, compound events involving simple events, and rules for calculating probabilities of independent and dependent compound events.
This document provides an overview of growth mindset concepts including:
- The difference between fixed and growth mindsets and how they influence priorities and attitudes.
- Ways to develop growth mindsets such as praising actions not abilities and balancing success with challenges.
- Research showing that previewing material can double learning progress and the importance of setting goals.
This document provides information about probability and introduces key probability concepts. It defines probability terms like event, certain, impossible and even chance. It introduces a probability scale from 0 to 1 to measure likelihood. Examples are provided to demonstrate calculating probability for different events and placing them on the scale. The document concludes by providing practice questions for students to apply their new probability knowledge.
Fabian Scarano - Preparing Your Team for the FutureTEST Huddle
EuroSTAR Software Testing Conference 2008 presentation on Preparing Your Team for the Future by Fabian Scarano. See more at conferences.eurostarsoftwaretesting.com/past-presentations/
This document contains 34 quantitative probability problems involving concepts like binomial, normal, and Poisson distributions. The problems cover topics like finding the probability of certain outcomes occurring based on given probabilities, estimating values from sample data, and calculating confidence intervals.
The document is a sample physics test with multiple choice and numerical questions covering various physics topics. It contains 8 objective questions, 4 reasoning questions, and 2 passages with 4 questions each related to the passages. It also includes 2 matrix matching questions and 2 numerical/subjective questions requiring numeric answers. The test examines concepts in mechanics, thermodynamics, electromagnetism, and other core areas of physics.
This Mathematics Learner's module discusses about the basic concepts of Probability and its strategies. It also teaches includes some examples about Probability.
This document discusses various concepts in probability, including empirical and theoretical probability, compound events, addition and multiplication rules, dependent and independent events, and counting principles. It provides examples to illustrate key probability concepts such as determining the probability of drawing certain cards or dice outcomes, compound events involving simple events, and rules for calculating probabilities of independent and dependent compound events.
This document provides an overview of growth mindset concepts including:
- The difference between fixed and growth mindsets and how they influence priorities and attitudes.
- Ways to develop growth mindsets such as praising actions not abilities and balancing success with challenges.
- Research showing that previewing material can double learning progress and the importance of setting goals.
This document provides information about probability and introduces key probability concepts. It defines probability terms like event, certain, impossible and even chance. It introduces a probability scale from 0 to 1 to measure likelihood. Examples are provided to demonstrate calculating probability for different events and placing them on the scale. The document concludes by providing practice questions for students to apply their new probability knowledge.
Fabian Scarano - Preparing Your Team for the FutureTEST Huddle
EuroSTAR Software Testing Conference 2008 presentation on Preparing Your Team for the Future by Fabian Scarano. See more at conferences.eurostarsoftwaretesting.com/past-presentations/
A review of the growth of the Israel Genealogy Research Association Database Collection for the last 12 months. Our collection is now passed the 3 million mark and still growing. See which archives have contributed the most. See the different types of records we have, and which years have had records added. You can also see what we have for the future.
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
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
-------------------------------------------------------------------------------
Find out more about ISO training and certification services
Training: ISO/IEC 27001 Information Security Management System - EN | PECB
ISO/IEC 42001 Artificial Intelligence Management System - EN | PECB
General Data Protection Regulation (GDPR) - Training Courses - EN | PECB
Webinars: https://pecb.com/webinars
Article: https://pecb.com/article
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For more information about PECB:
Website: https://pecb.com/
LinkedIn: https://www.linkedin.com/company/pecb/
Facebook: https://www.facebook.com/PECBInternational/
Slideshare: http://www.slideshare.net/PECBCERTIFICATION
This presentation includes basic of PCOS their pathology and treatment and also Ayurveda correlation of PCOS and Ayurvedic line of treatment mentioned in classics.
Introduction to AI for Nonprofits with Tapp NetworkTechSoup
Dive into the world of AI! Experts Jon Hill and Tareq Monaur will guide you through AI's role in enhancing nonprofit websites and basic marketing strategies, making it easy to understand and apply.
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.
How to Build a Module in Odoo 17 Using the Scaffold MethodCeline George
Odoo provides an option for creating a module by using a single line command. By using this command the user can make a whole structure of a module. It is very easy for a beginner to make a module. There is no need to make each file manually. This slide will show how to create a module using the scaffold method.
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.
This slide is special for master students (MIBS & MIFB) in UUM. Also useful for readers who are interested in the topic of contemporary Islamic banking.
Thinking of getting a dog? Be aware that breeds like Pit Bulls, Rottweilers, and German Shepherds can be loyal and dangerous. Proper training and socialization are crucial to preventing aggressive behaviors. Ensure safety by understanding their needs and always supervising interactions. Stay safe, and enjoy your furry friends!
বাংলাদেশের অর্থনৈতিক সমীক্ষা ২০২৪ [Bangladesh Economic Review 2024 Bangla.pdf] কম্পিউটার , ট্যাব ও স্মার্ট ফোন ভার্সন সহ সম্পূর্ণ বাংলা ই-বুক বা pdf বই " সুচিপত্র ...বুকমার্ক মেনু 🔖 ও হাইপার লিংক মেনু 📝👆 যুক্ত ..
আমাদের সবার জন্য খুব খুব গুরুত্বপূর্ণ একটি বই ..বিসিএস, ব্যাংক, ইউনিভার্সিটি ভর্তি ও যে কোন প্রতিযোগিতা মূলক পরীক্ষার জন্য এর খুব ইম্পরট্যান্ট একটি বিষয় ...তাছাড়া বাংলাদেশের সাম্প্রতিক যে কোন ডাটা বা তথ্য এই বইতে পাবেন ...
তাই একজন নাগরিক হিসাবে এই তথ্য গুলো আপনার জানা প্রয়োজন ...।
বিসিএস ও ব্যাংক এর লিখিত পরীক্ষা ...+এছাড়া মাধ্যমিক ও উচ্চমাধ্যমিকের স্টুডেন্টদের জন্য অনেক কাজে আসবে ...
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.
2024 State of Marketing Report – by HubspotMarius Sescu
https://www.hubspot.com/state-of-marketing
· Scaling relationships and proving ROI
· Social media is the place for search, sales, and service
· Authentic influencer partnerships fuel brand growth
· The strongest connections happen via call, click, chat, and camera.
· Time saved with AI leads to more creative work
· Seeking: A single source of truth
· TLDR; Get on social, try AI, and align your systems.
· More human marketing, powered by robots
ChatGPT is a revolutionary addition to the world since its introduction in 2022. A big shift in the sector of information gathering and processing happened because of this chatbot. What is the story of ChatGPT? How is the bot responding to prompts and generating contents? Swipe through these slides prepared by Expeed Software, a web development company regarding the development and technical intricacies of ChatGPT!
A review of the growth of the Israel Genealogy Research Association Database Collection for the last 12 months. Our collection is now passed the 3 million mark and still growing. See which archives have contributed the most. See the different types of records we have, and which years have had records added. You can also see what we have for the future.
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
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
-------------------------------------------------------------------------------
Find out more about ISO training and certification services
Training: ISO/IEC 27001 Information Security Management System - EN | PECB
ISO/IEC 42001 Artificial Intelligence Management System - EN | PECB
General Data Protection Regulation (GDPR) - Training Courses - EN | PECB
Webinars: https://pecb.com/webinars
Article: https://pecb.com/article
-------------------------------------------------------------------------------
For more information about PECB:
Website: https://pecb.com/
LinkedIn: https://www.linkedin.com/company/pecb/
Facebook: https://www.facebook.com/PECBInternational/
Slideshare: http://www.slideshare.net/PECBCERTIFICATION
This presentation includes basic of PCOS their pathology and treatment and also Ayurveda correlation of PCOS and Ayurvedic line of treatment mentioned in classics.
Introduction to AI for Nonprofits with Tapp NetworkTechSoup
Dive into the world of AI! Experts Jon Hill and Tareq Monaur will guide you through AI's role in enhancing nonprofit websites and basic marketing strategies, making it easy to understand and apply.
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.
How to Build a Module in Odoo 17 Using the Scaffold MethodCeline George
Odoo provides an option for creating a module by using a single line command. By using this command the user can make a whole structure of a module. It is very easy for a beginner to make a module. There is no need to make each file manually. This slide will show how to create a module using the scaffold method.
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.
This slide is special for master students (MIBS & MIFB) in UUM. Also useful for readers who are interested in the topic of contemporary Islamic banking.
Thinking of getting a dog? Be aware that breeds like Pit Bulls, Rottweilers, and German Shepherds can be loyal and dangerous. Proper training and socialization are crucial to preventing aggressive behaviors. Ensure safety by understanding their needs and always supervising interactions. Stay safe, and enjoy your furry friends!
বাংলাদেশের অর্থনৈতিক সমীক্ষা ২০২৪ [Bangladesh Economic Review 2024 Bangla.pdf] কম্পিউটার , ট্যাব ও স্মার্ট ফোন ভার্সন সহ সম্পূর্ণ বাংলা ই-বুক বা pdf বই " সুচিপত্র ...বুকমার্ক মেনু 🔖 ও হাইপার লিংক মেনু 📝👆 যুক্ত ..
আমাদের সবার জন্য খুব খুব গুরুত্বপূর্ণ একটি বই ..বিসিএস, ব্যাংক, ইউনিভার্সিটি ভর্তি ও যে কোন প্রতিযোগিতা মূলক পরীক্ষার জন্য এর খুব ইম্পরট্যান্ট একটি বিষয় ...তাছাড়া বাংলাদেশের সাম্প্রতিক যে কোন ডাটা বা তথ্য এই বইতে পাবেন ...
