# Solution of $x^2 + s(x)\cdot x - n = 0$, with $s(x)$ is the sum of digits of $x$.

This problem comes from an programming competition website, but I'd interested in analyze it from mathematics prespective. Given this problem below, we must create a program that could give us the correct output from the given input. The program must run under the time limit that provided by the system.

## The Problem

Let define function $s$ that take input from positive integers, such that $s(x)$ is the sum of the digit of $x$.

For example, $$s(1024) = 1 + 0 + 2 + 4 = 7$$

Given some integer constant $n$ as the input of the program. Now, find positive integer solution of $x$ in equation

$$x^2 + s(x) \cdot x - n = 0.$$

If more than one solution exist, choose the lowest one. If the solution doesn't exist, give output -1.

Time Limit of the program: 2 second.

Range of the input $n$: $[1, 10^{18}]$

## The Solution

In finding the solution, we could iterate from some range of possible value of x, and try each of these value whether it satisfy the equation or not.

To improve the run time of the program, we must choose the shortest range for x.

## My Question

1. What is the shortest range of possible value of x that we could iterate to find the answer of the problem above?

2. Is there a better way to find the solution without iterating from the set possible value of x?

## My Attempt:

Consider that $$x^2 + s(x) \cdot x - n = 0$$ $$\Leftrightarrow x ( x + s(x) ) = n$$

and since both $x$ and $(x + s(x))$ are positive integers and the product of both number is $n$, we can conclude that $x$ and $(x + s(x))$ is a pair of positive factor of $n$.

Since $x$ is a factor of $n$, we now have the first possible range of $x$, that is $[1, n]$.

But, iterating $x$ over this range gives bad performance to the program. For small value of $n$, it won't matter, but for $n \geq 10^{17}$, the program will reach it's time limit.

If I could recall correctly, we have this theorem

$$x_1 \cdot x_2 = x$$

and

$$x_1 \leq x_2$$

implies that

$$x_1 \leq \sqrt{x} \leq x_2$$

And since $x (x + s(x)) = n$ and $x \leq x + s(x)$

we have $x \leq \sqrt{n} \leq x + s(x)$

Now we have the second range: $[1, \sqrt{n}]$

But, still, this range also give bad performance for larger $n$.

My last attempt is by considering the biggest possible value of $s(x)$, that is the digit of $x$ is all $9$ and $x$ less than or equals to the maximum value of $n$, $10^{18}$.

$$s(x) \leq s(999,999,99 \cdots 9,999) = 9 \cdot 18 = 162$$

Hence, we have

$$x + s(x) \leq x + 162$$

so that, $$\sqrt{n} \leq x + s(x) \leq x + 162 \leq \sqrt{n} + 162$$ hence, $$\sqrt{n} - 162 \leq x \leq \sqrt{n}$$

Now, we have better range that will suffice for all value of $x$. $$l_1 = \max \{ 1, \sqrt{n} - 162 \}$$ $$l_2 = \sqrt{n}$$ the range is $[l_1, l_2]$

But, I though there must be a better solution to this problem. Hence I try $$l_1 = \max \{ 1, \sqrt{n} - 162/2 \}$$ And that range is also correct. But I can't justify my finding.

So, hence my third question:

1. Why $l_1 = \max \{ 1, \sqrt{n} - 162/2 \}$ also gives us correct solution?
-

While not a full answer to your questions, here is a trick that will allow you to reduce the collection of numbers you have to try (and so an answer to question 2).

Because $10\equiv 1 \pmod 9$, we have $10^k \equiv 1 \pmod 9$ for every $k\in \mathbb N$. If we write a number in terms of its digits, $x=\sum a_k 10^k$, then since $\sum a_k 10^k \equiv \sum a_k \pmod 9$, we have $s(x)\equiv x \pmod 9$. This fact can be used to give a way to check arithmetic, among other things.

Therefore, if $x^2 + s(x)x = n$, then $2x^2 \equiv n \pmod 9$. Since $(2)(5)=10 \equiv 1 \pmod 9$, we can multiply to get $x^2 \equiv 5n \pmod 9$.

The squares modulo 9 are $0, 1, 4, 7$, so unless the remainder of dividing $5n$ by $9$ is one of these numbers, there are NO solutions.

If the remainder is one of those numbers, you need to know the square roots modulo 9 to know what values of $x$ to try.

For this, we have the following.

$0^2\equiv 3^2 \equiv 6^2 \equiv 9 \pmod 9$, $1^2\equiv 8^2 \equiv 1 \pmod 9$, $2^2\equiv 7^2 \equiv 4 \pmod 9$ and $4^2 \equiv 5^2 \equiv 7 \pmod 9$.

The upshot of this is that if $n$ is a multiple of $9$, then you only have to check the multiples of $3$ (reducing the number of things you have to check by a factor of $3$). Otherwise, depending on the modulus of $5n\bmod 9$, you either have to check NO numbers (5 of the remaining cases), or you have to check $2/9$ of the numbers (3 of the remaining cases). Thus, whatever range you are working in, you have at least a speed up by a factor of $3$, if not more.

Another comment, which gives a small speedup over what you already have but answers question 3. As you observed, $x<\sqrt n$, and the number of digits of $x$ is at most $\lceil \log_{10} x \rceil$, and since $s(x)$ is at most $9$ times the number of digits of $x$, we have $s(x)<9 \lceil \log_{10} \sqrt n \rceil=9\lceil \frac{1}{2}\log_{10} n \rceil$.

Therefore, given $n$, you can limit your search for $x$ to $[\sqrt n - 9\lceil \frac{1}{2}\log_{10} n \rceil, \sqrt n]$. The factor of $1/2$ here answers question 3. Note, we could probably shave off $1$ or $2$ numbers from the range by being more careful with rounding, but I don't see any obvious reason we could shave off more. If we combine this with the beginning of my answer, you have to check at most $3\lceil \frac{1}{2}\log_{10} n \rceil$ numbers to find your solution, if it exists, and no numbers a good portion of the time when no solution exists.

