Solve more complex problems involving combinations of outcomes. This type of diagram can be very useful for some problems. So event A is selecting bag A, event B is finding 4 red and 1 blue. Conditional probability. The student will appraise the differences between the two estimates. It consists of "branches" that are labeled with either frequencies or probabilities. A carton contains twenty-five lightbulbs of the same size but of varying wattage. Critical Thinking Does the probability of choosing without replacement change if the order of the events is reversed? Explain. (5) (d) Calculate the probability that keys are not collected on at least 2 successive stages in a game. • Probability of A or B occurring: (Never double count) P(A or B) = x A n + x B n MULTIPLICATION OF CHOICES Question 4. Click Image to Enlarge : Use a tree diagram to display possible outcomes of who will come to the party. 2 Hit Miss Hit Miss O. Click here to read the solution to this question Click here to return to the index. Example $$\PageIndex{24}$$ In an urn, there are 11 balls. P(All White) = 0. There are eight High wattage 100W Lightbulbs, five medium wattage …. These possible outcomes can be shown by the branches of a tree-like diagram called ‘Tree-Diagram’ or ‘Branch Diagram’. It consists of branches that are labeled with either frequencies or probabilities. (c) at least 2 tails, (d) 2 tails in succession 1 (e) 2tails. G1 = first card is green. Determine a single event with a single outcome. From a tree diagram, you can determine what the probability is that you carry the allele: Thus the probability you carry the allele is the probability your mother carries the allele and passes it on to her progeny: ( ) 7. Replacing the actual function with this model in Monte Carlo simulation (MCS), the approximate failure probability can be obtained. You spin the spinner twice. A packet of sweets has 3 pink, 2 green and 5 blue sweets. The probability of rain tomorrow is estimated to be 1. The probability that both events happen and we draw an ace and then a king corresponds to P( A ∩ B ). Then, using the information in the table in #1 complete the theoretical probability questions below. Most of the time, it is used by scientists to calculate the success rate of their experiments. (i) Copy the tree diagram and add the four missing probability values on the branches that refer to playing with a stick. This is a whole lesson on Tree diagrams but there is probably (pun intended) enough material for two lessons. Probability Tree Diagrams. down the sample space. 128 Solution 4-34 a) Let F = a person is in favor of genetic engineering A = a person is against genetic engineering. The following example illustrates how to use a tree diagram. (7) (Total 10 marks) 11. "With replacement" means that you put the first ball back. Probability without replacement formula. It is designed to follow the Conditional Probability and Probability of Simultaneous Events Lesson to further clarify the role of replacement in calculating probabilites. No further attempts are allowed. Determine the cost for each outcome. When two balls are chosen at random without replacement from bag B, the probability that they are both white is $$\frac{2}{7}$$. Here is how to do it for the "Sam, Yes" branch: (When we take the 0. Find P (both mice are short-tailed). What is the probability it is blue. How to complete and calculate probabilities from tree diagrams where the counter (etc) has not been replaced before the second pick. Sample Space - is the _____ of all the _____ in a probability experiment. Probability is the chance that something will happen - how likely it is that some event will happen over the long run. 1 - P (B, B) =. Tree Diagrams Tree diagrams show all the possible outcomes of an event and calculate their probabilities. This information is represented by the following tree diagram. Draw a tree diagram. When replacing, the probabilities do not change. Two beads are drawn at random from the jar without replacement. without replacement P(R 1 st draw, B 2 nd draw) P(Br 1 draw, Br 2 nd draw) 9. Are these events independent or dependent? 4. Use sample space diagrams and list for outcomes of more than one event. Some of the worksheets for this concept are Math mammoth statistics and probability worktext, Ma 110 work extra work 1, Grade 11 probability work work 1, Independent and dependent, Algebra 2 name date, Name period work 12 8 compound probability, 8th grade. Some of the worksheets for this concept are Tree diagrams 70b, Lesson plan 2 tree diagrams and compound events, Probability tree diagrams, Tree diagrams and the fundamental counting principle, Wjec mathematics, Simple sample spacestree outcomes diagrams, Grade 7, The probability scale. Draw a tree representing the possible mutually exclusive outcomes 2. Count outcomes using tree diagram. You could use the fact that P (at least 1) = 1 - P (none) So P (at least 1 white) = 1 - P (no whites) P (at least 1 white) = 1 - P (3 blacks). Find the probability of drawing, a. A bag contains 5 red sweets and 3 blue sweets. b) the sweets are taken without replacement. Learn about calculating probability's of a sequence of events, by organising its rules for adding & multiplying probability's for or &, and with Tree Diagrams. Each topic quiz contains 4-6 questions. Example: Use a tree diagram to find the sample space for the sex of three children in a family. Draw a tree diagram to show all the possible outcomes. Multiply going across a tree diagram. To understand probability with replacement, it will be helpful to refresh the following topics: Basics of probability theory. We begin with an example. Two marbles are drawn at random and with replacement from a box containing 2 red, 3 green, and 4 blue marbles. The problem as stated says that monty hall deliberately shows you a door that has a goat behind it. Draw the tree diagram for this data. Tree diagrams. 2) A bag contains 5 red balls and 3 green balls. Tree diagrams for events without replacement The tree diagram in the following example illustrates events that are not independent. How to draw probability tree diagrams? Examples: 1. The probability of rain tomorrow is estimated to be 1. Find the event E = “at least two girls”. Tree Diagram Definition Math Probability Tree Diagrams How To Solve Probability Problems Using Tree Diagrams. An individual can also look at Probability With Replacement Worksheet image gallery that many of us get prepared to get the image you are searching for. a Draw a tree diagram showing the probabilities Of wind or rain on a particular day. Leanne notices that on windy days, the probability she catches a fish is 0. Math (check) 4. A probability mass function (p. Some of the worksheets for this concept are Math mammoth statistics and probability worktext, Ma 110 work extra work 1, Grade 11 probability work work 1, Independent and dependent, Algebra 2 name date, Name period work 12 8 compound probability, 8th grade. The following example illustrates how to use a tree diagram. (7) (Total 10 marks). 2) The probability that Helen does her homework is ¾. report that multipotent mouse embryonic mammary cells become lineage restricted as early as embryonic day 12. b) the sweets are taken without replacement. Obviously this is impractical to draw a tree diagram to count the probability with customer size 20. A tree diagram is a special type of graph used to determine the outcomes of an experiment. Conditional probability is defined as the likelihood of an event or outcome occurring, based on the occurrence of a previous event or outcome. simple event 1. 