তাই একজন নাগরিক হিসাবে এই তথ্য গুলো আপনার জানা প্রয়োজন ...।
বিসিএস ও ব্যাংক এর লিখিত পরীক্ষা ...+এছাড়া মাধ্যমিক ও উচ্চমাধ্যমিকের স্টুডেন্টদের জন্য অনেক কাজে আসবে ...
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.
2024 State of Marketing Report – by HubspotMarius Sescu
https://www.hubspot.com/state-of-marketing
· Scaling relationships and proving ROI
· Social media is the place for search, sales, and service
· Authentic influencer partnerships fuel brand growth
· The strongest connections happen via call, click, chat, and camera.
· Time saved with AI leads to more creative work
· Seeking: A single source of truth
· TLDR; Get on social, try AI, and align your systems.
· More human marketing, powered by robots
ChatGPT is a revolutionary addition to the world since its introduction in 2022. A big shift in the sector of information gathering and processing happened because of this chatbot. What is the story of ChatGPT? How is the bot responding to prompts and generating contents? Swipe through these slides prepared by Expeed Software, a web development company regarding the development and technical intricacies of ChatGPT!
Product Design Trends in 2024 | Teenage EngineeringsPixeldarts
The realm of product design is a constantly changing environment where technology and style intersect. Every year introduces fresh challenges and exciting trends that mold the future of this captivating art form. In this piece, we delve into the significant trends set to influence the look and functionality of product design in the year 2024.
How Race, Age and Gender Shape Attitudes Towards Mental HealthThinkNow
Mental health has been in the news quite a bit lately. Dozens of U.S. states are currently suing Meta for contributing to the youth mental health crisis by inserting addictive features into their products, while the U.S. Surgeon General is touring the nation to bring awareness to the growing epidemic of loneliness and isolation. The country has endured periods of low national morale, such as in the 1970s when high inflation and the energy crisis worsened public sentiment following the Vietnam War. The current mood, however, feels different. Gallup recently reported that national mental health is at an all-time low, with few bright spots to lift spirits.
To better understand how Americans are feeling and their attitudes towards mental health in general, ThinkNow conducted a nationally representative quantitative survey of 1,500 respondents and found some interesting differences among ethnic, age and gender groups.
Technology
For example, 52% agree that technology and social media have a negative impact on mental health, but when broken out by race, 61% of Whites felt technology had a negative effect, and only 48% of Hispanics thought it did.
While technology has helped us keep in touch with friends and family in faraway places, it appears to have degraded our ability to connect in person. Staying connected online is a double-edged sword since the same news feed that brings us pictures of the grandkids and fluffy kittens also feeds us news about the wars in Israel and Ukraine, the dysfunction in Washington, the latest mass shooting and the climate crisis.
Hispanics may have a built-in defense against the isolation technology breeds, owing to their large, multigenerational households, strong social support systems, and tendency to use social media to stay connected with relatives abroad.
Age and Gender
When asked how individuals rate their mental health, men rate it higher than women by 11 percentage points, and Baby Boomers rank it highest at 83%, saying it’s good or excellent vs. 57% of Gen Z saying the same.
Gen Z spends the most amount of time on social media, so the notion that social media negatively affects mental health appears to be correlated. Unfortunately, Gen Z is also the generation that’s least comfortable discussing mental health concerns with healthcare professionals. Only 40% of them state they’re comfortable discussing their issues with a professional compared to 60% of Millennials and 65% of Boomers.
Race Affects Attitudes
As seen in previous research conducted by ThinkNow, Asian Americans lag other groups when it comes to awareness of mental health issues. Twenty-four percent of Asian Americans believe that having a mental health issue is a sign of weakness compared to the 16% average for all groups. Asians are also considerably less likely to be aware of mental health services in their communities (42% vs. 55%) and most likely to seek out information on social media (51% vs. 35%).
AI Trends in Creative Operations 2024 by Artwork Flow.pdfmarketingartwork
Creative operations teams expect increased AI use in 2024. Currently, over half of tasks are not AI-enabled, but this is expected to decrease in the coming year. ChatGPT is the most popular AI tool currently. Business leaders are more actively exploring AI benefits than individual contributors. Most respondents do not believe AI will impact workforce size in 2024. However, some inhibitions still exist around AI accuracy and lack of understanding. Creatives primarily want to use AI to save time on mundane tasks and boost productivity.
Organizational culture includes values, norms, systems, symbols, language, assumptions, beliefs, and habits that influence employee behaviors and how people interpret those behaviors. It is important because culture can help or hinder a company's success. Some key aspects of Netflix's culture that help it achieve results include hiring smartly so every position has stars, focusing on attitude over just aptitude, and having a strict policy against peacocks, whiners, and jerks.
PEPSICO Presentation to CAGNY Conference Feb 2024Neil Kimberley
PepsiCo provided a safe harbor statement noting that any forward-looking statements are based on currently available information and are subject to risks and uncertainties. It also provided information on non-GAAP measures and directing readers to its website for disclosure and reconciliation. The document then discussed PepsiCo's business overview, including that it is a global beverage and convenient food company with iconic brands, $91 billion in net revenue in 2023, and nearly $14 billion in core operating profit. It operates through a divisional structure with a focus on local consumers.
Content Methodology: A Best Practices Report (Webinar)contently
This document provides an overview of content methodology best practices. It defines content methodology as establishing objectives, KPIs, and a culture of continuous learning and iteration. An effective methodology focuses on connecting with audiences, creating optimal content, and optimizing processes. It also discusses why a methodology is needed due to the competitive landscape, proliferation of channels, and opportunities for improvement. Components of an effective methodology include defining objectives and KPIs, audience analysis, identifying opportunities, and evaluating resources. The document concludes with recommendations around creating a content plan, testing and optimizing content over 90 days.
How to Prepare For a Successful Job Search for 2024Albert Qian
The document provides guidance on preparing a job search for 2024. It discusses the state of the job market, focusing on growth in AI and healthcare but also continued layoffs. It recommends figuring out what you want to do by researching interests and skills, then conducting informational interviews. The job search should involve building a personal brand on LinkedIn, actively applying to jobs, tailoring resumes and interviews, maintaining job hunting as a habit, and continuing self-improvement. Once hired, the document advises setting new goals and keeping skills and networking active in case of future opportunities.
A report by thenetworkone and Kurio.
The contributing experts and agencies are (in an alphabetical order): Sylwia Rytel, Social Media Supervisor, 180heartbeats + JUNG v MATT (PL), Sharlene Jenner, Vice President - Director of Engagement Strategy, Abelson Taylor (USA), Alex Casanovas, Digital Director, Atrevia (ES), Dora Beilin, Senior Social Strategist, Barrett Hoffher (USA), Min Seo, Campaign Director, Brand New Agency (KR), Deshé M. Gully, Associate Strategist, Day One Agency (USA), Francesca Trevisan, Strategist, Different (IT), Trevor Crossman, CX and Digital Transformation Director; Olivia Hussey, Strategic Planner; Simi Srinarula, Social Media Manager, The Hallway (AUS), James Hebbert, Managing Director, Hylink (CN / UK), Mundy Álvarez, Planning Director; Pedro Rojas, Social Media Manager; Pancho González, CCO, Inbrax (CH), Oana Oprea, Head of Digital Planning, Jam Session Agency (RO), Amy Bottrill, Social Account Director, Launch (UK), Gaby Arriaga, Founder, Leonardo1452 (MX), Shantesh S Row, Creative Director, Liwa (UAE), Rajesh Mehta, Chief Strategy Officer; Dhruv Gaur, Digital Planning Lead; Leonie Mergulhao, Account Supervisor - Social Media & PR, Medulla (IN), Aurelija Plioplytė, Head of Digital & Social, Not Perfect (LI), Daiana Khaidargaliyeva, Account Manager, Osaka Labs (UK / USA), Stefanie Söhnchen, Vice President Digital, PIABO Communications (DE), Elisabeth Winiartati, Managing Consultant, Head of Global Integrated Communications; Lydia Aprina, Account Manager, Integrated Marketing and Communications; Nita Prabowo, Account Manager, Integrated Marketing and Communications; Okhi, Web Developer, PNTR Group (ID), Kei Obusan, Insights Director; Daffi Ranandi, Insights Manager, Radarr (SG), Gautam Reghunath, Co-founder & CEO, Talented (IN), Donagh Humphreys, Head of Social and Digital Innovation, THINKHOUSE (IRE), Sarah Yim, Strategy Director, Zulu Alpha Kilo (CA).
Trends In Paid Search: Navigating The Digital Landscape In 2024Search Engine Journal
The search marketing landscape is evolving rapidly with new technologies, and professionals, like you, rely on innovative paid search strategies to meet changing demands.
It’s important that you’re ready to implement new strategies in 2024.
Check this out and learn the top trends in paid search advertising that are expected to gain traction, so you can drive higher ROI more efficiently in 2024.
You’ll learn:
- The latest trends in AI and automation, and what this means for an evolving paid search ecosystem.
- New developments in privacy and data regulation.
- Emerging ad formats that are expected to make an impact next year.
Watch Sreekant Lanka from iQuanti and Irina Klein from OneMain Financial as they dive into the future of paid search and explore the trends, strategies, and technologies that will shape the search marketing landscape.