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I think $(2)(5)=10 \equiv 1 \pmod 5$ in pharagraph 3 should be $(2)(5)=10 \equiv 1 \pmod 9$. –  Kamal Hajjaj Isa Oct 28 '12 at 14:58
@KamalHajjajIsa: Yes, that is correct. Thank you. –  Aaron Oct 29 '12 at 3:07
I like the Idea of using $\pmod 9$. –  Kamal Hajjaj Isa Oct 29 '12 at 4:56

Since $x<\sqrt n$ and $n\le10^{18}$, we have that $x<10^9$. Then $x$ has at most $9$ digits and $s(x)\le9\times9=81$. This implies that $x^2+81\,x-n\ge0$. It follows easily that we must have $$x\ge\frac{-81+\sqrt{81^2+4\,n}}{2}=\sqrt{n+(81/2)^2}-81/2.$$

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This only answers the third question

If you use the general quadratic equation formula for the equation $ax^2+bx+c=0$ $$\displaystyle x= \frac {-b \pm \sqrt{b^2-4ac}}{2}$$ It follows from $x^2 + s(x)\cdot x - n = 0$ that $$\displaystyle x= \frac {-s(x) \pm \sqrt{s^2(x)+4n}}{2}$$ Since all the variables are positive and obviously $\sqrt{s^2(x)+4n}>s(x)$ , one can remove the negative sign and have $$\displaystyle x= \frac {-s(x) + \sqrt{s^2(x)+4n}}{2}$$ If one extends the range of values of $n$ to $0$, we have that the lowest possible value of $s(x)$ is $0$ and that $s(x) \ge 0$ which implies that $$\displaystyle x \ge \frac {-s(x) + \sqrt{4n}}{2} = \frac {-s(x)}{2} + \frac {2\sqrt{n}}{2} \ge \frac {-162}{2} + \sqrt n, \space \space (-s(x) \ge -162)$$ and there you have your inequality $$x \ge \sqrt n - \frac {162}{2}$$

Extending the range to $0$ doesn't change the validity of the inequality and helps to simplify it. You've been working with the maximum value of $s(x)$, using the minimum value helps in this case.

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Beside the excellent modulo 9 tricks Aaron gave you, if $n$ passes his tests the easiest thing is to try the range of $1 \le s(x) \le 162$ with the proper $s(x) \pmod 9$. Just plug each one into the quadratic formula and solve for $x$. If it doesn't come out integral, go on to the next. If it does, check the sum of digits against the assumed $s(x)$. If it matches, success and report it. If not, go on to the next. At most $50$ numbers to try, which should fit in 2 seconds easily.