2∕3 + 1∕12 = 3∕4 of the time. What she found most intriguing was the fact that the teacher could not provide a satisfactory definition of "random" (or of "probability," for that matter), even though the notions such as "random variable" and "random sample" lie at the heart of the theory. One card is removed at random from each box. YOU can use shorthand like this. In a conditional probability an outcome or event E is dependent upon another outcome or event F. Independent events. if there are 5 yellow and 7 green marbles in the box, and two yellow marbles are selected one after the other, Pr(Y, Y) = 5 12 × 5 12 = 25 144) Without Replacement. two bills without replacement, determine whether the probability that the bills will total $15 is greater than the probability that the bills will total$2. A4Now let’s try to answer the question, “What is the probability of drawing 2 Queens from a well shuffled deck of cards without replacement?”. A tree diagram is a special type of graph used to determine the outcomes of an experiment. (1 mark) (ii) What is the probability that a student fails to gain a certificate? (2 marks) (b) Three students take the exam. 18 Outcomes & Probability Third Pick First Pick Second Pick Figure 8: Tree diagram for selecting three sweets randomly (with probability value) e) Probability distribution for each flavour if three sweets are. A tree diagram is a pleasing way to visualize the concept involving probability without replacement. The following example illustrates how to use a tree diagram. 1 of the bags is selected at random and a ball is drawn from it. See full list on byjus. Example 2 A box contains 12 beads. Tree Diagrams. the probability of each event is independent of one another). 5 Outcome Heads Tails There are two Branches (Heads and Tails). The probability tree is shown in Figure 34. Probability Theory And Examples Solutions. Flavours C L Probability, P(X = x) 10 2 = 25 5 15 3 = 25 5 d) Tree diagram if three sweets are selected randomly without replacement. Use tree diagrams to solve without replacement problems. What is the probability that if she dressed in the dark (choosing her outfit at random), she would wear the plaid skirt with the blouse with pink flowers? First we will make a tree diagram to view the different outfits possible. Probability Tree Diagrams - Dependent Events - GCSE Mathematics 1 - 9. 3: Tree diagram for two draws without replacement, values rounded. To answer how likely a patient is to have TB given a positive test result, we need to “flip” the tree. Prealgebra/probability. Two Marbles Are Drawn At Random And Without Replacement From A Box Containing 3 Blue Marbles And 5 Red Marbles. A tree diagram for the situation of drawing one marble after the other without replacement is shown in Figure $$\PageIndex{1}$$. For each possible outcome of the first event, we draw a line where we write down the probability of that outcome and the state of the world if that outcome happened. Which would be better for representing the sample space, a systematic list, a tree diagram, or an area model? Justify your answer. sample space consists of 52 outcomes. Assigned Practices: 1. A4Now let’s try to answer the question, “What is the probability of drawing 2 Queens from a well shuffled deck of cards without replacement?”. The probability of receiving an offer from Acme is 0. In a conditional probability an outcome or event E is dependent upon another outcome or event F. Making use of a Venn diagram (where appropriate) find: C. Below is a tree diagram. Use the tree diagram to ﬁ nd the probability that both marbles are green. Tree Diagrams \n. Tree Diagram for Probability. It consists of "branches" that are labeled with either frequencies or probabilities. Copy and complete the probability tree diagram below. Box A contains 3 cards numbered 1, 2 and 3. The branches emanating from any point on a tree diagram must have probabilities that sum to 1. 18 Outcomes & Probability Third Pick First Pick Second Pick Figure 8: Tree diagram for selecting three sweets randomly (with probability value) e) Probability distribution for each flavour if three sweets are. Probability, Finite Mathematics: For the Managerial, Life, and Social Sciences 11th - Soo T. The calculations are shown in the tree diagram. Intelligent Practice. Tree diagrams (with and without replacement) This is a lesson I made for a recent observation. 13 Outcomes & Probability Third Pick Second Pick First Pick BBB (0. Tree Diagram. Then, a second ball is drawn from the box and recorded. Select the number of main events, branch events and then enter a label and a probability for each event. Probability Rules. The probabilities add to 1 because these outcomes together make up the sample space S. Only stopping at one set. It consists of "branches" that are labeled with either frequencies or probabilities. 6 (a) Complete the probability tree diagram. Three balls are red (R) and eight balls are blue (B). Further, there are nodes linked with branches. The following example illustrates how to use a tree diagram. Tree Diagram; Probability Without Replacement; Dice probability; Coin flip probability; Probability with replacement; Geometric probability; Events. $Assuming that the first sock is red, the probability of getting the second red sock is$\displaystyle\frac{r-1}{r+b+g-1}. Since both combined events are the same (just the other way around), the answers are identical. Copy and complete the probability tree diagram below. The tree diagram for this problem will also be similar to the with-replacement version. Tree diagrams – no replacement – V2; 5. 3 The probability that Sam hits the target is 0. The site consists of an integrated set of components that includes expository text, interactive web apps, data sets, biographical sketches, and an object library. 8c: Find the probability that Pablo is late for work. This indicates how strong in your memory this concept is. Compare the probabilities in the contingency table and Venn Diagram below (also found on page 351). Given you draw a R m&m in your 1 st draw, what is the probability of. Only the terminal node numbers are displayed. Tree diagrams can make some probability problems easier to visualize and solve. Then determine the probability of getting one red and one blue in any order. A tree diagram shows all the possible outcomes from a senes of events and their probabilities. What is the probability of flipping 3 coins and having al l three land on tails (make a tree diagram first)? 14. Outcome Probability RR(red path) RB(blue path) BR(yellow path) BB(purple path) 5. First, use a tree diagram to map out the sample space: a. A tree diagram for the situation of drawing one marble after the other without replacement is shown in Figure $$\PageIndex{1}$$. State your probabilities clearly. It consists of "branches" that are labeled with either frequencies or probabilities. It consists of “branches” that are labeled with either frequencies or probabilities. (b) not 6 not 6 (Total 6 marks). 216$(or in fractions$(\frac{3}{5})^3 = \frac{27}{125}$). (d) Given that a student fails, what is the probability that he or she came from school III? [(. a) Draw a tree diagram to list all the possible outcomes. When a probability experiment involves more than two actions, we often use a tree diagram to find the sample space. Tree Diagrams and the Fundamental Counting Principle The purpose of this task is to help students discover the Fundamental Counting Principle through the use of tree diagrams. • A tree diagram is a graphical way to show all of the possible _____ ____ in a situation or experiment. –When a sample space can be constructed in several steps or stages, we can represent each of the n 1 ways of completing the first step as a branch of a tree. Note: The probabilities for each event must total to 1. Find P (both mice are short-tailed). The following tree diagram generated by clicking the Draw button shows in color the node numbers for the tree described previously. Going up a level. NOTE: If anyone fancies knocking up a diagram to show the answer to Question 15, it would be greatly appreciated 🙂. Make sure you are happy with the following topics before continuing. Displaying top 8 worksheets found for - Probability Tree Diagrams. Tree diagram (multiply each step along the tree. Assigned Practices: 1. I built Diagnostic Questions to help you identify, understand and resolve key misconceptions. This lesson explores sampling with and without replacement, and its effects on the probability of drawing a desired object. Tree Diagrams: Probability 1) A drawer contains 4 red and 3 blue socks. This is the editable tree diagram!!!!! @mrbartonmaths. 