If you’re looking to assess your paid search strategy and design an industry-aligned plan for 2024, then this webinar is for you.
5 Public speaking tips from TED - Visualized summarySpeakerHub
From their humble beginnings in 1984, TED has grown into the world’s most powerful amplifier for speakers and thought-leaders to share their ideas. They have over 2,400 filmed talks (not including the 30,000+ TEDx videos) freely available online, and have hosted over 17,500 events around the world.
With over one billion views in a year, it’s no wonder that so many speakers are looking to TED for ideas on how to share their message more effectively.
The article “5 Public-Speaking Tips TED Gives Its Speakers”, by Carmine Gallo for Forbes, gives speakers five practical ways to connect with their audience, and effectively share their ideas on stage.
Whether you are gearing up to get on a TED stage yourself, or just want to master the skills that so many of their speakers possess, these tips and quotes from Chris Anderson, the TED Talks Curator, will encourage you to make the most impactful impression on your audience.
See the full article and more summaries like this on SpeakerHub here: https://speakerhub.com/blog/5-presentation-tips-ted-gives-its-speakers
See the original article on Forbes here:
http://www.forbes.com/forbes/welcome/?toURL=http://www.forbes.com/sites/carminegallo/2016/05/06/5-public-speaking-tips-ted-gives-its-speakers/&refURL=&referrer=#5c07a8221d9b
ChatGPT and the Future of Work - Clark Boyd Clark Boyd
Everyone is in agreement that ChatGPT (and other generative AI tools) will shape the future of work. Yet there is little consensus on exactly how, when, and to what extent this technology will change our world.
Businesses that extract maximum value from ChatGPT will use it as a collaborative tool for everything from brainstorming to technical maintenance.
For individuals, now is the time to pinpoint the skills the future professional will need to thrive in the AI age.
Check out this presentation to understand what ChatGPT is, how it will shape the future of work, and how you can prepare to take advantage.
The document provides career advice for getting into the tech field, including:
- Doing projects and internships in college to build a portfolio.
- Learning about different roles and technologies through industry research.
- Contributing to open source projects to build experience and network.
- Developing a personal brand through a website and social media presence.
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- Practicing interviews through mock interviews and whiteboarding coding questions.
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A brief introduction to DataScience with explaining of the concepts, algorithms, machine learning, supervised and unsupervised learning, clustering, statistics, data preprocessing, real-world applications etc.
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Time Management & Productivity - Best PracticesVit Horky
Here's my presentation on by proven best practices how to manage your work time effectively and how to improve your productivity. It includes practical tips and how to use tools such as Slack, Google Apps, Hubspot, Google Calendar, Gmail and others.
The six step guide to practical project managementMindGenius
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Beginners Guide to TikTok for Search - Rachel Pearson - We are Tilt __ Bright...
Iit class xii_maths_probability
1. 000000000.
, AT H E M AT I C S
M +
,S T U D Y M A T E R I A L+
, +
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, +
, +
, PROBABILITY +
, +
, IIT-JEE +
, +
, +
, +
, +
NARAYANA INSTITUTE OF CORRESPONDENCE COURSES
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2. 2004 NARAYANA GROUP
This study material is a part of NARAYANA INSTITUTE OF CORRESPONDENCE COURSES for IIT-JEE, 2008-09. This is meant
for the personal use of those students who are enrolled with NARAYANA INSTITUTE OF CORRESPONDENCE COURSES, FNS
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without prior information and written permission of the institute.
3. PREFACE
Dear Student,
Heartiest congratulations on making up your mind and deciding to be an engineer to serve the society.
As you are planning to take various Engineering Entrance Examinations, we are sure that this STUDY PACKAGE is
going to be of immense help to you.
At NARAYANA we have taken special care to design this package according to the Latest Pattern of IIT-JEE, which
will not only help but also guide you to compete for IIT-JEE, AIEEE other State Level Engineering Entrance
Examinations.
The salient features of this package include :
! Power packed division of units and chapters in a scientific way, with a correlation being there.
! Sufficient number of solved examples in Physics, Chemistry Mathematics in all the chapters to motivate the
students attempt all the questions.
! All the chapters are followed by various types of exercises, including Objective - Single Choice Questions,
Objective - Multiple Choice Questions, Comprehension Type Questions, Match the Following, Assertion-Reasoning
Subjective Questions.
These exercises are followed by answers in the last section of the chapter including Hints Solutions wherever
required. This package will help you to know what to study, how to study, time management, your weaknesses and
improve your performance.
We, at NARAYANA, strongly believe that quality of our package is such that the students who are not fortunate
enough to attend to our Regular Classroom Programs, can still get the best of our quality through these packages.
We feel that there is always a scope for improvement. We would welcome your suggestions feedback.
Wish you success in your future endeavours.
THE NARAYANA TEAM
ACKNOWLEDGEMENT
While preparing the study package, it has become a wonderful feeling for the NARAYANA TEAM to get the
wholehearted support of our Staff Members including our Designers. They have made our job really easy through
their untiring efforts and constant help at every stage.
We are thankful to all of them.
THE NARAYANA TEAM
4. CONTENTS
PROBABILITY
C O N T E N T S
1. Theory
2. Solved Problems
(i) Subjective Problems
(ii) Single Choice Problems
Multiple Choice Problems
(iii) Comprehension Type Problems
Matching Type Problems
Assertion-Reason Type Problems
3. Assignments
(i) Subjective Questions
(ii) Single Choice Questions
(iii) Multiple Choice Questions
(iv) Comprehension Type Questions
Matching Type Questions
Assertion-Reason Type Questions
(v) Problems Asked in IIT-JEE
4. Answers
5. Mathematics : Probability NARAYANA
INSTITUTE OF CORRESPONDENCE COURSES
PROBABILITY
IIT- JEE SYLLABUS
Addition and multiplication rules of Probability, Conditional probability, Independence
of events, Computation of probability of events using permutations and combinations.
CONTENTS INTRODUCTION
♦ Basic Definition
From the time immemorial, Human Life has
♦ Probability of an event been full of uncertainties. In our everyday life,
we very often make guesses and use
♦ Odds in favour and against an event statements like - probably it will rain today,
possibility of India to win this Cricket World
♦ Addition theorem of probability cup is more than any other team, chances of
congress coming to power this year are strong.
♦ Algebra of events In the above statements, the words probably,
certainly, possibility, chances etc. convey the
♦ Conditional probability
sense of uncertainty or certainty about the
occurrence of some event. The word ‘Probability’
♦ Multiplication theorem of probability
or ‘chance’ of any event measures one’s degree
♦ Total probability theorem of belief about the occurrence of that event.
Ordinarily, it may appear that there cannot be
♦ Baye’s theorem any exact measurement for these uncertainties
but in probability theory, we do have methods
♦ Binomial distribution for repeated experiments for measuring the degree of certainty or
uncertainty of events in terms of numbers lying
♦ Geometrical probability between 0 and 1, provided certain conditions
are satisfied. A probability of 1 means 100%
chance of occurrence of an event which
obviously is the maximum chance.
1
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Mathematics : Probability
1. BASIC DEFINITIONS
RANDOM EXPERIMENT
An experiment which can result in more than one outcome and so, whose outcome cannot be predicted
with certainty, is called Random experiment.
Point to be Noted : An experiment whose outcome is known in advance, is not a random experiment.
For example, when a ball is thrown upward, it will surely fall downward. So it is not a random
experiment.
Examples:
(i) Tossing of a coin may result in head or tail.
(ii) Throwing of a die may result in anyone of six outcomes {1, 2, 3, 4, 5, 6}
SAMPLE SPACE
The sample space of a random experiment is defined as the set of all possible outcomes of the
experiment. Sample space is usually represented by ‘S’. So since sample space consists of all the
possibilities, sample space will surely occur in any experiment.
Examples:
(i) Tossing of a coin results in either a head or a tail turning up. Let H denote the occurrence of
head and T denote the occurrence of tail.
Then, sample space S = {H, T}
(ii) Throwing of a die will result in anyone of six outcomes 1, 2, 3, 4, 5 or 6.
So sample space = {1, 2, 3, 4, 5, 6}
(iii) Tossing of two coins results in the sample space S = {HH, HT, TH, TT}
DISCRETE SAMPLE SPACE
A sample space having finite number of sample points, is said to be ‘Discrete Sample Space’.
SAMPLE POINT
Each possible outcome of a random experiment is called a sample point. For a point H T are
sample point or sample elements.
EVENT
The event is a subset of a sample space.
Points to be Noted
(i) φ is also a subset of S which is called an impossible event.
(ii) Sample space plays the same role in ‘probability’ as does the universal set in ‘Set Theory’.
Example:
In the throwing of a die, sample space is S = {1, 2, 3, 4, 5, 6}
A = {1, 3, 5}, B = {2, 4, 6}, C = {1, 2, 3, 4} , all being the subsets of S, are called events. Here A is
the event of occurrence of an odd number, B is the event of occurrence of an even number, and C is
the event of occurrence of a number less than 5.
SIMPLE EVENT OR ELEMENTARY EVENT
An event which cannot be further split is a simple event.
2
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COMPOUND EVENT
An event consisting of more than one sample points is called Compound Event.
Examples:
(i) If a die is throwing, then sample space is S = {1, 2, 3, 4, 5, 6}
A = {1, 3, 5}, B = {2, 4, 6} and C = {1, 2, 3, 4} are compound events.
TRIAL
Each performance of the random experiment is called a Trial.