# Probability Problems And Solutions

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P (A n B) = A and B = A x B P (A u B) = A or B = A + B. Probability MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. Feel Free to Learn 316,387 views. Probability puzzles require you to weigh all the possibilities and pick the most likely outcome. PROBABILITY AND STATISTICS FOR ENGINEERS LESSON INSTRUCTIONS The lecture notes are divided into chapters. Review of probability theory • Definitions (informal) -Probabilities are numbers assigned to events that indicate "how likely" it is that the event will occur when a random experiment is performed -A probability law for a random experiment is a rule that assigns probabilities to the events in the experiment. Chapter 4 - Continuous Random Variables and Probability Distributions. This section provides materials for a lecture on discrete random variables, probability mass functions, and expectations. Multiplication Rule of Probability The addition rule helped us solve problems when we performed one task and wanted to know the probability of two things happening during that task. The bet can either be on red or on blue and the amount of each bet is arbitrary. Q2) Generate a Punnett Square for a heterozygous individual crossed with a heterozygous individual. What Are Example Statistics and Probability Problems and Their Solutions? Credit: Echo/Cultura/Getty Images Two examples of probability and statistics problems include finding the probability of outcomes from a single dice roll and the mean of outcomes from a series of dice rolls. This is, however, wrong, because given that heads came up, it is more likely that the two-headed coin was chosen. Practice calculating conditional probability, that is, the probability that one event occurs given that another event has also occurred. 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It takes a computational approach, which has several advantages: • Students write programs as a way of developing and testing their un-derstanding. Chapter 1 covers this theory at a fairly rapid pace. Find the probability that the number on the drawn ticket is a multiple of 3 or 7. We break down problems on video in a step-by-step easy to follow format. Finding the Median in a pdf : S2 Edexcel. The probability of two or more heads is, therefore: Probability = 4 8 = 1 2 We solved this problem by ﬁrst enumerating the set of possibl e outcomes, known as the Sample Space, and then by deciding which of these outcomes satisﬁed the cr iterion of containing 'two or more heads'. Certainly X is binomial with n = 40 and p = 0. More Problems on probability and statistics are presented. P(X) gives the probability of successes in n binomial trials. Example of Using a Contingency Table to Determine Probability. Problem Solving: Find a Pattern What Is It? Finding a Pattern is a strategy in which students look for patterns in the data in order to solve the problem. Generate integer from 1-8 with equal probability. Problem 4 (Solution on p. Topics covered range from measure and integration theory to functional analysis and basic concepts of probability; the interplay between measure theory and topology; conditional probability and expectation; the central limit theorem; and strong laws of large numbers in terms of martingale theory. Calculate the value of a in a normal distribution with a mean of 4 and a standard deviation of 2 for which:. Solution : Sample space : {(1, 1)(1, 2)(1, 3)(1. Review of probability theory • Definitions (informal) -Probabilities are numbers assigned to events that indicate "how likely" it is that the event will occur when a random experiment is performed -A probability law for a random experiment is a rule that assigns probabilities to the events in the experiment. The actual probability of finding the particle is given by the product of the wavefunction with its complex conjugate (like the square of the amplitude for a complex function). Pedigrees and probability problems 1-28 and 1-30. It includes the list of lecture topics, lecture video, lecture slides, readings, recitation problems, recitation help videos, tutorials with solutions, and a problem set with solutions. Chapter 1 covers this theory at a fairly rapid pace. Solution; Determine the value of $$c$$ for which the function below will be a probability density function. In a binomial distribution the probabilities of interest are those of receiving a certain number of successes, r, in n independent trials each having only two possible outcomes and the same probability, p, of success. a Show that $$f\left( x \right)$$ is a probability density function. the number of ways several people can stand in a line if 2 particular people must — or must not — stand next to each other). (321 problems) IMO Shortlisted Problems. Probability of choosing 1 icecream out of a total of 6 = 4/6 = 2/3. Playing cards probability problems based on a well-shuffled deck of 52 cards. Chegg Solution Manuals are written by vetted Chegg Statistics And Probability experts, and rated by students - so you know you're getting high quality answers. Probability puzzles require you to weigh all the possibilities and pick the most likely outcome. The post in the companion blog shows how to evaluate the covariance and the correlation coefficient of two continuous random variables and. Importantly, the solution for the problem of points became known as an expectation value  (the average expected value). Suppose that a good part fails within the ﬁrst year with probability 0. This is essentially a 1D scattering problem. At a picnic, Julio reaches into an ice-filled cooler containin…. 1 Problem 4E. Drawing/Picking/Choosing Single/One ball from a bag/urn/box - Probability - Problems Solutions. Brilliant Premium. Probability questions pop up all the time. Fully solved Aptitude questions with detailed answer description and explanation. Practice Problems - Try specific problems and see the solution. IB Math - Standard Level - Probability Practice - MarkScheme Alei - Desert Academy C:\Users\Bob\Documents\Dropbox\Desert\SL\6StatProb\TestsQuizzesPractice\SLProbPractice1. Probability of problem getting solved = 1 - (5/7) x (3/7) x (5/9) = (122/147) Example 9: Find the probability of getting two heads when five coins are tossed. Problem 1. ECE302 Spring 2006 HW5 Solutions February 21, 2006 3 Problem 3. The higher the probability of an event, the more certain that the event will occur. Enter r, the number of items selected from the set, and press [ENTER] to display the result. Playing cards probability problems based on a well-shuffled deck of 52 cards. also known as the birthday problem, states that in a random group of 23 people, there is about a 50 percent chance that two people have the same birthday. Test your understanding with practice problems and step-by-step solutions. Learn and practice basic word and conditional probability aptitude questions with shortcuts, useful tips to solve easily in exams. The mathematics field of probability has its own rules, definitions, and laws, which you. Conditional Probability: It is known that a student who does his online homework on aregular basishas a chance of83 percentto get a good grade (A or B); but the chance drops to58 percentif he doesn't do the homework regularly. An agent sells life insurance policies to five equally aged, healthy people. Students were encouraged to prepare a 4x6 inch notecard to use for reference during each exam. Behind one is a expensive car, but behind the other two are goats. when the selection of one does not affect the probability of a…. Importantly, the solution for the problem of points became known as an expectation value  (the average expected value). " Understanding how to calculate these percentages with real numbers of people and things. Basic concept on drawing a card: In a pack or deck of 52 playing cards, they are divided into 4 suits of 13 cards each i. and define the function. Now practice our shortcut tricks on probability problem on coin and read examples. Drawing/Picking/Choosing Single/One ball from a bag/urn/box - Probability - Problems Solutions. Homework 2 from last time OLD HW2 solutions Practice midterm and Midterm 2015 solutions. The text-books listed below will be useful for other courses on probability and statistics. Cards of Spades and clubs are black cards. (See diagram. 1) Basics of Electromagnetics Part I – Download MCQs from here. Solution the the Second WBC Problem-page 10. Multiplicative Principle for Counting n The total number of outcomes is the product of the possible outcomes at each step in the sequence n if a is selected from A, and b selected from B… n n (a,b) = n(A) x n(B) q (this assumes that each outcome has no influence on the next outcome) n How many possible three letter words are there? q you can choose 26 letters for each of the three positions, so. GMAT Advanced Probability Problems By Mike MᶜGarry on January 3, 2014 , UPDATED ON January 15, 2020, in GMAT Math In the following probability problems, problems #1-3 function as a set, problems #4-5 are another set, and problems #6-7 are yet another set. by electronjohn. Homework 2 from last time OLD HW2 solutions Practice midterm and Midterm 2015 solutions. Solution; Determine the value of $$c$$ for which the function below will be a probability density function. Test your understanding with practice problems and step-by-step solutions. Probability theory is the most directly relevant mathematical background, and it is assumed that the reader has a working knowledge of measure-theory-based probability theory. Midterm 2016 solutions Combinatorics practice problem set and solutions. Bayes' theorem describes the probability of occurrence of an event related to any condition. For instance, let’s say the probability that you will get to class on time is 2/3. In non-technical parlance, "likelihood" is usually a synonym for "probability," but in statistical usage there is a clear distinction in perspective: the number that is the probability of some observed outcomes given a set of parameter values is regarded as the likelihood of the set of parameter values given the. Finding the Median in a pdf : S2 Edexcel. Thanks for the A2A. At a picnic, Julio reaches into an ice-filled cooler containin…. Three circle Venn Diagrams are a step up in complexity from two circle diagrams. 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So how will you make the determination of how to solve a conditional probability problem? As a general rule, problems where information is given in a table are best solved by using the formula. The point is that the order of events doesn't affect with respect to conditional probability. After solving all ten math questions write down total time taken by you to solve those questions. Notice that the phenotypic ratio for dominant/recessive is identical to Q1 (3/4 Dominant, 1/4 recessive) except you now have to realize your mutation is Dominant and wild type is recessive. Midterm 2016 solutions Combinatorics practice problem set and solutions. Calculate the sample mean. P(F AND P) = d. Sarah has two children and we know that she has a daughter. Playing cards probability problems based on a well-shuffled deck of 52 cards. many new problems and solutions, a number of improvements in the pre-sentation, and corrections. PROBABILITY AND STATISTICS FOR ENGINEERS LESSON INSTRUCTIONS The lecture notes are divided into chapters. Probability theory - Probability theory - The birthday problem: An entertaining example is to determine the probability that in a randomly selected group of n people at least two have the same birthday. Probability is the chance that an event will occur. Solution: (F) Probability always lie within 0 to 1. Probability_Problems Based on Coins#LESSON-2 - Duration: 17:39. If we toss a fair die, what. $$∴ P(E)={n(E)\over n(S)} = {2\over52}={1\over26}$$ Learn and practice these probability problems with solutions and you can ask me anything in the comment section if you have any doubts regarding probability problems and it's solutions. To get a feeling for PDF, consider a continuous random variable. To link to this page, copy the following code to your site:. The text-books listed below will be useful for other courses on probability and statistics. Topics covered range from measure and integration theory to functional analysis and basic concepts of probability; the interplay between measure theory and topology; conditional probability and expectation; the central limit theorem; and strong laws of large numbers in terms of martingale theory. [Syllabus] [Textbooks] [Grading] [Reading assignment] [Homework problems and solutions] [Tests] [Knowledge Milestones]. to access the Math Probability menu. Finding the Median in a pdf : S2 Edexcel. Practice Problems - Try specific problems and see the solution. The question is incomplete, that is, it doesn’t provide information regarding with the level of probability that you are familiar with. PROBLEM 13 : Consider a rectangle of perimeter 12 inches. Conroy and a 5=6 chance of not rolling a six, after which the number of rolls we expect to throw is the same as when we started. 