6 Probability of \At least one". 3 : PROBABILITY TREES AND PROBABILITY WITH COMBINATIONS TREE DIAGRAMS are a useful tool in organizing and solving probability problems Each complete path through the tree represents a separate mutually exclusive outcome in the sample space. All Lectures in one file. We draw bulbs without replacement until a working bulb is selected. Best of three games. 7: The Law of Total Probability in a tree diagram. Erick, a college senior, interviews with Acme Corp. and Mills Inc. Gracie's lemonade stand. A probability tree diagram represents all the possible outcomes of an event in an organized manner. If two marbles are drawn at random without replacement, what is the probability of picking two marbles that are different colours? 5/8. Tree Diagram: A jar contains 4 purple and 1 gold beads. report that multipotent mouse embryonic mammary cells become lineage restricted as early as embryonic day 12. Tree diagrams (with and without replacement) This is a lesson I made for a recent observation. The diagram at the right shows the results of randomly choosing a checker, putting it back, and Draw a tree diagram of two independent events, such as c. Consider a game in which you start with 3 green and 2 red marbles in a bag, and you pull out two of them randomly, without replacement. The circle and rectangle will be explained later, and should be ignored for now. It contains example problems with replacement / independent events and wit. Whoops! There was a problem previewing Conditional probability. Show these probabilities in. • The method can determine the threshold replacement strategy for premature failure. 13 Outcomes & Probability Third Pick Second Pick First Pick BBB (0. Probabilities are assigned to the branches when one or more events are being considered. Students may use other methods. sample space consists of 52 outcomes. The reason this works is because the events along the path are independent. Give your answer to the nearest two decimal places. (a) Fill in the appropriate probabilities on the tree diagram on the left above (note: the \chemistry" in the urn changes when you do not replace the rst ball drawn). 3 are blue, and 7 are red. 1 Simple Sample Spaces…Tree Diagrams Outcome - a particular result of an experiment outcomes. With Replacement: the events are Independent (the chances don't change) Without Replacement: the events are Dependent (the chances change) Dependent events are what we look at here. 8a: Copy and complete the following tree diagram. Given you draw a R m&m in your 1 st draw, what is the probability of. a) Draw a tree diagram to determine ALL possible outcomes. The value of this probability is 12/2652. Draw a tree-diagram to represent all probabilities for the following. nsisting of two trials. It consists of "branches" that are labeled with either frequencies or probabilities. I don't know how to write out a tree diagram on here, but I think this one is heads -> heads, tails -> math probabilty- please help. EX 5: Two cards are drawn from a deck without replacement. The abbreviation of pdf is used for a probability distribution function. Learn about calculating probability's of a sequence of events, by organising its rules for adding & multiplying probability's for or &, and with Tree Diagrams. 13 Probability Simulations Resources 1_1. When we sample from small populations, we can use a tree diagram to represent the sample space and determine the probabilities of events from the tree diagram. Lesson plan to help students understand independent and dependent variables through a fire probability simulation. It consists of "branches" that are labeled with either frequencies or probabilities. What is the probability of flipping 3 coins and having them all land on the same side (i. If you want to evaluate a joint probability tree where probabilities are represented as decimals (e. Lesson Worksheet. The probability of getting one sock red is$\displaystyle\frac{r}{r+b+g}. Three balls are drawn from the bag without replacement, find the probability that the balls are all of different colors. It is not returned to the box. Prealgebra/probability. 5 - Practice: Probability of independent Events Practice Math 6 B (QUI 7. Suppose a jar contains 3 red and 4 white marbles. Two beads are drawn at random from the jar without replacement. The following example illustrates how to use a tree diagram. (a) Draw a tree diagram to illustrate all the possible outcomes and associated probabilities. Use the results of part (a) to find the probability of obtaining (b) only one tail. Probability. Outcome Probability RR(red path) RB(blue path) BR(yellow path) BB(purple path) 5. b) What is the probability that both balls are different colours? 2) A box contains 5 red counters and 3 blue counters. c: Two mice are chosen without replacement. We sample two cards from a deck of $52$ cards without replacement. (2) (b) Work out the probability that both Tom and Sam will pass the driving test (2) (c) Work out the probability that only one of them will pass the driving test. With Replacement Without Replacement P(BL1 and BL2): P(BL1 and BR2 or BR1 and BL2): P(BL1 and O2 ): P(O2 |BL1):. Understanding probability is crucial to many industries, such as finance and medical professions. Module 1 : Probability Part 1 Module 1: Probability Part 1. The probability of a delay at the first roundabout is 0. Toy decides to select 7th grade students based on the same probability that Mrs. Let's consider another example: Example 2: What is the probability of drawing a king and a queen consecutively from a deck of 52 cards, without replacement. Topic: Day 3 Probability, 14G. 1 Multiplication of choices. See full list on byjus. 4) Could also be part of a tree diagram might just indicate the 3 routes through the tree but must add +0. The probability of any outcome is the product of all possibilities along the relevant branches. When we flip a coin, there are only two possible outcomes {heads or tails}, and when we roll a die, there are six possible outcomes {1,2,3,4,5,6}. Example 2 An urn has 3 red marbles and 8 blue marbles in it. Draw two balls, one at a time, with replacement. Draw a tree diagram for this problem. This can be an event, such as the probability of rainy weather, or. Only stopping at one set. 2857, so the answer is 0. 5(a) In the space below, draw a probability tree diagram to represent this information [3 marks] 5(b) Calculate the probability that one red and one green ball are taken from the bag. Find the probability of: Stopping at both sets of lights. Probability tree diagrams - multiply probabilities along the branches and add probabilities in columns. p(A n B) = 0. A tree diagram is a special type of graph used to determine the outcomes of an experiment. Without independence, the probability of a $$B_2$$ branch is affected by the $$B_1$$ that precedes it. What is the probability of picking a green and then a purple skittle. Probability of Independent Events A bag contains 7 red marbles 6 green marbles 5 yellow marbles and 2 orange marbles. If you want a complete lesson, a Tarsia jigsaw, or a fun and engaging lesson activity, then you have come to the right place!. simple event 1. The student will appraise the differences between the two estimates. Tree Diagrams A tree diagram is a way of seeing all the possible probability 'routes' for two (or more) events. The following tree diagram generated by clicking the Draw button shows in color the node numbers for the tree described previously. Step 1: Draw the Probability Tree Diagram and write the probability of each branch. 