EQUALLY LIKELY EVENTS
A set of events is said to be equally likely if taking into consideration all the relevent factors, there is
no reason to expect one of them in preference to the others i.e. in simple words, the events are
equally likely to occur. Obviously, the chances of occurrence of equally likely events are equal and so
their probabilities are equal.
Mathematically:
Examples :
(i) In case of tossing a fair coin, occurrence of head and tail are equally likely events
(ii) In the throwing of an unbiased die, all the six faces 1, 2, 3, 4, 5 and 6 are equally likely to
come up.
MUTUALLY EXCLUSIVE EVENTS
A set of events is said to be mutually exclusive if the occurrence of one of them rules out the
occurrence of any of the remaining events i.e. in simple words, no two out of the set of events can
occur simultaneously.
In set theoretic notation, two events A1 and A2 are mutually exclusive if
A1 ∩ A 2 = φ
For more than two events,
In set theoretic notation, events A1, A2, ......., Am are mutually exclusive if
A i ∩ A j = φ for all pairs (i, j) satisfying 1 ≤ i, j ≤ m , where i ≠ j
Examples:
(i) In the tossing of a coin, the events {H} and {T} are mutually exclusive events since head
and tail can’t occur simultaneously in any performance of the experiment.
(ii) In the throwing of a die, the events E1 = {1, 3, 5} and E2 = {2, 4, 6} are mutually exclusive
events since an odd number and an even number can’t occur simultaneously in any
performance of the experiment.
EXHAUSTIVE EVENTS
A set of events is said to be exhaustive if atleast one of them must necessarily occur on each
performance of the experiment i.e. in simple words, if all the events collectively, cover all the possible
outcomes of the experiment, then they are called exhaustive events.
In set theoretic notation, events A1, A2, ......, Am are called exhaustive
m
If A1 ∪ A 2 ∪ ..... ∪ A m = S i.e. ∪A
i =1
i =S
Mathematically, events A1, A2, ....., Am are called exhaustive.
Example:
In the throwing of a die, the events {1, 2}, {2, 3, 4}, {3, 4, 5, 6} and {1, 3, 6} are exhaustive system
of events.
Since {1, 2} ∪ {2, 3, 4} ∪ {3, 4, 5, 6} ∪ {1, 3, 6} = {1, 2, 3, 4, 5, 6} = S
3
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Mathematics : Probability
INDEPENDENT EVENTS
Two or more events are said to be independent if occurrence or non-occurrence of any of them does
not affect the probability of occurrence or non-occurrence of other events.
Examples:
(i) Consider the drawing of two balls one after the other, with replacement from an urn containing
two or more balls of varying colours. Then, the two draws are independent of each other and any
two events defined on two draws (one on each draw) will be independent.
But if the two balls were drawn one by one, without replacement, then probability of occurrence
of any event defined on second draw would depend upon the result of first draw. So any two
events defined on two draws (one on each draw) would have been dependent.
(ii) Consider the drawing of two cards one after the other, with replacement from a pack of 52
playing cards. Then, the two draws are independent of each other and any two events defined on
two draws (one on each draw) will be independent.
But if the two cards were drawn one by one, without replacement, then probability of occurrence
of any event defined on 2nd draw would depend upon the result of first draw. So any two events
defined on two draws (one on each draw) would have been dependent.
RELATION BETWEEN MUTUALLY EXCLUSIVE AND INDEPENDENT EVENTS
Mutually exclusive events will never be independent and independent events will never be mutually
exclusive as explained below.
Mutually exclusive events are those events in which occurrence of one of them rules out the occurrence
of other. It shows clearly that the chances of their occurrences or non-occurrences depend upon
each other and so the events cannot be independent. Similarly independent events are the events
which don’t depend upon each other but mutually exclusive events affect each other. So independent
events can’t be mutually exclusive.
2. PROBABILITY OF AN EVENT
CLASSICAL OR MATHEMATICAL DEFINITION OF PROBABILITY
If there are n equally likely, mutually exclusive and exhaustive events and m of which are favourable
to the event E,
n(E)
then probability of occurrence of event E = P(E) =
n(S)
number of cases favourable to event E m
= =
Total number of cases n
It may be observed here that 0 ≤ m ≤ n
m
⇒ 0≤
≤ 1 ⇒ 0 ≤ P(E) ≤ 1
n
The number of cases unfavourable to event E = n – m
n−m m
⇒ P(E c ) = = 1 − = 1 − P(E)
n n
C
So P(E) + P(E ) = 1
n
If E is sure event, P(E) = =1
n
4
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0
If E is impossible event, P(E) = =0
n
So 0 ≤ P(E) ≤ 1
Example :
In the tossing of a coin, sample space S = {H, T} ⇒ n(S) = 2.
n(H) 1
Then, P(H) = =
n(S) 2
n(T) 1
Similarly, P(T) = =
n(S) 2
So occurrence of head and tail are equally likely events in the tossing of a coin.
3. ODDS IN FAVOUR AND AGAINST AN EVENT
If a cases are favourable to the event A and b cases are not favourable to the event A that is
number of favourable cases a
favourable to the A , then P(A) = =
number of total cases a+b
number of non favourable cases b
and P(A) = =
number of total cases a+b
P(A)
Odds in favour of event A = = a:b
P(A)
P(A)
Odds in against of event A = = b:a [P(A) + P (A) = 1]
P(A)
Illustration 1: Find the probability of the event A if (i) odds in favour of event A are 5 : 7 (ii) odds
against A are 3 : 4.
Solution : (i) Odds in favour of event A are 5 : 7. Let P(A) = p
∴ p:1–p=5:7
p 5
⇒ = ⇒ 7p = 5 − 5p
1− p 7
5
⇒ 12p = 5 ⇒ p=
12
5
∴ P(A) =
12
(ii) Odds against event A are 3 : 4. Let P(A) = p.
∴ 1–p:p=3:4
1− p 3
⇒ = ⇒ 4 – 4p = 3p
p 4
4
⇒ 7p = 4 ⇒ p=
7
4
∴ P(A) =
7
5
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Mathematics : Probability
Illustration 2: A bag contains 5 white, 7 black and 8 red balls. A ball is drawn at random. Find the
probability of getting:
(i) red ball (ii) non-white ball
Solution : Number of white balls = 5
Number of black balls = 7
Number of red balls = 8
∴ Total number of balls = 5 + 7 + 8 = 20.
(i) Let R = event of getting red ball
n(R) 8 2
∴ P(red ball) = P(R) = = =
n(S) 20 5
(ii) Let W = event of getting white ball
number of non − white balls 7 + 8 15 3
∴ P(non-white ball) = P(W) = = = =
total number of balls 20 20 4
4. ADDITION THEOREM OF PROBABILITY
If A and B are two events related with an experiment then the probability that either of the events
will occur (or at least one of the events will occur) is
P(A ∪ B) = P(A) + P(B) − P(A ∩ B) where (A ∩ B) is the event defined as the event that both A
and B are occuring.
Note : If A and B are mutually exclusive events then P(A ∩ B) = 0
∴ for mutually exclusive events
P(A ∪ B) = P(A) + P(B)
Illustration 3: A pair of dice is rolled once. Find the probability of throwing a total of 8 or 10.
Solution : In this case, sample space S consists for 6 × 6 = 36 equally likely simple events of the type
(x, y) where x, y ∈ {1, 2, 3, 4, 5, 6}
i.e. S = {(x, y) : x, y ∈ {1, 2, 3, 4, 5, 6}}.
Let E1: ‘a total of 8’ and E2: ‘a total of 10’, then E1 and E2 are mutually exclusive events as
the sum of number on the uppermost faces of the two dice in a single throw cannot be 8 and
10 simultaneously.
Now E1 = {(2, 6), (4, 4), (6, 2), (3, 5), (5, 3)} and E2 = {(4, 6), (5, 5), (6, 4)}.
∴ Required probability = P(E1 ∪ E 2 )
5 3 8 2
= P(E1) + P(E2) = + = =
36 36 36 9
Illustration 4 : Find the probability of 4 turning up for at least once in two tosses of a fair die.
Solution : Here S = {(1, 1), (1, 2), ........, (6, 5), (6, 6)}.
Let A = event of getting 4 on the first die
and B = event of getting 4 on the second die.
∴ A = {(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)}
and B = {(1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4)}.
6
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n(A) 6 1 n(B) 6 1
∴ P(A) = = = and P(B) = = =
n(S) 36 6 n(S) 36 6 .
The events A and B are not m.e. because the sample (4, 4) is common to both.
1
∴ A ∩ B = {(4, 4)} ∴ P(A ∩ B) =
36
By addition theorem, the required probability of getting four at least once is
1 1 1 11
P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = + − = .
6 6 36 36
Note : Probabilities of mutually exclusive, exhaustive and equally likely events:
Let E1, E2, ..., En be n mutually exclusive, exhaustive and equally likely events.
Then P(E1 ∪ E 2 .... ∪ E n ) = 1 (since exhaustive events)
But P(E1 ∪ E 2 .... ∪ E n ) = P(E1 ) + P(E 2 ) + ..... + P(E n ) (since mutually exclusive events)
⇒ P(E1) + P(E2) + .... + P(En) = 1
Also P(E1) = P(E2) = ..... = P(En) (Since equally likely events)
1
⇒ n P(Ei) = 1, i ∈ {1, 2, 3, .....n} ⇒ P(E i ) =
n
So P(E1) = P(E2) = .... = P(En) = 1/n
1
So probability of each of n mutually exclusive, exhaustive and equally likely events is .
n
EXERCISE – 1
1. An honest die is thrown randomly. A person requires either multiple of 2 or multiple of 3. What
is the probability in his favour?