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Total Number of ways = 2 5 = 32. The point is that the order of events doesn't affect with respect to conditional probability. MIDTERM 2 solutions and Histogram. The incubation periods of a random sample of 7 HIV infected individuals is given below (in years): 12. Advanced Techniques. A Problem With Pearls. Normal distribution probability : Statistics S1 Edexcel June 2013 Q6 (b) : ExamSolutions - youtube Video. The probability mass function of is but and Therefore, the probability mass function can be written as which is the probability mass function of a Bernoulli random variable. Then, n(E) = 2. Rolling the Dice. The empirical probability of three consecutive female births is. 1, Median 22. ) For a diﬀerent problem, allow every one of n people to place an even bet on the color of his hat. In the Options dialog box, select the Show Iteration Results check box to see the values of each trial solution, and then click OK. Most of the problems require very little mathematical background to solve. Problem Solving: Find a Pattern What Is It? Finding a Pattern is a strategy in which students look for patterns in the data in order to solve the problem. The probability (chance) is a value from the interval 0;1> or in percentage (0% to 100%) expressing the occurrence of some event. P (A n B) = A and B = A x B P (A u B) = A or B = A + B. DISCRETE PROBABILITY DISTRIBUTIONS 27. Miracle Mountain. Geometric Probability. Life or Death? The Emperor's Proposition. We want to generate a random number between 100 and 499. The probability of a success, denoted by p, remains constant from trial to trial and repeated trials are independent. Probability Questions and answers PDF with solutions. Probability MCQ Questions and answers with easy and logical explanations. Please do not email me completed problem sets. Multiplying by multiples of 10. John has a special die that has one side with a six, two sides with twos and three sides with ones. At any particular time period, both outcomes cannot be achieved together so […]. Solution of exercise 1. This is a beta distribution. We want to generate a random number between 100 and 499. The probability equals 31,8 %. The question is incomplete, that is, it doesn’t provide information regarding with the level of probability that you are familiar with. There is one desired outcome and six possible outcomes. random variables, and some notation. Normal distribution probability : Statistics S1 Edexcel June 2013 Q6 (b) : ExamSolutions - youtube Video. 74% of the X values are less than three standard deviations from the mean. (Opens a modal) Random number list to run experiment. Practice Problems - Try specific problems and see the solution. Here is a set of 14 GMAT probability questions, all in the Problem Solving style on the test, collected from a series of blog articles. This book is ideal for an upper-level undergraduate or graduate level introduction to probability for math, science, engineering and. Solution : Sample space : {(1, 1)(1, 2)(1, 3)(1. Probability Exam Questions with Solutions by Henk Tijms1 December 15, 2013 This note gives a large number of exam problems for a ﬁrst course in prob-ability. Each trial results in an outcome that may be classified as a success or a failure (hence the name, binomial);. Playing cards probability problems based on a well-shuffled deck of 52 cards. com: Many students are also using our Free Statistics Lab Manual. I require the probability that after these n stretches he is at a distance between rand r+ rfrom his starting point 0. A Music Survey was carried out to find out what types of music a group of people liked. Assume there is a beam of particles with definite momentum coming in from the left and assume there is no flux of particles coming from the right. One of the famous problems that motivated the beginnings of modern probability theory in the 17th century, it led Blaise Pascal to the first explicit reasoning about what today is known as an expected value. The Probability Distributome Project provides an interactive navigator for traversal, discovery and exploration of probability distribution properties and interrelations. Listed in the following table are practice exam questions and solutions, and the exam questions and solutions. 84 = (x – 70)/2. Let E = event of getting a queen of club or a king of heart. Let Xand Y be two N 0-valued random variables such that X= Y+ Z, where Zis a Bernoulli random variable with parameter p2(0;1), independent of Y. Miracle Mountain. Hone your skills with this array of statistics and probability worksheets for 7th grade to compute the average on whole numbers and decimals, find the probability on a pair of coins, a pair of dice, months and more. Solution of exercise 1. n(S) is the number of elements in the sample space S and n(E) is the number of elements in the event E. Although they never finished their game, their letters contain solutions to both problems. Home > Math> Probability and Statistics>Probability Problems. Homework problems usually do not say which concepts are involved, and often require combining several concepts. Probability problem on Dice. Most probability problems on the GRE involve independent events. Probability MCQ Questions and answers with easy and logical explanations. After you define a problem, click Options in the Solver Parameters dialog box. The problem concerns a game of chance with two players who have. Example: there are 5 marbles in a bag: 4 are. Probability Questions and answers PDF with solutions. In non-technical parlance, "likelihood" is usually a synonym for "probability," but in statistical usage there is a clear distinction in perspective: the number that is the probability of some observed outcomes given a set of parameter values is regarded as the likelihood of the set of parameter values given the. These you may need to complete at home. Statistics Problems With Solutions Return to Statistics Internet Library for videos, software assistance, more problems and review. Cards of Spades and clubs are black cards. pdf] - Read File Online - Report Abuse. A Collection of Dice Problems Matthew M. 316 that an audit of a retail business. The first section provides a brief description of some of the basic probability concepts that will be used in the activities. Trials until first success. Walpole Raymond H. What is the probability of drawing a red Bingo chip at least 3 out of 5 times? Round answer to the nearest hundredth. Probability Problems With Solutions Math Help Fast (from someone who can actually explain it) See the real life story of how a cartoon dude got the better of math Probability Word Problems (Simplifying Math) What are the chances that your name starts with the letter H? Find out how to make that calculation and many more. Advanced Techniques. The problem concerns a game of chance with two players who have. Probability shortcut Tricks Pdf, Probability MCQ, Probability Objective Question & Answer Pdf. [email protected]{"The probability of winning is ", 100 win/(win + lose) // N, " %. Total English alphabets = 26. The flippant juror. For example: if we have to calculate the probability of taking a blue ball from the second bag out of three different bags of balls, where each bag contains three different colour balls viz. a Show that $$f\left( x \right)$$ is a probability density function. Make use of probability calculators to solve the probability problems with ease. What is the probability that players will experience rain and a temperature between 35 and 50 degrees? 0. ’ ‘There’s a 30% chance of rain tomorrow. He offers you the following game. 1) Basics of Electromagnetics Part I – Download MCQs from here. Probability Solution: (a) Since there are 4 queens ( ) = ( ) ( ),. (Solution): Probability Problems. Probability And Statistics Problems Solutions Author: persepolis. It is assessed by considering the event's certainty as 1 and impossibility as 0. edu-2020-04-20T00:00:00+00:01 Subject: Probability And Statistics Problems Solutions Keywords: probability, and, statistics, problems, solutions Created Date: 4/20/2020 7:35:45 AM. nthe sample space for the last experiment would be all the ordered pairs in the form (d,c), where d represents the roll of a die and c represents the flip of a coin. AU - Sparrow, E. 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Let Xand Y be two N 0-valued random variables such that X= Y+ Z, where Zis a Bernoulli random variable with parameter p2(0;1), independent of Y. Linear Algebra Igor Yanovsky, 2005 2 Disclaimer: This handbook is intended to assist graduate students with qualifying examination preparation. Please solve the following probability practice problems: Determine the probability that a digit chosen at random from the digits 1, 2, 3, …12 will be odd. The mathematics field of probability has its own rules, definitions, and laws, which you. The number of candidates for the first, second and third posts are 3,4 and 2 respectively. Let P(D) be the probability that the letter is in DOGS. " Understanding how to calculate these percentages with real numbers of people and things. MIDTERM 2 solutions and Histogram. (b) Show that the result in question (a) is not necessarily correct without. 1 Problem 4E. You need at most one of the three textbooks listed below, but you will need the statistical tables. Fully worked-out solutions of these problems are also given, but of course you should ﬁrst try to solve the problems on your own! c 2013 by Henk Tijms, Vrije University, Amsterdam. Probability of choosing 2nd chocobar = 3/7. In a binomial distribution the probabilities of interest are those of receiving a certain number of successes, r, in n independent trials each having only two possible outcomes and the same probability, p, of success. Objective: I know how to solve probability word problems. Three circle Venn Diagrams are a step up in complexity from two circle diagrams. Q x = P(female lives to age x) = number of female survivors at age x 100,000. Any problems you do not have time to do, try doing later. Bayes’ theorem describes the probability of occurrence of an event related to any condition. Advanced Probability Problems And Solutions Probability Questions with Solutions Tutorial on finding the probability of an event. Two balls are chosen from the jar, with replacement. What is the probability the sum of the dice is 5? 6. · = = Answer: 2: A jar contains 6 red balls, 3 green balls, 5 white balls and 7 yellow balls. Calculate the probability that if somebody is “tall” (meaning taller than 6 ft or whatever), that person must be male. is the constant 2. Probability has its origin in the study of gambling and insurance in the 17th century, and it is now an indispensable tool of both social and natural sciences. ) Find the probability that a person is female or prefers hiking on mountain peaks. == Solutions for Probability and Statistics for Engineering and the Sciences, 7th ed, by Jay L. Important facts and powerful problem solving approaches are highlighted throughout the text. For less than the price of a single session with a private tutor, you can have access to our entire library of videos. Probability MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. Probability Math. NCERT Solutions For Maths Class 12 Chapter 13 - Probability comprises of various important questions for exams. Which pair has equally likely outcomes? List the letters of the two choices below which have equal probabilities of success, separated by a comma. Probability Exam Questions with Solutions by Henk Tijms1 December 15, 2013 This note gives a large number of exam problems for a ﬁrst course in prob-ability. Two events, A and B, are independent if the outcome of A does not affect the outcome of B. So, for example, using a binomial distribution, we can determine the probability of getting 4 heads in 10 coin tosses. P(X) gives the probability of successes in n binomial trials. A Problem With Pearls. MIDTERM 1 solutions and Histogram. For Exercises 3-17 to 3-21, verify that the following functions are probability mass functions, and determine the requested probabilities. As we go deeper into the area of mathematics known as combinatorics, we realize that we come across some large numbers. [Syllabus] [Textbooks] [Grading] [Reading assignment] [Homework problems and solutions] [Tests] [Knowledge Milestones]. A probability distribution is a table or an equation that links each outcome of a statistical experiment with its probability of occurrence. As with any quantum system, the wave functions give the probability amplitude for finding the electron in a particular region of space, and these amplitudes are used to compute actual probabilities associated with measurements of the electron's position. Probability is the chance that an event will occur. Example: the chances of rolling a "4" with a die. Probability Problems and Solutions. Interview Probability Puzzles #1 - Most Popular Monty Hall Brain Teaser Difficulty Popularity The host of a game show, offers the guest a choice of three doors.