2 Sam's throw 0. Which tree diagram shows the correct probabilities for this situation?. Play this game to review Probability. [1] Probability is quantified as a number between 0 and 1 (where 0 indicates impossibility and 1 indicates certainty). The root node is 1. Using a tree diagram or another suitable method, calculate the theoretical probability distribution for the number of blue marbles in a sample of 3 marbles selected from the urn, with replacement. 01) Calculate probabilities from Venn diagrams and tables ( Video 8. Tree Diagram Definition Math Bestmaths. The probability of each outcome is written on its branch. [3] All the discs are replaced in the bag. Probability. For example, Figure 1 is an illustration of conditional probability. Solution for Use Bayes' theorem or a tree diagram to calculate the indicated probability. These possible outcomes can be shown by the branches of a tree-like diagram called ‘Tree-Diagram’ or ‘Branch Diagram’. Geometric / geometric: If X 1 and X 2 are geometric random variables with probability of success p 1 and p 2 respectively, then min(X 1, X 2) is a geometric random variable with probability of success p = p 1 + p 2 – p 1 p 2. In this explainer, we will learn how to use tree diagrams to calculate conditional probabilities. An experiment consists of rolling a red die and a green die and noting the result of each roll. The probability that the first marble is red and the second white. The following tree diagram shows the probabilities when a coin is tossed two times. All outcomes must be shown from each node. Teach your students how to complete and find probabilities from tree diagrams both with and without replacement. Tree diagrams can make some probability problems easier to visualize and solve. (a) Fill in the appropriate probabilities on the tree diagram on the left above (note: the \chemistry" in the urn changes when you do not replace the rst ball drawn). There are 4 blue marbles, and 2 red marbles. 1 Multiplication of choices. 1 while on non-windy days the probability she [3 marks] catches a fish is 0. This is a complete lesson on probability trees that extends previous learning on tree diagrams to include selection without replacement. probability of getting an A. Downloadable version. 392) Two cards are drawn without replacement from a 52-card deck. Example: Probability of tossing a coin. There are two versions of random sampling: sampling with replacement and sampling without replacement. Use a tree diagram to calculate conditional probability. (1) (a) (b) Complete the tree diagram. Probability of drawing a king = 4/52 = 1/13. arrow_back Back to Tree Diagrams - conditional / without replacement Tree Diagrams - conditional / without replacement: Lessons. Transcript. A couple plan to have exactly three children. Some of the worksheets for this concept are Math mammoth statistics and probability worktext, Ma 110 work extra work 1, Grade 11 probability work work 1, Independent and dependent, Algebra 2 name date, Name period work 12 8 compound probability, 8th grade. a) Draw a tree diagram to determine ALL possible outcomes. Tree Diagram-Barron’s P. probability simulation two -way table sample space S = {H, T} tree diagram probability model replacement event P(A) complement AC disjoint mutually exclusive event Venn diagram union (or) intersection (and) conditional probability independent events general multiplication rule general addition rule. This type of diagram can be very useful for some problems. The probability of any outcome is the product of all possibilities along the relevant branches. (a) Construct a tree diagram and list the sample space. Draw a tree diagram to represent the probabilities in each case. You can choose from a blue, purple, red, or green mat and a metal or wood frame. I'm going to show you an example of modified tree diagram to solve the following question. In a group of 10 students taking the exam, there are 3 who have prepared very well, 4 well, 2 moderately well and one poorly. Least Squares Regression and Correlation. with replacement b. An online probability tree calculator for you to generate the probability tree diagram. Another counter is taken at random. Complete the probability tree that. Tree diagrams -used when given probabilities are sequential in nature 67 of the students th tic s. Draw a Venn Diagram and then find the probability of receiving an offer from either Acme Corp. Tree diagrams can make some probability problems easier to visualize and solve. Tree diagrams are useful for solving probability problems with more than one stage. Q9: A bag contains 2 black balls and 8 white balls. Age range: 14-16. (2) Jan 10. To start with, instead of looking for a matching pair, let's find the probability that both socks are red. 1 26 3 12 2 13 3 12 2 13. What is the probability that: a) a purple marble is chosen from the cup? b) a green marble is chosen? We must make a tree diagram to show this process of ﬁrst choosing the container and. Lesson Worksheet. Divide the number of events by the number of possible outcomes. Two sweets are drawn at random (i) with replacement and (ii) without replacement. First, use a tree diagram to map out the sample space: a. Check your tree against mine. You will learn how to find the probability of single or combined events using tree or sample space diagrams. For example, for the experiment "toss a coin three times and record the results from each toss", we could draw the following tree diagram. (Level 7) One ball is drawn from the bag, then another without replacement. If you want to evaluate a joint probability tree where probabilities are represented as decimals (e. Then, students draw a tree diagram for. The circle and rectangle will be explained later, and should be ignored for now. b Find the probability of selecting: i a blue marble followed by a white marble (B, W) ii 2 blue marbles iii exactly one blue marble c If the experiment was repeated with replacement, fi nd the answers to each question in part b. Tree Diagram: A jar contains 4 purple and 1 gold beads. Two sweets are selected without replacement. YOU can use shorthand like this. probability simulation two -way table sample space S = {H, T} tree diagram probability model replacement event P(A) complement AC disjoint mutually exclusive event Venn diagram union (or) intersection (and) conditional probability independent events general multiplication rule general addition rule. The probability that the weather is fine on any day is " +. Assign probabilities to outcomes and determine probabilities for events. replacement (meaning eat the skittle, then take another). (test negative but actually have the disease). to find the probability of event C or event D happening add probabilities down the tree. Since both combined events are the same (just the other way around), the answers are identical. When working with conditional probabilities, it is helpful to use a tree diagram to illustrate the probability of the different outcomes. calculate and interpret conditional probabilities through representation using expected frequencies with two-way tables, tree diagrams and Venn diagrams. Students may use other methods. doc 1_Double_Spinners. There are 12 possible outfits for the student to wear. G2 = second card is green. Are they dependent or independent events? 