2. A and B are solving a problem of mathematics. The probability of A to solve is 1/2 and probability
of B to solve is 1/3, respectively. What is the probability that problem is solved?
5. ALGEBRA OF EVENTS
(i) P( A) + P( A c ) = 1
(ii) P ( A − B ) = P ( A ∩ B) = P ( A ∪ B ) − P ( B) = P ( A ) − P ( A ∩ B )
(iii) P (B − A ) = P (B ∩ A ) = P ( A ∪ B ) − P ( A ) = P (B ) − P ( A ∩ B )
(iv) P ( A∆ B ) = P ( A ∪ B ) − P ( A ∩ B )
(v) P( A∆B) = 2P( A ∪ B) − ( P( A) + P(B) )
(vi) P ( A∆B) = P ( A ) + P (B) − 2P ( A ∩ B)
(vii) ( )
P A ∩ B = P(S) − P( A ∪ B) = 1 − P( A ∪ B)
(viii) A ⊆ B ⇒ P ( A ) ≤ P (B )
7
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Mathematics : Probability
(ix) P(exactly two of A, B, C occur)
= P ( A ∩ B ) + P (B ∩ C) + P (C ∩ A ) − 3P ( A ∩ B ∩ C)
(x) P(at least two of A, B, C occur)
= P ( A ∩ B ) + P (B ∩ C) + P (C ∩ A ) − 2P ( A ∩ B ∩ C)
Illustration 5: Three critics review a book. Odds in favour of the book are 5 : 2, 4 : 3 and 3 : 4 respectively
for the three critics. Find the probability that majority are in favour of the book.
Solution : Let the critics be E1, E2 and E3. Let P(E1), P(E2) and P(E3) denotes the probabilities of the
critics E1, E2 E3 to be in favour of the book. Since the odds in favour of the book for the
critics E1, E2 E3 are 5 : 2, 4 : 3 and 3 : 4 respectively,
5 5 4 4 3 3
∴ P(E1 ) = = ; P(E 2 ) = = and P(E 3 ) = =
5+2 7 4+3 7 3+ 4 7
Clearly, the event of majority being in favour = the event of at least two critics being in favour.
∴ The required probability
= P(E1E 2 E 3 ) + P(E1E 2 E3 ) + P(E1E 2 E 3 ) + P(E1E 2 E3 )
= P(E1 ).P(E 2 ).P(E 3 ) + P(E1 ) P(E 2 ) P(E 3 ) + P(E1 ) P(E 2 ) P(E 3 ) + P(E1 )P(E 2 )P(E 3 )
{ ∵ E1, E2 E3 are independent}
5 4 3 5 4 3 5 4 3 5 4 3 1
= . . 1 − + 1 − . . . .1 − . + . . = 3 [80 + 24 + 45 + 60]
7 7 7 7 7 7 7 7 7 7 7 7 7
209
=
343
Illustration 6: The probability that atleast one of A and B occurs is 0.6. If A and B occur simultaneously
with probability 0.2. Find P(A) + P(B) .
Solution : Given P(A ∪ B) = 0.6
and P(A ∩ B) = 0.2
Now P(A) + P(B) = 1 − P(A) + 1 − P(B)
= 2 – [P(A) + P(B)] = 2 – [P (A) + P(B) – P (A ∩ B) ] – P (A ∩ B)
= 2 − P(A ∪ B) − P(A ∩ B) = 2 – 0.6 – 0.2 = 1.2.
Illustration 7: The probability of two events A and B are 0.25 and 0.50 respectively. The probability
of their simultaneous occurrence is 0.14. Find the probability that neither A nor B
occurs.
Solution : We have P(A) = 0.25, P(B) = 0.50 and P(A ∩ B) = 0.14
By addition theorem,
P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.25 + 0.50 – 0.14 = 0.61
Now, P(neither A nor B) = P(A ∩ B) = P(A ∪ B)* = 1 − P(A ∪ B) = 1 − 0.61 = 0.39
8
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1 2
Illustration 8: A and B are two non-mutually exclusive events. If P ( A) = , P ( B) = and
4 5
1
P ( A ∪ B) = , find the values of P ( A ∩ B) and P ( A ∩ B c ) .
2
1 2 1
Solution : We have P(A) = , P(B) = , P(A ∪ B) =
4 5 2
By addition theorem, P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
1 1 2
∴ = + − P(A ∩ B)
2 4 5
1 2 1 2 + 8 − 10 3
∴ P(A ∩ B) = + − = =
4 5 2 20 20
We have, (A ∩ Bc ) ∩ (A ∩ B) = A ∩ (Bc ∩ B) = A ∩ φ = φ
∴ The event A ∩ Bc and A ∩ B are mutually exclusive and
(A ∩ Bc ) ∪ (A ∩ B) = A ∩ (Bc ∪ B) = A ∩ S = A (S is the sample space)
∴ By addition theorem, P(A) = P(A ∩ B ) + P(A ∩ B)
c
1 3 1 3 1
∴ = P(A ∩ Bc ) + or P(A ∩ B ) = −
c
=
4 20 4 20 10
EXERCISE – 2
1. Let A and B be two events defined on a sample space. Given P(A) = 0.4, P(B) = 0.80 and
( )
P A ∩ B = 0.10. Find (i) P A ∪ B ( ) ( ) (
(ii) P A ∪ B ∪ A ∩ B
)
, P ( A ∩ B ) = , P ( B ) = . Determine P(A) and P(B).
5 1 1
2. Given P ( A ∪ B ) =
6 3 2
6. CONDITIONAL PROBABILITY
If A and B are two events, then the conditional probability of event A given that event B has already
( )
occured P A B is defined as P A B = ( )
P(A ∩ B)
P(B)
Similarly, the probability of B given that A has already occured will be P B A =( )
P(A ∩ B)
P(A)
Therefore, P(A ∩ B) = P(A)P B ( A ) = P(B) ⋅ P ( A B)
Illustration 9: If a pair of dice is thrown and it is known that sum of the numbers is even, then find the
probability that the sum is less than 6.
Solution : Let A be the given event and let B be the event, whose probability is to be found. Then
B P ( B ∩ A ) 4 / 36 2
Required probability P = P(A) = 18 / 36 = 9 .
A
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7. MULTIPLICATION THEOREM OF PROBABILITY
If A and B are any two events, then
A
P ( A ∩ B ) = P (B). P if P (B ) ≠ 0
B
Similarly, P ( A ∩ B) = P ( A ). P
B if P ( A ) ≠ 0
A
Generalized form of Multiplication Theorem
Suppose A1, A2, ........, Am be m events such that P ( A1 ∩ A 2 ∩ ..... ∩ A m ) ≠ 0, then
A A3 A4
P( A1 ∩ A 2 ∩ ..... ∩ A m ) = P( A1 ) P 2 P P ....
A1 A1 A 2 A1 A 2 A 3
Am
.. . P
A 1 A 2 .. .. A m −1
Illustration 10: P1, P2, . . . , P8 are eight players participating in a tournament. If i j, then Pi will win
the match against Pj. Players are paired up randomly for first round and winners of
this round again paired up for the second round and so on. Find the probability that P4
reaches in the final.
Solution : Let A1 be the event that in the first round the four winners are P1, P4, Pi, Pj, where i ∈ {2, 3},
j ∈ {5, 6, 7}and let A2 be the event that out of the four winners in the first round, P1 and P4
reaches in the final.
The event A1 will occur, if P4 plays with any of P5, P6, P7 or P8 (say with P6) and P1, P2 and
P3 are not paired with P5, P7 and P8. Further A2 will occur if P1 plays with Pj.
A2
The required probability = P ( A1 ∩ A 2 ) = P(A1). P A .
1
6
4× − 3
( 2 )3 3
× 1 4
= = .
8 4 35
( 2 )4 4 ( 2 )2 2
(Here we have used the concept of division into groups).
Note : For mutually independent and pairwise independent events.
Let A and B be events associated with a random experiment. The events A and B are
independent if and only if P(A ∩ B) = P(A) ⋅ P(B) . Since A and B are mutually independent
( )
events P B A = P(B) or P A B = P(A) ( )
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Let A and B be independent events.
P(A ∩ B) P(A ∩ B)
∴ P(A ∩ B) = P(B) = P(A / B) P(B) ∵ P(A / B) =
P(B) P(B)
= P(A) P(B) (∵ P(A / B) = P(A) )
∴ P(A ∩ B) = P(A) P(B) .
Conversely, let P(A ∩ B) = P(A) P(B) .
P(A ∩ B) P(A) P(B)
∴ P(A / B) = = = P(A)
P(B) P(B)
P(B ∩ A) P(A ∩ B) P(A) P(B)
and P(B / A) = = = = P(B)
P(A) P(A) P(A)
∴ P(B/A) = P(A) and P(B/A) = P(B).
∴ A and B are independent events.
So P(A ∩ B) = P(A) P(B) is the necessary and sufficient condition for the events A and B
to be independent.
Illustration 11: A and B are two independent events. The probability that both A and B occur is 1/6 and the
probability that neither of them occurs is 1/3. Find the probability of the occurrence of A.