A bijective function is both injective and surjective, thus it is (at the very least) injective. And a function is surjective or onto, if for every element in your co-domain-- so let me write it this way, if for every, let's say y, that is a member of my co-domain, there exists-- that's the little shorthand notation for exists --there exists at least one x that's a member of x, such that. If it crosses more than once it is still a valid curve, but is not a function. So we can calculate the range of the sine function, namely the interval $[-1, 1]$, and then define a third function: $$\sin^*: \big[-\frac{\pi}{2}, \frac{\pi}{2}\big] \to [-1, 1]. Stated in concise mathematical notation, a function f: X → Y is bijective if and only if it satisfies the condition for every y in Y there is a unique x in X with y = f(x). Hence every bijection is invertible. A function f : A -> B is said to be onto function if the range of f is equal to the co-domain of f. How to Prove a Function is Bijective without Using Arrow Diagram ? The inverse is conventionally called \arcsin. A function that is both One to One and Onto is called Bijective function.$$ Now this function is bijective and can be inverted. This is equivalent to the following statement: for every element b in the codomain B, there is exactly one element a in the domain A such that f(a)=b.Another name for bijection is 1-1 correspondence (read "one-to-one correspondence).. Functions that have inverse functions are said to be invertible. Mathematical Functions in Python - Special Functions and Constants; Difference between regular functions and arrow functions in JavaScript; Python startswith() and endswidth() functions; Hash Functions and Hash Tables; Python maketrans() and translate() functions; Date and Time Functions in DBMS; Ceil and floor functions in C++ Each value of the output set is connected to the input set, and each output value is connected to only one input value. As pointed out by M. Winter, the converse is not true. My examples have just a few values, but functions usually work on sets with infinitely many elements. Infinitely Many. Thus, if you tell me that a function is bijective, I know that every element in B is “hit” by some element in A (due to surjectivity), and that it is “hit” by only one element in A (due to injectivity). The function f is called as one to one and onto or a bijective function, if f is both a one to one and an onto function. Ah!...The beautiful invertable functions... Today we present... ta ta ta taaaann....the bijective functions! The figure shown below represents a one to one and onto or bijective function. Some types of functions have stricter rules, to find out more you can read Injective, Surjective and Bijective. Definition: A function is bijective if it is both injective and surjective. More clearly, f maps distinct elements of A into distinct images in B and every element in B is an image of some element in A. A function is invertible if and only if it is a bijection. In mathematics, a bijective function or bijection is a function f : A → B that is both an injection and a surjection. In essence, injective means that unequal elements in A always get sent to unequal elements in B. Surjective means that every element of B has an arrow pointing to it, that is, it equals f(a) for some a in the domain of f. Question 1 : And I can write such that, like that. Below is a visual description of Definition 12.4. Curve, but is not true, surjective and bijective functions have stricter rules, to find out more can... Represents a one to one and onto or bijective function is both and! Of the output set is connected to only one input value a one to and!!... the beautiful invertable functions... Today we present... ta ta ta taaaann! 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# Prove that $\frac{a_1^2}{a_1+a_2}+\frac{a_2^2}{a_2+a_3}+ \cdots \frac{a_n^2}{a_n+a_1} \geq \frac12$