1) You roll two dice. When two balls are chosen at random without replacement from bag B, the probability that they are both white is $$\frac{2}{7}$$. Tree diagrams can make some probability problems easier to visualize and solve. Now, for the conditional probability we want to view that 3∕4 as if it was 1 whole, which we achieve by multiplying by its reciprocal, namely 4∕3. tree diagram. Disjoint Events (Revisited) Drawing with and without Replacement Making a Picture –Venn Diagrams, Probability Tables, and Tree Diagrams. Draw two marbles, one at a time, this time without replacement from the urn. The first step to solving a probability problem is to determine the probability that you want to calculate. (a) Draw a tree diagram to represent all the possible paths that the mouse could take. 7E-19 Three desperados A, B and C play Russian roulette in which they take turns pulling the trigger of a six-cylinder revolver loaded with one bullet. We will then draw two balls from the chosen box, without replacement and with equal probability on those remaining. Let's consider another example: Example 2: What is the probability of drawing a king and a queen consecutively from a deck of 52 cards, without replacement. The probability that it is on time on any day is 0. a Draw a tree diagram showing all outcomes and probabilities. If we select 4 computers at random from the distribution center (with replacement) what is the probability that at least 1 of the computers is a tablet computer? P. A tree diagram is a special type of graph used to determine the outcomes of an experiment. two bills without replacement, determine whether the probability that the bills will total $15 is greater than the probability that the bills will total$2. 18 Outcomes & Probability Third Pick First Pick Second Pick Figure 8: Tree diagram for selecting three sweets randomly (with probability value) e) Probability distribution for each flavour if three sweets are. We pick a card, write down what it is, then put it back in the deck and draw again. Probability & Tree Diagrams. There is a total of 3 colour sequences through which we end up with 1 of. MEMORY METER. Related Topics: algebra, dependent, independent, input, output, probability, variable. Find the probability that: a) Both Adam and Beth hit with their first dart b) At least one of them hits with their first dart B hit P(hit, hit) = 0. It consists of “branches” that are labeled with either frequencies or probabilities. Homework x1. The number of "Male and Smoke" divided by the total = 19/100 = 0. T and H (in any order)? 3. Draw a tree diagram representing the results. Draw a probability tree to show this situation and find the probability that the kicker is sucessful with the penalty. Use two-way tables to calculate conditional. Marbles are drawn vice with replacement What green) A 1 о в Bliss. (a) P(C\A) = 0. Assigned Practices: 1. From the tree diagram, we can see that there is a total of 8 different possible outcomes. and Mills Inc. Click Image to Enlarge : Use a tree diagram to display possible outcomes of who will come to the party. In other cases, different problem. Just like a tree, tree diagrams branch out and can become quite intricate. The probability that the woman we draw is not married is, by the complement rule, P(not married) = 1 – P(married) = 1 – 0. LC Probability & Statistics; LC Functions & Calculus; LC HL Self-Directed Quizzes; LC OL Self-Directed Quizzes; Junior Cycle Assessments Show sub menu. For that to happen you need the 1st card to be a Queen and the second card to be a Queen. (c) at least 2 tails, (d) 2 tails in succession 1 (e) 2tails. Try these multiple choice questions. In an urn, there are 11 balls. Probability tree diagrams. Given that attorneys must frequently make decisions in environments of uncertainty, probability can be a useful skill for law students to learn. Tree diagrams – no replacement – V2; 5. 1 while on non-windy days the probability she [3 marks] catches a fish is 0. The correct answer is A. What is a tree diagram? Why is it helpful?. Conditional probability, and Bayes' Theorem, are important sub-topics. Table of Contents. with replacement P(R 1 st draw, B 2 nd draw) P(Br 1 draw, Br 2 nd draw) b. Subsection 3. Using a tree diagram, find the probability that the second marble is red, given that the first one is red. The student will appraise the differences between the two estimates. It's just the whole space, in fact, to the probability of and Crime and B, what's the probability of A and B in this sequel to the probability of me. For n prizes and boxes, you end up pruning nn — n! branches. com for - Lessons and worksheets suitable for the 9 - 1 GCSE Specification - A-Level teaching resources for Core. Rules of Differentiation. In this tree diagrams worksheet, students solve and complete 2 different problems First, they draw a tree diagram for selecting two marbles with replacement and find the other probabilities. a) Tree diagram for the experiment. Plan Objectives 1 To find the. Each branch is a possible outcome and is labelled with a probability. A tree diagram is a special type of graph used to determine the outcomes of an experiment. Play this game to review Mathematics. A disc is chosen at random from the bag and the colour is noted. "With replacement" means that you put the first ball back in the. Draw a tree diagram of the situation. Age range: 14-16. A second sweet is then removed. It consists of "branches" that are labeled with either frequencies or probabilities. To help understand this, let’s first recall the formula for conditional probability. OLVER EDUCATION. 21 shows these four outcomes and their probabilities. To find the P(QQQ), we find the probability of drawing the first queen which is 4/52. An experiment consists of rolling a red die and a green die and noting the result of each roll. Finally, find the probability of ending at each gate. A tree diagram is a special type of graph used to determine the outcomes of an experiment. The correct answer is A. a) Tree diagram for the experiment. Forexample,ifyoufliponefaircoin, S ={ H , T }where. b) How many outcomes are in the sample space? Exercise 7. Two marbles are selected without replacement. Tree Diagrams can be used to represent the total possible outcomes when you have 2 or more events. Whoops! There was a problem previewing Conditional probability. JC Number; JC Geometry & Trigonometry; JC Algebra & Functions; JC Probability & Statistics; JC Unifying Strand; TY Ideas; TY Analytics Module; Make-a-Mock; Branches; Newsletters; Events; Teacher. Draw a tree diagram to represent the probabilities in each case. Hint: The 4 digits in this probability add to 18. Unit 11 Day 3 Conditional Probability (3). 7, P (A') =. I introduce you to new notation which I would encourage you to do as it will help with conditional probability at. Draw a tree representing the possible mutually exclusive outcomes 2. Draw a probability tree showing the possible outcomes. Learning Outcomes As a result of studying this topic, students will be able to • understand and use the following terminology: trial, outcome, set of all possible outcomes, relative frequency, event, theoretical probability, If you require probability tree diagram worksheets with answers, or probability maths questions and answers you can. A girls' choir is choosing a concert uniform. Determine the probability of getting 2 heads in two successive tosses of a balanced coin. 🚨 Claim your spot here. So we are calculating 99% of 10% which is 0. Let us take note that two cards, one at a time, are drawn at random from the box. Assign students to choose four of their shirts and four pairs of pants. 5(a) In the space below, draw a probability tree diagram to represent this information [3 marks] 5(b) Calculate the probability that one red and one green ball are taken from the bag. WITH REPLACEMENT: Find the probability of drawing three queens in a row, with replacement. Find the probability of the first marble being green and the second marble being yellow. Tree diagrams and conditional probability. 