Solution : Let P(A) = x and P(B) = y. Since A and B are independent,
P(A ∩ B) = P(A) P(B) = xy = 1/ 6, ...(i)
and P(A '∩ B') = P(A ') P(B') = (1 − x) (1 − y) = 1/ 3 ...(ii)
It should be noted that x 0, y 0 ...(iii)
Solving (i) and (ii), we get x = 1/3 or x = 1/2.
REMARK
Three events A, B and C are said to be mutually independent if,
P ( A ∩ B ) = P(A). P(B), P ( A ∩ C ) = P(A). P(C), P ( B ∩ C ) = P(B). P(C) and
P ( A ∩ B ∩ C ) = P(A). P(B). P(C)
These events would be said to be pair–wise independent if,
P ( A ∩ B ) = P(A). P(B), P ( B ∩ C ) = P(B). P(C) and P ( A ∩ C ) = P(A). P(C).
Illustration 12: A lot contains 50 defective and 50 non-defective bulbs. Two bulbs are drawn at random,
one at a time with replacement. The events A, B and C are defined as under:
A = {The first bulb is defective}, B = {The second bulb is non-defective}
C = {The two bulbs are either both defective or both non-defective}
Catogorize the events A, B and C to be pairwise independent or mutually independent.
1 1 1 1 1 1 1 1 1
Solution : P(A) = ×1 = , P(B) = 1× = and P(C) = × + × =
2 2 2 2 2 2 2 2 2
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P ( A ∩ B) = P (the first bulb in defective and the second bulb is non-defective)
1 1 1
= × = .
2 2 4
1 1 1
P ( B ∩ C ) = P(both the bulbs are non-defective) = × =
2 2 4
1 1 1
P ( C ∩ A ) = P (both the bulbs are defective) = × =
2 2 4
P ( A ∩ B ∩ C ) = P (φ ) = 0
As P ( A ∩ B ) = P(A).P(B), P ( B ∩ C ) = P ( B ).P ( C ) and P ( C ∩ A ) = P(C).P(A) , the
events
A, B and C are pairwise independent.
Since P ( A ∩ B ∩ C ) = 0 ≠ P(A).P(B).P(C) , A, B and C are not mutually independent.
If an event is independent of itself, then identify the event. Think about it mathematically as
well as logically also.
8. TOTAL PROBABILITY THEOREM
Let E1, E2, ..., En be n mutually exclusive and exhaustive events and event A is such that it can occur
with any of the events E1, E2, E3, ....... En then the probability of the occurrence of event A can be
given as P(A) = P(A ∩ E1 ) + P(A ∩ E 2 ) + .......... + P(A ∩ E n )
∴ P(A) = P(E1 ) ⋅ P A + P(E 2 ) ⋅ P A + ........... + P(E n ) ⋅ P A or ;
E1 E2 En
n
P(A) = ∑ P(E )P A E
i =1
i i
Illustration 13: In a certain city only 2 newspapers A and B are published. It is known that 25% of the
city population reads A and 20% reads B while 8% reads both A and B. It is also
known that 30% of those who read A but not B look into advertisement and 40% of
those who read B but not A look into advertisements while 50% of those who read both
A and B look into advertisements. What is the percentage of the population who read
an advertisement?
Solution : Let P(A) and P(B) be the percentage of the population in a city who read newspapers A and
B respectively.
25 1 , P(B) = 20 = 1 and 8 2
∴ P(A) = = P(A ∩ B) = =
100 4 100 5 100 25
∴ Percentage of those who read A but not B
1 2 17
= P(A ∩ B) = P(A) − P(A ∩ B) = − = = 17%
4 25 100
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Similarly,
3
P(B ∩ A) = P(B) − P(A ∩ B) = = 12%
25
∴ Percentage of those who read advertisements
= 30% of P(A ∩ B) + 40% of P(B ∩ A) + 50% of P(A ∩ B)
30 17 40 3 50 2 139
= × + × × × = = 13.9%
100 100 100 25 100 25 1000
Hence the percentage of the population who read an advertisement is 13.9%.
Illustration 14: Two sets of candidates are competing for the positions on the board of directors of a
company. The probabilities that the first and second sets will win are 0.6 and 0.4
respectively. If the first set wins, the probability of introducing a new product is 0.8,
and the corresponding probability, if the second set wins is 0.3. What is the probability
that the new product will be introduced?
Solution : Let A1 (A2) denotes the event that first (second) set wins and let B be the event that a new
product is introduced.
∴ P(A1) = 0.6, P(A2) = 0.4
B B
P = 0.8, P = 0.3
A1 A2
B B
P(B) = P ( B ∩ A1 ) + P ( B ∩ A 2 ) = P ( A1 ).P + P (A2 ) P .
A1 A2
= 0.6 × 0.8 + 0.4 × 0.3 = 0.6 .
Illustration 15:In a multiple choice question there are four alternative answers of which one or more
than one is correct. A candidate will get marks on the question only if he ticks the correct
answers. The candidate decides to tick answers at random. If he is allowed up to three
chances to answer the question, find the probability that he will get marks on it.
Solution : The total number of ways of ticking one or more alternatives out of 4 is
4
C1 + 4 C2 + 4 C3 + 4 C4 = 15 .
Out of these 15 combinations only one combination is correct. The probability of ticking the
1 14 1 1
alternative correctly at the first trial is that at the second trial is = and that
15 15 14 15
14 13 1 1
at the third trial is = . Thus the probability that the candidate will get
15 14 13 15
1 1 1 1
marks on the question if he is allowed upto three trials is + + = .
15 15 15 5
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Illustration 16: Three groups A, B and C are competing for positions on the Board of Directors of a
company. The probabilities of their winning are 0.5, 0.3, 0.2 respectively. If the Group
A wins, the probability of introducing a new product is 0.7 and the corresponding
probabilities for groups B and C are 0.6 and 0.5 respectively. Find the probability that
the new product will be introduced.
Solution : Let A, B, C denote the events of getting positions on the Board of directors by groups A, B
and C respectively.
Since P(A) + P(B) + P(C) = 0.5 + 0.3 + 0.2 = 1, so the events A, B and C are exhaustive.
Let the events of introducing the new product be denoted by D.
Then given that
P(D/A) = 0.7, P(D/B) = 0.6 and P(D/C) = 0.5.
P(D) = P(D ∩ A) + P(D ∩ B) + P(D ∩ C)
= P(A) P(D/A) + P(B) P(D/B) + P(C) P(D/C)
= 0.5 × 0.7 + 0.3 × 0.6 + 0.2 × 0.5
Illustration 17: A speaks truth in 75% of the cases and B in 80% of the cases. In what percentage of
cases are they likely to contradict each other in stating the same fact?
Solution : Let E1 denote the event that A speaks the truth and E2 the event that B speaks the truth,
Then
75 3 3 1
P(E1 ) = = so that P(E1 ) = 1 − =
100 4 4 4
80 4 4 1
P(E 2 ) = = so that P(E 2 ) = 1 − =
100 5 5 5
Let E be the event that A and B contradict each other. Obviously A and B will contradict
each other if one of them speaks the truth and the other does not.
∴ E = E1E 2 + E1E 2 when E1 E 2 and E1 E 2 are mutually exclusive events.
Now P(E1 E 2 ) = the probability that A speaks the truth and B does not.
3 1 3
= P(E1). P( E 2 ) = × = .
4 5 20
Similarly,
1 4 1
P(E1 E 2 ) = P( E1 ). P(E2) = × =
4 5 5
Since E1 E 2 and E1 E 2 are mutually exclusive events, ∴ we have
P(E) = P(E1 E 2 + E1 E 2 ) = P(E1 E 2 ) + P(E1 E 2 )
3 1 35
= + = .
20 5 100
Hence in 35% cases A and B will contradict each other.
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EXERCISE – 3
1. There are 2 groups of subjects, one of which consists of 5 science subjects and 3 engineering
subjects and other consists of 3 science and 5 engineering subjects. An unbiased die is cast. If
the number 3 or 5 turns up a subject is selected at random from first group, otherwise the
subject is selected from second group. Find the probability that an engineering subject is selected.
2. An anti aircraft gun can take a maximum of four shots at an enemy plane moving away from it.
The probability of hitting the plane at first, second, third and fourth shotes are 0.4, 0.3, 0.2 and
0.1 respectively. What is the probability that the gun hits the plane.
3. A, B and C in order toss a coin. The first one to throw a head wins. What are their respective
chances of winning? Assume that the game continue indefinitely.
9. BAYE’S THEOREM
Let E1, E2, ..., En be n mutually exclusive and exhaustive events and event A is such that it can occur
with any of the events E1, E2, E3, ....... En then given that event A has already occurred then the
probability that it has occurred with event Ei is actually the conditional probability P E i .
A
E P(Ei ∩ A)
P i = (from conditional probability)
A P(A)
P(A), by total probability theorem, is given as
A A A
P(A) = P(E1 ) P + P(E 2 ) P + ........ + P(E n ) P
E1 E2 En
E P(Ei ∩ A) = P(Ei )P(A / Ei )
So P i = (using multiplication theorem)
A P(A) P(A)
P(E i ) P(A / E i )
= (using total probability theorem)
A A A
P(E1 ) P + P(E 2 ) P + ....... + P(E n ) P
E1 E2 En
E P(E ) P(A / Ei )
∴ P i = n i
A A
∑ P(Ei ) P E
i
i=1
Illustration 18: Box I contains 2 white and 3 red balls and box II contains 4 white and 5 red balls. One
ball is drawn at random from one of the boxes and is found to be red. Find the probability
that it was from box II.