Let $$a_1, a_2, a_3, \dots , a_n$$ be positive real numbers whose sum is $$1$$. Prove that $$\frac{a_1^2}{a_1+a_2}+\frac{a_2^2}{a_2+a_3}+ \ldots +\frac{a_n^2}{a_n+a_1} \geq \frac12\,.$$

I thought maybe the Cauchy and QM inequalities would be helpful. But I can't see how to apply it. Another thought (might be unhelpful) is that the sum of the denominators on the left hand side is $$2$$ (the denominator on the right hand side). I would really appreciate any hints.

• Apply Cauchy with one term being left hand side, second term being $((a_1+a_2) + (a_2+a_3) + \cdots)$ – user27126 Apr 28 '13 at 7:02
• Thanks so much I've solved it, should I close the question, or post the solution I just found, with your hint? – John Marty Apr 28 '13 at 7:05
• it would be good if you post a complete solution and accept it as the answer. – user27126 Apr 28 '13 at 7:06

Thanks to Sanchez for giving me a hint to solve this. Here is a full solution.

By the Cauchy-Schwarz inequality we have:

$${\frac{a_1^2}{a_1+a_2}+\frac{a_2^2}{a_2+a_3}+ \cdots \frac{a_n^2}{a_n+a_1}=\frac{a_1^2}{(\sqrt{a_1+a_2})^2}+\frac{a_2^2}{(\sqrt{a_2+a_3})^2}+ \cdots+ \frac{a_n^2}{(\sqrt{a_n+a_1})^2} \geq \frac{1}{a_1+\cdots + a_n+a_1+ \cdots + a_n}\left(\frac{a_1 \cdot \sqrt{a_1+a_2}}{\sqrt{a_1+a_2}} + \frac{a_2 \cdot \sqrt{a_2+a_3}}{\sqrt{a_2+a_3}}+ \cdots + \frac{a_n \cdot \sqrt{a_n+a_1}}{\sqrt{a_n+a_1}}\right)\\=\frac{a_1+a_2+a_3+ \cdots a_n}{{2(a_1+a_2+a_3+ \cdots a_n)}}=\frac12}$$

as required. (We know that $a_1+a_2+a_3+ \cdots +a_n=1$)

• thanks, I couldn't solve it before your hint, thanks so much. – John Marty Apr 28 '13 at 7:34
• I made an attempt to improve readability. Is it OK? – Pedro Tamaroff Jul 6 '13 at 0:37

Here's another solution.

Observe that $$\sum \frac{a_i^2 - a_{i+1}^2} { a_i+ a_{i+1}} = \sum a_i - a_{i+1} = 0,$$

Hence $\sum \frac{a_{i}^2}{a_i+a_{i+1}} = \sum \frac{a_{i+1}^2}{a_i+a_{i+1}}$, and we just need to show that

$$\sum \frac{a_i^2 + a_{i+1}^2}{a_i + a_{i+1} } \geq 1.$$

This makes the inequality much more symmetric, and easier to manipulate. In particular,

$$\sum \frac{a_i^2 + a_{i+1}^2}{a_i + a_{i+1} } \geq \sum \frac{1}{2} (a_i + a_{i+1}) = 1$$

The following modified form of the CBS inequality is often helpful:

Lemma. If $x_i \in \Bbb{R}$ and $a_i > 0$, then $$\frac{x_{1}^{2}}{a_{1}} + \cdots + \frac{x_{n}^{2}}{a_{n}} \geq \frac{(x_{1} + \cdots + x_{n})^{2}}{a_{1}+\cdots+a_{n}}.$$

The proof is straightforward using the CBS inequality, following the methodology exactly John Marty used.

Applying this, we have

$$\frac{a_{1}^2}{a_{1}+a_{2}} + \cdots + \frac{a_{n}^2}{a_{n}+a_{1}} \geq \frac{(a_{1} + \cdots + a_{n})^{2}}{2(a_{1}+\cdots+a_{n})} = \frac{1}{2}.$$


# Question to calculating probability

The question:

79% of the drivers in a city always fasten their seatbelt when driving, and everyday some drivers receive a ticket(for various reasons...). If a driver has fastened their seatbelt there is a chance of 7% that the driver receives a ticket. If a driver does not fasten their seatbelt there is a 22% chance that he will receive a ticket. If a driver receives a ticket what is the probability that he had fastened his seatbelt?

My assumption is that 7/29 of the people who receive a ticket had fastened their seatbelt and 22/29 did not. So I just calculated (7/29) * (0,79) + (22/29) * (0,21) and I got 0.35 but that seems to be wrong.

• Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. – José Carlos Santos Jan 18 '18 at 16:13
• Okay thanks for the advise. My assumption is that 7/29 of the people who receive a ticket had fastened their seatbelt and 22/29 did not. So i just calculated (7/29) * (0,79) + (22/29) * (0,21) and i got 0.35 – Simon Jan 18 '18 at 16:18
• Edit your question instead of posting it as a comment. – José Carlos Santos Jan 18 '18 at 16:19
• This is a prefect time for Bayes' Law! – Bram28 Jan 18 '18 at 16:28

Let $S$ and $T$ be the events "seatbelt fasten" and "receive ticket" respectively.

• $P(S) = 0.79$
• $P(T \mid S) = 0.07$
• $P(T \mid S^C) = 0.22$

Use Bayes' formula to find $P(S \mid T)$.

\begin{align} P(S \mid T) &= \frac{P(S \cap T)}{P(T)} \\ &= \frac{P(T \mid S) P(S)}{P(T \mid S) P(S)+P(T \mid S^C) P(S^C)} \\ &= \frac{0.07 \cdot 0.79}{0.07 \cdot 0.79 + 0.22 \cdot (1 - 0.79)} \\ &= \frac{79}{145} \end{align}

If a driver receives a ticket what is the probability that he had fastened his seatbelt is $79/145$.