3: Tree diagram for two draws without replacement, values rounded. Tree diagrams are a tool to organize outcomes and probabilities around the structure of the data. The probability that Pat will arrive late is 0. 1 = 1\$ To find the probability of any path, multiply the probabilities on the corresponding branches. Example 1. (a) Draw a tree diagram and from it write. 1 Sampling without replacement Example 3 A box contains 4 red marbles and 3 white ones. 1 Randomness, Probability, and Simulation (pp. A tree diagram is a special type of graph used to determine the outcomes of an experiment. Then a second marble is chosen at random. One ball is drawn from the bag, then another without replacement. a) Draw a tree diagram for this experiment b) Find the probability that at least one of the two persons favors genetic engineering. Replacement and Probability. A Tree Diagram and Sample Space A tree diagram is a graphic representation of the step by step competion of an experiment showing all possible results of each step. It is generally drawn from a starting point on the left and then branches to the right with each possilbe outcome shown as the end of the next branch or bridge. So, the probability that the student doesn't know the answer AND answers correctly is. Find and create gamified quizzes, lessons, presentations, and flashcards for students, employees, and everyone else. If it does rain, Mudlark will start favourite in the horse race, with probability of winning. Probability Trees RAG. I missed out tree diagrams without replacement. You spin the spinner once. Two are taken from the box, without replacement. Intuitive conditional probability seemingly not working. tree diagram. Two marbles are chosen without replacement. In this video, we will learn how to use tree diagrams to calculate conditional probabilities. In an urn, there are 11 balls. Let's consider another example: Example 2: What is the probability of drawing a king and a queen consecutively from a deck of 52 cards, without replacement. Question 1: Find the probability that a player selects two red counters. Draw a tree diagram representing the results. Using a tree diagram, find the probability that the second marble is red, given that the first one is red. Ron has a bag containing 3 green pears and 4 red pears. b) How many outcomes have a sum of the 2 numbers greater than or equal. iii: Find the probability that both biscuits are plain. Find the probability that: both. How easy is it? Simply open one of the tree diagram templates included, input your information and let SmartDraw do the rest. A tree diagram is a special type of graph used to determine the outcomes of an experiment. I built Diagnostic Questions to help you identify, understand and resolve key misconceptions. So don’t let your student become confused by probability, our probability activities are probably the best resources available. eBook: Read p. Probability: Venn Diagrams and Two-Way Tables. Draw a tree diagram to represent the probabilities in each case. If the weather is fine, the probability that Carlos is late arriving at school is !!*. b) How many outcomes are in the sample space? Exercise 7. 392) Two cards are drawn without replacement from a 52-card deck. "With replacement" means that you put the first ball back. Tree diagrams can make some probability problems easier to visualize and solve. This probability is found by using the tree like a probability distribution table, simply identify the leaves that have this event (F), and then sum their probabilities. The probability that it will be windy on a particular day is 0. Make a tree diagram for an experiment that consists of two trials. To answer how likely a patient is to have TB given a positive test result, we need to “flip” the tree. [2] [3] The higher the probability of an event, the more certain we are that the event will occur. Determine the following geometric. MEMORY METER. It consists of "branches" that are labeled with either frequencies or probabilities. • Tree diagram - a diagram which can be helpful in illustrating possible outcomes of an experiment. Example: Probability of tossing a coin. Draw a probability tree showing the possible outcomes. The following example illustrates how to use a tree diagram. Since it's with replacement the first time i'm drawing, the probability would be 2 9 and the second time would also be 2 9 which would be 4 81. Find the probability that: both. SEE MORE : 9. Visit weteachmaths. I don't know how to write out a tree diagram on here, but I think this one is heads -> heads, tails -> math probabilty- please help. (1 mark) (ii) What is the probability that a student fails to gain a certificate? (2 marks) (b) Three students take the exam. Tree diagrams can make some probability problems easier to visualize and solve. The breakdown of the lot size and the sample size in the numerator and denominator of (3. one green ball and one blue ball. Tree Diagrams A tree diagram is a way of seeing all the possible probability 'routes' for two (or more) events. Resource type: Lesson (complete) 4. Therefore, in a family of three children, the probability of having three girls is 1 out of 8. It consists of "branches" that are labeled with either frequencies or probabilities. With Replacement Without Replacement P(BL1 and BL2): P(BL1 and BR2 or BR1 and BL2): P(BL1 and O2 ): P(O2 |BL1):. 34, and the probability of selecting a black marble on the first draw is 0. (This path has been drawn on the tree diagram with arrows. the form of a tree diagram or table Æ express the probability of an event as a fraction, a decimal, and a percent independent events • results for which the outcome of one event has no eff ect on the outcome of another event • ruler probability • the likelihood or chance of an event occurring Determining Probabilities Using Tree Diagrams. 5 (since the probability of getting a heads on the first flip is 0. These explanations and tutorials will help you find the probability of all sorts of events, from rolling a number on a die to winning the lottery. We sample two items from the box without replacement. If A and B are independent (that is, the occurrence of a specific one of these two events does not influence the probability of the other event), then. Since nn grows much, much faster, than n!, Z’s algorithm becomes prohibitively tedious in a hurry. Questions 28 – 29 refer to the following probability tree diagram which shows tossing an unfair coin FOLLOWED BY drawing one bead from a cup containing 3 red (R), 4 yellow (Y) and 5 blue (B. ii: Write down the value of b. 368 #6 A plumbing contractor obtains 60% of her boiler circulators from a company whose defect rate is. (iii) the product of the two numbers is at least 5. (2) Jan 10. 7, and the probability that Jamie will pass. 🚨 Claim your spot here. It contains example problems with replacement / independent events and wit. The possibilities are: 4 H, 3 H and 1 T (in various orders), 2 H and 2 T (in various orders), 1 H and 3 T (in various orders), or 4 T. Tree Diagrams •Sample spaces can also be described graphically with tree diagrams. This item is taken from IGCSE Mathematics (0580) Paper 43 of May/June 2013. This site is the homepage of the textbook Introduction to Probability, Statistics, and Random Processes by Hossein Pishro-Nik. If it is fine he only has a 1 in 20 chance of winning. The number of "Male and Smoke" divided by the total = 19/100 = 0. Submitted by Hannah Yates on 6 March 2017. Draw a tree diagram showing the possible outcomes. 13 Outcomes & Probability Third Pick Second Pick First Pick BBB (0. Construct two tree diagrams (one for with replacement and the other for without replacement) showing the drawing of two M&Ms, one at a time, from the M&Ms you were given, as recorded in the table above.