Solution : Let A denote the event that the drawn ball is red
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Let A1 ≡ The event that box I is selected and let A2 ≡ The event that box II is selected
I II
2w 4w
3R 5R
A
P ( A 2 ).P
1 5
.
A2 A2 2 9 25
∴ P = = 1 5 1 3=
A A A . + . 32
P ( A1 ) P + P ( A 2 ).P 2 9 2 5
A1 A2
Illustration 19: Let A and B be two independent witnesses in a case. The probability that A will speak
the truth is x and the probability that B will speak the truth is y. A and B agree in a
xy
certain statement. Show that the probability that the statement is true is .
1 − x − y + 2 xy
Solution : Let E1 be the event that both A and B speak the truth, E2 be the event that both A and B tell
a lie and E be the event that A and B Agree in a certain statement. Let C be the event that
A speaks the truth and D be the event that B speaks the truth.
∴ E1 = C ∩ D and E2 = C′ ∩ D′ . P(E1) = P ( C ∩ D ) = P(C) P(D) = xy and
P(E2) = P (C′ ∩ D′ ) = P ( C′ ) P ( D′ ) = (1 – x) (1 – y) = 1 – x – y + xy
E
Now P = probability that A and B will agree when both of them speak the truth = 1
E1
E
and P E = probability that A and B will agree when both of them tell a lie = 1.
2
E1
Clearly, be the event that the statement is true.
E
E P ( E ).P ( E / E
1 1 ) xy.1 xy
∴ P E = P ( E ).P ( E / E ) + P ( E ) P ( E / E ) = xy /1 1 x y xy .1 = 1 x y 2xy
1
1 1 2 2 +( − − + ) − − +
Illustration 20: In a factory machines A, B, C manufacture 15%, 25%, 60% of the total production of
bolts respectively. Of the bolts manufactured by machine A, 4%; by B, 2% and by C, 3%
are defective respectively. A bolt is drawn at random and is found to be defective. What
is the probability that it was produced by (i) machine A, (ii) machine B, (iii) machine C?
Solution : Let A, B and C be the events of production of bolt on machines A, B and C respectively and
D be the event for a bolt to be defective. Then
15 4
P(A ∩ D) = ×
100 100
25 2
P(B ∩ D) = ×
100 100
60 3
P(C ∩ D) = ×
100 100
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290
∴ P(D) = P(A ∩ D) + P(B ∩ D) + P(C ∩ D) =
100 × 100
We have to find the probabilities
A B C
P , P , P
D D D
15 4
×
A P(A ∩ D) 6
Now P = = 100 100 =
D P(D) 290 29
100 × 100
50
B P(B ∩ D) 100 × 100 5
P = = =
D P(D) 290 29
100 × 100
180
C P(C ∩ D) 100 × 100 18
P = = =
D P(D) 290 29
100 × 100
10. BINOMIAL DISTRIBUTION FOR REPEATED EXPERIMENTS
If the probability of success of an event in one trial is p, and that of its failure is q so that p + q = 1,
then the probability of exactly r successes in n trials of the concerned experiment is
n
Cr pr qn–r. i.e. (r + 1)th term in the expansion of (q + p)n.
Case – I
Probability of success r times out of n total trials
= P(r) = nCr pr qn–r
Case – II
Probability of success at least r times out of total n trials
= P(≥ r) = nCr pr qn–r + nCr+1 pr+1 qn–r–1 + .... + nCn pn
Case – III
Probability of success at most ‘r’ times out of total n tirals
= P(≤ r) = nC0 qn + nC1 p1 qn–1 + nC2 p2 qn–2 + ..... + nCr pr qn–r
Illustration 21: Find the minimum number of tosses of a pair of dice, so that the probability of getting
the sum of the numbers on the dice equal to 7 on atleast one toss, is greater than 0.95.
(Given log102 = 0.3010, log103= 0.4771).
Solution : n(S) = 36
Let E be the event of getting the sum of digits on the dice equal to 7, then n(E) = 6.
6 1 5
P(E) = = = p , then P ( E′ ) = q =
36 6 6
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probability of not throwing the sum 7 in first m trails = qm .
m
5
∴ P(at least one 7 in m throws) = 1 – qm = 1 – .
6
m
5
According to the question 1 − 0.95
6
m
5
⇒ 0.05
6
⇒ m {log10 5 − log10 6} log10 1 − log10 20
∴ m 16.44
Hence, the least number of trails = 17.
Illustration 22: A coin is tossed n times: What is the chance that the head will present itself an odd
number of times?
Solution : In one throw of a coin, the number of possible ways is 2 since either head (H) or tail (T) may
appear. In two throws of a coin, the total no. of ways in 2 × 2 = 22 , since corresponding to
each way of the first coin there are 2 ways of second. Similarly in three throws of a coin, the
number of ways is 23 and thus in n throws, the number of total ways = 2n.
The favourable no. of ways = the no. of ways in which head will occur once or thrice or 5
times, and so on.
= n C1 + n C3 + n C5 ...... = 2n −1
2n −1 1
Hence required probability P = = .
2n 2
Illustration 23: In five throws with a single die, what is the chance of throwing (a) three aces exactly
(b) three aces at least?
Solution : Let p be the chance of throwing an ace and q be the chance of not throwing an ace in a
single throw with one die then p = 1/6 and q = 5/6.
Now the chances of throwing no ace, one ace, two ace etc. in five throws with a single die
are the first, second, third terms etc. in the binomial expansion
(p + q)5 = q 5 + 5 C1q 4 p + 4 C2 q3 p 2 + 5 C3q 2 p3 + 5 C4 q1 p 4 + p5
Hence (a) chance of 3 aces exactly
= 5 C3q 2 p3 = 10(5/6)2 (1/6)3 = 125/3888.
(b) Chance of throwing three aces at least
= 5 C3q 2 p3 + 5 C4 q p4 + p5
= 10(5/6)2 (1/6)3 + 5(5/6) (1/6)4 + (1/6)5
= (1/6)5 [250 + 25 + 1] = 23/648.
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EXERCISE – 4
1. Numbers are selected at random, one at a time, from the two digit numbers
00, 01, 02, . . . , 99 with replacement. An event E occurs if and only if the product of the two
digits of a selected number is 18. If four numbers are selected, find the probability that the event
E occurs at least 3 times.
2. Find the least number of times must a fair die be tossed, in order to have a probability of at least
91
, of getting atleast one six.
216
3. A die is thrown 7 times. What is the probability that an odd number turns up (i) exactly 4 times
(ii) atleast 4 times.
11. GEOMETRICAL PROBABILITY
If the number of points in the sample space is infinite, then we can not apply the classical definition
of probability. For instance, if we are interested to find the probability that a point selected at random
in a circle of radius r, is nearer to the centre then the circumference, we can not apply the classical
definition of probability. In this case we define the probability as follows :
Measure of the favourable region
P= ,
Measure of the sample space
where measure stands for length, area or volume depending upon whether S is one-dimensional, two
– dimensional or three – dimensional region.
Thus the probability, that the chosen point is nearer to the centre than the circumference =
area of the region of the favourable point π ( r / 2 )
2
1
= = = .
area of the circle πr 2 4
Illustration 24: Two numbers x ∈ R and y ∈ R are selected such that x ∈ [0, 4 ] and y ∈ [0, 4 ] .
Find the probability that the selected numbers satisfy y 2 ≤ x .
Solution : n(S) = area of the square OABC = 4 × 4 = 16 .
The selected numbers x, y will satisfy y 2 ≤ x if the point (x, y) is an interior point of the
parabola y2 = x.
∴ n(E) = The area of that portion of the square Y
which falls in the interior of the parabola y2 = x y=4
C B (4, 4)
(0, 4)
4
16 .
= ∫ xdx =
3 x=0 x=4
0
n (E ) 16 / 3 1 X
= = . O y = 0 A (4, 0)
Required probability =
n (S ) 16 3
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Illustration 25:Two point P, Q are taken at random on a straight line OA of lenght a, show that the
a−b
2
chance that PQ b, where b a is .
a
Solution : The points are as likely to fall in the order O, A, C, P as in the order O, C, A, P. We may
therefore suppose that C is to the right of A.
Draw OP′ at right angles to OP and equal to it. Complete the figure as in the diagram, where
OL = AB′ = b
If δx is small, the number of cases in which the distance of A from O lies between x and
x + δx and C is in AP, is represented by δx . AP i.e. by the area of the shaded rectangle.
Of these, the favourable cases are those in which C lies in BP, and their number is represented
by the upper part of the shaded rectangle cut off by LM. Hence the total number of cases is
represented by area of the triangle of OPP′ , and the total number of favourable cases by the
of the triangle LMP ′ ,
∆ LMP′ a − b
2
∴ the required chance = =
∆ OPP′ a
■■
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SOLVED EXAMPLES
SECTION - I
SUBJECTIVE QUESTIONS
Problem 1: A coin is tossed m + n times (m n). Show that the probability of at least m consecutive
n+2
heads come up is .