You can't assume $\frac{7}{29}$ of people who receive a ticket fastened their seatbelt. Consider an extreme case, where the percentage of people who fasten their seatbelt is much higher than $79\%$ -- suppose it is $100\%$. Then clearly $100\%$ of people who receive a ticket fastened their seatbelt. If $0\%$ of people fasten their seatbelt, then $0\%$ who receive a ticket fastened their seatbelt. If the percentage who fasten is somewhere in between -- e.g., $79\%$ -- then the percentage who fastened given that they received a ticket is somewhere in between, but we can't just say it's $\frac{7}{29}$.

Are you familiar with Bayes' Theorem? That's probably the best way to approach this problem.


What is the value of $\left(\log_{21}(3)\right)^2+\log_{21}(147)\log_{21}(1323)$?

What is the value of $$\left(\log_{21}(3)\right)^2+\log_{21}(147)\log_{21}(1323)$$ ?

$$1)1\qquad\qquad2)2\qquad\qquad3)3\qquad\qquad4)4$$

To solve this question I tried writing the expression as:

$$\left(\log_{21}(3)\right)^2+\log_{21}(7^2\times3)\log_{21}(7^2\times3^3)$$ $$=\left(\log_{21}(3)\right)^2+(2\log_{21}7+\log_{21} 3)(2\log_{21}7+3\log_{21}3)$$

I don't know how to continue.

• Don't be so eager to separate the $\log7$ from the $\log3$. The most potent course of simplification you have here is $\log21=1$. Jul 1 at 10:23
• @Arthur Thanks! The question looks horrible for a timed exam and I became a little nervous! Jul 1 at 10:29
• Denote $x=\log_{21}3$ and $y=\log_{21}7$. Then, $147=21\times 7$ and $1323=21^2\times 3$; the original expression is then $x^2+(1+y)(2+x)=x(x+y)+(x+y)+y+2$; iteratively use $x+y=1$ to simplify. Jul 1 at 10:47
• @Prasun Biswas: the computation is simpler if one notices that $147=21^2/3$ and $1323=21^2 \times 3$. Jul 1 at 10:51
• @Mindlack: Nice! Indeed, that is far more simpler because we wouldn't need the $y$, we have $x^2+(2-x)(2+x)=x^2+4-x^2=4$ Somehow, I missed that. +1 Jul 1 at 10:55

Here’s a quick-and-dirty solution based on the fact that one of the answers must be correct.

All the quantities are positive. Since $$1323>441=21^2$$, its base-$$21$$ logarithm is greater than $$2$$. As $$147 \geq 21 \times 5$$, its base-$$21$$ logarithm is at least $$1.5$$. This means that the sum is greater than $$2 \times 1.5$$ so it must be $$4$$.

• I don't get the second point regarding $147 \geq 21 \times 5$. Could you elaborate? Jul 1 at 11:04
• @user71207 $147\ge21\times5\ge21\sqrt{21}$ And the fact that $147\ge21^{\frac32}$ means $\log_{21}(147)\ge\frac32$ Jul 1 at 11:09

$$\log_{21}1323=\log_{21}(3^3\cdot 7^2)=\log_{21}3+\log_{21}21^2=2+\log_{21}3$$
$$\log_{21}147=\log_{21}21+\log_{21}7=1+\log_{21}7$$

$$(\log_{21}3)^2+\log_{21}147\cdot\log_{21}1323 \\=(\log_{21}3)^2+(1+\log_{21}7)(2+\log_{21}3) \\=\log_{21}3(\log_{21}3+\log_{21}7)+\log_{21}3+2\log_{21}7+2 \\=\log_{21}3(\log_{21}3+\log_{21}7)+(\log_{21}3+\log_{21}7)+\log_{21}7+2 \\=\log_{21}3+1+\log_{21}7+2 \\=4$$

• Hi, the problem in this post and also this one that I asked were from entrance exam in the country I'm living (average time for solving each question is about 90 seconds). You solved both of them very well and I think you are very good at solving these questions! So can you please suggest some books or name of exam that you think are good for an exam that contains these type of questions? Thanks in advanced! Jul 27 at 19:58
• I mainly use books that are meant for olympiads, so I don't know many books that are meant for solving questions fast. Jul 28 at 6:28

Alternate:

Let $$L(x)$$ denote $$\log_{21}(x).$$

Then $$[L(3)]^2 + [L(147) \times L(9 \times 147)]$$

$$= [L(3)]^2 + ~\langle ~L(147) \times ~\{ [2 \times L(3)] + L(147) ~\} ~\rangle$$

$$= [L(3)]^2 + [L(147)]^2 + [2 \times L(3) \times L(147)]$$

$$= [L(3) + L(147)]^2.$$

At this point, you have that $$3 \times 147 = 21^2$$,

so $$[L(3) + L(147)] = 2.$$


# Why is $SST=SSE + SSR$? (One variable linear regression)

Note: $SST$ = Sum of Squares Total, $SSE$ = Sum of Squared Errors, and $SSR$ = Regression Sum of Squares. The equation in the title is often written as:

$$\sum_{i=1}^n (y_i-\bar y)^2=\sum_{i=1}^n (y_i-\hat y_i)^2+\sum_{i=1}^n (\hat y_i-\bar y)^2$$

Pretty straightforward question, but I am looking for an intuitive explanation. Intuitively, it seems to me like $SST\geq SSE+SSR$ would make more sense. For example, suppose point $x_i$ has corresponding y-value $y_i=5$ and $\hat y_i=3$, where $\hat y_i$ is the corresponding point on the regression line. Also assume that the mean y-value for the dataset is $\bar y=0$. Then for this particular point i, $SST=(5-0)^2=5^2=25$, while $SSE=(5-3)^2=2^2=4$ and $SSR=(3-0)^2=3^2=9$. Obviously, $9+4<25$. Wouldn't this result generalize to the entire dataset? I don't get it.

Adding and subtracting gives \begin{eqnarray*} \sum_{i=1}^n (y_i-\bar y)^2&=&\sum_{i=1}^n (y_i-\hat y_i+\hat y_i-\bar y)^2\\ &=&\sum_{i=1}^n (y_i-\hat y_i)^2+2\sum_{i=1}^n(y_i-\hat y_i)(\hat y_i-\bar y)+\sum_{i=1}^n(\hat y_i-\bar y)^2 \end{eqnarray*} So we need to show that $\sum_{i=1}^n(y_i-\hat y_i)(\hat y_i-\bar y)=0$. Write $$\sum_{i=1}^n(y_i-\hat y_i)(\hat y_i-\bar y)=\sum_{i=1}^n(y_i-\hat y_i)\hat y_i-\bar y\sum_{i=1}^n(y_i-\hat y_i)$$ So, (a) the residuals $e_i=y_i-\hat y_i$ need to be orthogonal to the fitted values, $\sum_{i=1}^n(y_i-\hat y_i)\hat y_i=0$, and (b) the sum of the fitted values needs to be equal to the sum of the dependent variable, $\sum_{i=1}^ny_i=\sum_{i=1}^n\hat y_i$.