A triangular pyramid is constructed using spherical balls so that each ball rests on exactly three balls of the next lower level.

Then, we calculate the number of paths leading from the apex to each position: A path starts at the apex and progresses downwards to any of the three spheres directly below the current position. Consequently, the number of paths to reach a certain position is the sum of the numbers immediately above it (depending on the position, there are up to three numbers above it).

The result is Pascal’s pyramid and the numbers at each level n are the coefficients of the trinomial expansion $(x + y + z)^n$. How many coefficients in the expansion of $(x + y + z)^{200000}$ are multiples of $10^{12}$?

## Solution Using the Multinomial Theorem

The generalization of the binomial theorem is the multinomial theorem. It says that multinomials raised to exponents can be expanded using the formula
$(x_1+x_2+\cdots+x_m)^n=\sum_{{k_1+k_2+\cdots+k_m=n}\atop{0\le k_i\le n}}\left({n}\atop{k_1,k_2,\ldots,k_m}\right)\prod_{1\le t\le m}x_t^{k_t}$
where
$\left({n}\atop{k_1,k_2,\ldots,k_m}\right)=\frac{n!}{k_1!k_2!\cdots k_m!}.$
Of course, when m=2 this gives the binomial theorem. The sum is taken over all partitions $k_1+k_2+\cdots+k_m=n$ for integers $k_i$. If n=200000 abd m=3, then the terms in the expansion are given by
$\left({200000}\atop{k_1,k_2,k_3}\right)x_1^{k_1}x_2^{k_2}x_3^{k_3}=\frac{200000!}{k_1!k_2!k_3!}x_1^{k_1}x_2^{k_2}x_3^{k_3}$
where $k_1+k_2+k_3=200000$. It’s worth pointing out that each of the coefficients is an integer, and thus has a unique factorization into products of prime integers. Of course, there’s no way that we’re going to calculate these coefficients. We only need to know when they’re divisible by $10^{12}$. Thus, we only need to consider how many factors of 2 and 5 are involved.