2m+ 2
Solution : Let H, T and S be the events “head turns up”, “tail turns up” and “head or tail turns up”
1
Then P(H) = P(T) = and P(S) = 1
2
Since the given event is “at least m consecutive heads turn up”, therefore in any favorable
out come there are m consecutive heads and the rest are any of head or tail
Consider the events
1 n 1
A1 = H, H, H,. . .,$ ,S,S,S,. . .,S
! # ! $
H
# with P(A1) = .1 = m
m times n times
2 m
2
1 1 n −1 1
A2 = T, H, H, H,. . .,$ ,S,S,S,. . .,S with P(A2) = . m .1 = m+1
! # ! $
H
#
m times n −1times
2 2 2
1 1 n −2 1
A3 = S,T, H, H, H,. . .,$ ,S,S,S,. . .,S with P(A3) = 1. . m .1
! # ! $
H
# =
m +1
m times n − 2 times
2 2 2
1 1 1
. . . An + 1 = S,S,S,. . .,S , T, H, H, H,. . .,$ With P(An + 1) = 1n – 1. . m = m+1
!#$ ! #
H
n −1 times
m times
2 2 2
The given event is A1 ∪ A 2 ∪ A 3 ∪ A n +1 . As A1, A2, A3, . . ., An + 1 are pair – wise
mutually exclusive.
The required probability
1 1 1 1
= P(A1) + P(A2) + P(A3) + . . . + P(An + 1) = m
+ m+1 + m+1 + . . . + m+1
2 2 2
! # 2 $
1 n 2+n n − times
= + m+1 = m+1 .
2m 2 2
21
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Mathematics : Probability
Problem 2: There are four six faced dice such that each of two dice bears the numbers 0, 1, 2, 3, 4 and
5 and the other two dice are ordinary dice bearing numbers 1, 2, 3, 4, 5 and 6. If all the four
dice are thrown, find the probability that the total of numbers coming up on all the dice is 10.
Solution : Total number of sample points in the sample space = 64 = 1296
Number of sample points in favour of the event
= Coefficient of x10 in the expansion of (1 + x + x2 + . . . + x5)2 (x + x2 + . . . + x6)2
= Coefficient of x10 in the expansion of x2(1 + x + x2 + . . . + x5)4
= Coefficient of x8 in the expansion of (1 + x + x2 + . . . + x5)4
4
1− x6
= Coefficient of x in the expansionof
8
1− x
= Coefficient of x8 in the expansion of (1 – x6)4 (1 – x)–4
4× 5 2 4× 5× 6 3
= Coefficient of x8 in the expansion of (1 – 4x6) 1 + 4x + x + x + ...
2! 3!
= 1 ×11C8 − 4 ×5 C2 = 125 .
125
∴ Required probability = .
1296
Problem 3: If m things are distributed among ‘a’ men and ‘b’ women, show that the probability that the
1 (b + a ) − (b − a )
m m
number of things received by men is odd, is .
2
( b + a )m
a
Solution : A particular thing is received by a man with probability p = and by a woman with
a+b
b
probability q = . If distributing a single object is an experiment, then this experiment is
a+b
repeated m time. The required probability = m C1 .p. qm – 1 + mC3.p3.qm – 3+mC5.p5.qm – 5 + . . .
1 (b + a ) − (b − a )
m
1 b−a
m m
(q + p )m − (q − p )m
= = 2 1 − b + a = .
2
2
( b + a )m
Problem 4: Let p be the probability that a man aged x years will die within a year. Let A1, A2, . . . , An
be n men each aged x years. Find the probability that out of these n men A1 will die with in
a year and is first to die.
Solution : P(no one among A1, A2 . . . , An dies within a year) = (1 – p)n
P (at least one among A1, A2, . . ., An dies within a year) = 1 – (1 – p)n
1
1 − (1 − p ) .
n
P(A1dies within a year and is first to die) =
n
22
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Problem 5 : A box contains 2 fifty paise coins, 5 twenty five paise coins and a certain fixed number
N(≥ 2) of ten and five paise coins. Five the probability that the total value of these 5 coins is
less than one rupee and fifty paise.
Solution : Total number of ways of drawing 5 coins from
N + 7 coins = N + 7C5. ...(i)
Let E be the event that the value of 5 coins is less one rupee fifty paisa. Then E '(value of 5
coins ≥ 1.50) has the following cases
50 p. 25 p. 5 or 10 p. ways
(2) (5) (N 2)
I 2 3 – 1. 5C 3
5
II 2 2 1 C 3. NC 1
2
III 1 4 – C 1.5C 4
Total ways = 20 + 10N = 10(N + 2) ...(ii)
10(N + 2)
∴ P(E) = 1 − ( N + 7)
C5
Problem 6: 5 girls and 10 boys sit at random in a row having 15 chairs numbered as 1 to 15, Find the
probability that end seats are occupied by the girls and between any two girls odd number of
boys sit.
Solution : There are four gaps in between the girls where the boys can sit. Let the number of boys in
these gaps be 2a + 1, 2b + 1, 2c + 1, 2d + 1, then
2a + 1 + 2b + 1 + 2c + 1 + 2d + 1 = 10
or, a+b+c+d=3
The number of solutions of above equation
= coefficient of x3 in (1 – x)–4 = 6C3 = 20
Thus boys and girls can sit in 20 × 10! × 5 ways.
20 × 10! × 5!
Hence the required probability =
15!
Problem 7: A chess game between two grandmasters A and B is won by whoever first wins A total of
2 games. A’s chances of winning, drawing or losing a particular game are p, q and r,
respectively. The games are independent and p + q + r = 1. Show that the probability that A
wins the match after (n + 1) games (n ≥ 1) is
p2[nqn – 1 + n(n – 1) rqn – 2]
Use this result to show that the probability that A wins the match is
p 2 (p + 3r)
(p + r)3
Find the probability that there is no winner.
Solution : A can win the match after (n + 1) games (n ≥ 1) in the following two mutually exclusive ways:
(i) A wins exactly one of the first n games and draws the remaining (n – 1), or
23
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Mathematics : Probability
(ii) A wins exactly one of the first n games, loses exactly one of the first n games and draws
the remaining (n – 2).
We have P(i) = nP1pqn – 1 and P(ii) = nP2pqn – 2r. Thus, the probability that A wins at the end
of the (n + 1)th game is
p(nP1pqn – 1 + nP2pqn – 2r) = p2[nqn – 1 + n(n – 1) rqn – 2]
The probability that A wins the match is
∞ ∞ ∞
∑ p2 [nq n −1 + n(n − 1)rq n −2 ] = p2
n =1
∑ nqn −1 + p2 r
n =1
∑ n(n − 1)q
n =1
n−2
We know that
∞
1
∑q
n =0
n
=
1− q
(0 p 1)
Differentiating both sides w.r.t q, we get
∞ ∞
1 2
∑ nq
n =1
n −1
=
(1 − q)2
and ∑ n(n − 1)q
n =1
n −2
=
(1 − q)3
Thus, the probability that A wins the match is
1 2p 2 r p2 2p 2 r
p2 . + = + [∵ p + q + r = 1]
(1 − q)2 (1 − q)3 (p + r) 2 (p + r)3
p 2 (p + r) + 2p 2 r p 2 (p + 3r)
= =
(p + r)3 (p + r)3
Similarly, the probability that B wins the match is
r 2 (r + 3p)
(r + p)3
p 2 (p + 3r) r 2 (r + 3p)
∴ P(A wins) + P(B wins) = +
(p + r)3 (r + p)3
p3 + 3p 2 r + 3pr 2 + r 3 (p + r)3
= = =1
(p + r)3 (p + r)3
Problem 8 : (i) If four squares are chosen at random on a chess board, find the probability that they lie
on a diagonal line.
(ii) If two squares are chosen at random on a chess board, what is the probability that they
have exactly one corner in common?
(iii) If nine squares are chosen at random on a chess board, what is the probability that they
form a square of size 3 × 3 ?
Solution : (i) Total no. of ways = 64C4
The chess board can be divided into two parts by a diagonal line BD. Now, if we begin
to select four squares from the diagonal P1Q1, P2Q2,...., BD, then we can find no. of
squares selected
= 2(4C4 + 5C4 + 6C4) = 182
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and similarly no. of squares for the diagonals chosen parallel to AC = 182
∴ Total favourable ways = 364
364
∴ required probability = 64
C4
(ii) Total ways = 64 × 63
Now if first square is in one of the four corners, the second square can be chosen in just
one way = (4) (1) = 4
If the first square is one of the 24 non-corner squares along the sides of the chess
board, the second square can be chosen in two ways = (24) (2) = 48.
Now, if the first square is amy of the 36 remaining squares, the second square can be
chosen in four ways = (36) (4) = 144
∴ favourable ways = 4 + 48 + 144 = 196
196 7
∴ required probability = =
64 × 63 144
(iii) Total ways = 64C9
A chess board has 9 horizontal and 9 vertical lines. We see that a square of size 3 × 3
can be formed by choosing four consecutive horizontal and vertical lines.
Hence favourable ways = (6C1) (6C1) = 36
36
∴ Required probability = 64
C9
Problem 9: If n positive integers taken at random are multiplied together, show that the probability that
5n − 4 n
the last digit of the product is 5 is
10n
10n − 8n − 5n + 4n
and that the probability of the last digit being 0 is
10n
Solution : Let n positive integers be x1, x2, ...., xn. Let a = x1.x2....xn.
Let S be the sample space, since the last digit in each of the numbers, x1, x2,..., xn can be any
one of the digits 0, 1, 3, ..., 9 (total 10)
∴ n(S) = 10n
Let E2 and E2 be the events when the last digit in a is 1, 3, 5, 7 or 9 and 1, 3, 7 or 9
respectively
∴ n(E1) = 5n and n(E2) = 4n
and let E be the event that the last digit in a is 5.
∴ n(E) = n(E1) – n(E2) = 5n – 4n
Hence required probability
n(E1 ) 5n − 4n
P(E) = =
n(S) 10n
25
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