Actually, I think (a) is easier to show in matrix notation for general multiple regression of which the single variable case is a special case: \begin{eqnarray*} e'X\hat\beta &=&(y-X\hat\beta)'X\hat\beta\\ &=&(y-X(X'X)^{-1}X'y)'X\hat\beta\\ &=&y'(X-X(X'X)^{-1}X'X)\hat\beta\\ &=&y'(X-X)\hat\beta=0 \end{eqnarray*} As for (b), the derivative of the OLS criterion function with respect to the constant (so you need one in the regression for this to be true!), aka the normal equation, is $$\frac{\partial SSR}{\partial\hat\alpha}=-2\sum_i(y_i-\hat\alpha-\hat\beta x_i)=0,$$ which can be rearranged to $$\sum_i y_i=n\hat\alpha+\hat\beta\sum_ix_i$$ The right hand side of this equation evidently also is $\sum_{i=1}^n\hat y_i$, as $\hat y_i=\hat\alpha+\hat\beta x_i$.

This is just the Pythagorean theorem!

Hence,

$$Y'Y=(Y-X\hat{\beta})'(Y-X\hat{\beta})+(X\hat{\beta})'X\hat{\beta}$$

or

$$SST=SSE+SSR$$

• Aug 2 '19 at 19:52

(1) Intuition for why $SST = SSR + SSE$

When we try to explain the total variation in Y ($SST$) with one explanatory variable, X, then there are exactly two sources of variability. First, there is the variability captured by X (Sum Square Regression), and second, there is the variability not captured by X (Sum Square Error). Hence, $SST = SSR + SSE$ (exact equality).

(2) Geometric intuition

Some of the total variation in the data (distance from datapoint to $\bar{Y}$) is captured by the regression line (the distance from the regression line to $\bar{Y}$) and error (distance from the point to the regression line). There's not room left for $SST$ to be greater than $SSE + SSR$.

(3) The problem with your illustration

You can't look at SSE and SSR in a pointwise fashion. For a particular point, the residual may be large, so that there is more error than explanatory power from X. However, for other points, the residual will be small, so that the regression line explains a lot of the variability. They will balance out and ultimately $SST = SSR + SSE$. Of course this is not rigorous, but you can find proofs like the above.

Also notice that regression will not be defined for one point: $b_1 = \frac{\sum(X_i -\bar{X})(Y_i-\bar{Y}) }{\sum (X_i -\bar{X})^2}$, and you can see that the denominator will be zero, making estimation undefined.

Hope this helps.

--Ryan M.

When an intercept is included in linear regression(sum of residuals is zero), $$SST=SSE+SSR$$.

prove $$\begin{eqnarray*} SST&=&\sum_{i=1}^n (y_i-\bar y)^2\\&=&\sum_{i=1}^n (y_i-\hat y_i+\hat y_i-\bar y)^2\\&=&\sum_{i=1}^n (y_i-\hat y_i)^2+2\sum_{i=1}^n(y_i-\hat y_i)(\hat y_i-\bar y)+\sum_{i=1}^n(\hat y_i-\bar y)^2\\&=&SSE+SSR+2\sum_{i=1}^n(y_i-\hat y_i)(\hat y_i-\bar y) \end{eqnarray*}$$ Just need to prove last part is equal to 0: $$\begin{eqnarray*} \sum_{i=1}^n(y_i-\hat y_i)(\hat y_i-\bar y)&=&\sum_{i=1}^n(y_i-\beta_0-\beta_1x_i)(\beta_0+\beta_1x_i-\bar y)\\&=&(\beta_0-\bar y)\sum_{i=1}^n(y_i-\beta_0-\beta_1x_i)+\beta_1\sum_{i=1}^n(y_i-\beta_0-\beta_1x_i)x_i \end{eqnarray*}$$ In Least squares regression, the sum of the squares of the errors is minimized. $$SSE=\displaystyle\sum\limits_{i=1}^n \left(e_i \right)^2= \sum_{i=1}^n\left(y_i - \hat{y_i} \right)^2= \sum_{i=1}^n\left(y_i -\beta_0- \beta_1x_i\right)^2$$ Take the partial derivative of SSE with respect to $$\beta_0$$ and setting it to zero. $$\frac{\partial{SSE}}{\partial{\beta_0}} = \sum_{i=1}^n 2\left(y_i - \beta_0 - \beta_1x_i\right)^1 = 0$$ So $$\sum_{i=1}^n \left(y_i - \beta_0 - \beta_1x_i\right)^1 = 0$$ Take the partial derivative of SSE with respect to $$\beta_1$$ and setting it to zero. $$\frac{\partial{SSE}}{\partial{\beta_1}} = \sum_{i=1}^n 2\left(y_i - \beta_0 - \beta_1x_i\right)^1 x_i = 0$$ So $$\sum_{i=1}^n \left(y_i - \beta_0 - \beta_1x_i\right)^1 x_i = 0$$ Hence, $$\sum_{i=1}^n(y_i-\hat y_i)(\hat y_i-\bar y)=(\beta_0-\bar y)\sum_{i=1}^n(y_i-\beta_0-\beta_1x_i)+\beta_1\sum_{i=1}^n(y_i-\beta_0-\beta_1x_i)x_i=0$$ $$SST=SSE+SSR+2\sum_{i=1}^n(y_i-\hat y_i)(\hat y_i-\bar y)=SSE+SSR$$

Here is a great graphical representation of why SST = SSR + SSE.

• How can we see from this picture that predicted values and residuals are orthogonal? Mar 4 at 12:17

If a model predicts $$3$$ and the residual is $$2$$ because the actual value is $$5$$, it doesn't look like variance is decomposing, since $$3^2 + 2^2 \neq 5^2$$.

If you only have one data point, your model would fit it perfectly and the residual would be zero, so you can't get that case by itself. There have to be multiple data points.

With multiple data points, the residuals won't all be positive.

If the model predicts 3, a residual of $$+2$$ and $$-2$$ should be equally likely. The balance of increases and decreases gives a nice cancellation:

$$\frac{(3+2)^2 + (3-2)^2}{2} = \frac{25+1}{2} = 13 = 9 + 4 = 3^2 + 2^2$$

This property, that $$(a+b)^2 + (a-b)^2 = 2(a^2 + b^2)$$, is what makes variance decomposition work, for uncorrelated components.