First, we’ll create a function $p(n,d)$ that outputs how many factors of $d$ are included in $n!$. We have that
$p(n,d)=\left\lfloor\frac{n}{d}\right\rfloor+\left\lfloor\frac{n}{d^2}\right\rfloor+\left\lfloor\frac{n}{d^3}\right\rfloor+ \cdots+\left\lfloor\frac{n}{d^r}\right\rfloor,$
where $d^r$ is the highest power of $d$ dividing $n$. For instance, there are 199994 factors of 2 in 200000!. Since we’re wondering when our coefficients are divisible by $10^{12}=2^{12}5^{12}$, we’ll be using the values provided by $p(n,d)$ quite a bit for $d=2$ and $d=5$. We’ll store two lists:
$p2=[p(i,2)\text{ for }1\le i\le 200000]\quad\text{and}\quad p5=[p(i,5)\text{ for }1\le i\le 200000].$
For a given $k_1,k_2,k_3$, the corresponding coefficient is divisible by $10^{12}$ precisely when
$p2[k_1]+p2[k_2]+p2[k_3]<199983\ \text{and}\ p5[k_1]+p5[k_2]+p5[k_3]<49987.$
That is, this condition ensures that there are at least 12 more factors of 2 and 5 in the numerator of the fraction defining the coefficients.

Now, we know that $k_1+k_2+k_3=200000$, and we can exploit symmetry and avoid redundant computations if we assume $k_1\le k_2\le k_3$. Under this assumption, we always have
$k_1\le\left\lfloor\frac{200000}{3}\right\rfloor=66666.$
We know that $k_1+k_2+k_3=200000$ is impossible since 200000 isn't divisible by 3. It follows that we can only have (case 1) $k_1=k_2 < k_3$, or (case 2) $k_1 < k_2=k_3$, or (case 3) $k_1 < k_2 < k_3$.

In case 1, we iterate $0\le k_1\le 66666$, setting $k_2=k_1$ and $k_3=200000-k_1-k_2$. We check the condition, and when it is satisfied we record 3 new instances of coefficients (since we may permute the $k_i$ in 3 ways).

In case 2, we iterate $0\le k_1\le 66666$, and when $k_1$ is divisible by 2 we set $k_2=k_3=\frac{200000-k_1}{2}$. When the condition holds, we again record 3 new instance.

In case 3, we iterate $0\le k_1\le 66666$, and we iterate over $k_2=k_1+a$ where $1\le a < \left\lfloor\frac{200000-3k_1}{2}\right\rfloor$. Then $k_3=200000-k_1-k_2$. When the condition holds, we record 6 instances (since there are 6 permutations of 3 objects).

## Cython Solution

I’ll provide two implementations, the first written in Cython inside Sage. Then, I’ll write a parallel solution in C.

%cython

import time
from libc.stdlib cimport malloc, free

cdef unsigned long p(unsigned long k, unsigned long d):
cdef unsigned long power = d
cdef unsigned long exp = 0
while power <= k:
exp += k / power
power *= d
return exp

cdef unsigned long * p_list(unsigned long n, unsigned long d):
cdef unsigned long i = 0
cdef unsigned long * powers = <unsigned long *>malloc((n+1)*sizeof(unsigned long))
while i <= n:
powers[i] = p(i,d)
i += 1
return powers

run_time = time.time()

# form a list of number of times each n! is divisible by 2.
cdef unsigned long * p2 = p_list(200000,2)

# form a list of number of times each n! is divisible by 5.
cdef unsigned long * p5 = p_list(200000,5)

cdef unsigned long k1, k2, k3, a
cdef unsigned long long result = 0

k1 = 0
while k1 <= 66666:
# case 1: k1 = k2 < k3
k2 = k1
k3 = 200000 - k1 - k2
if 199982 >= (p2[k1]+p2[k2]+p2[k3]) and 49986 >= (p5[k1]+p5[k2]+p5[k3]):
result += 3
# case 2: k1 < k2 = k3
if k1 % 2 == 0:
k2 = (200000 - k1)/2
k3 = k2
if 199982 >= (p2[k1]+p2[k2]+p2[k3]) and 49986 >= (p5[k1]+p5[k2]+p5[k3]):
result += 3
# case 3: k1 < k2 < k3
a = 1
while 2*a < (200000 - 3*k1):
k2 = k1 + a
k3 = 200000 - k1 - k2
if 199982 >= (p2[k1]+p2[k2]+p2[k3]) and 49986 >= (p5[k1]+p5[k2]+p5[k3]):
result += 6
a += 1
k1 += 1

free(p2)
free(p5)

elapsed_run = round(time.time() - run_time, 5)

print "Result: %s" % result
print "Runtime: %s seconds (total time: %s seconds)" % (elapsed_run, elapsed_head)

When executed, we find the correct result relatively quickly.

Result: 479742450
Runtime: 14.62538 seconds (total time: 14.62543 seconds)

## C with OpenMP Solution

#include <stdio.h>
#include <stdlib.h>
#include <malloc.h>
#include <omp.h>

/*****************************************************************************/
/* function to determine how many factors of 'd' are in 'k!'                 */
/*****************************************************************************/
unsigned long p(unsigned long k, unsigned long d) {
unsigned long power = d;
unsigned long exp = 0;
while (power <= k) {
exp += k/power;
power *= d;
}
return exp;
}

/*****************************************************************************/
/* create a list [p(0,d),p(1,d),p(2,d), ... ,p(n,d)] and return pointer      */
/*****************************************************************************/
unsigned long * p_list(unsigned long n, unsigned long d) {
unsigned long i;
unsigned long * powers = malloc((n+1)*sizeof(unsigned long));
for (i=0;i<=n;i++) powers[i] = p(i,d);
return powers;
}

/*****************************************************************************/
/* main                                                                      */
/*****************************************************************************/
int main(int argc, char **argv) {
unsigned long k1, k2, k3, a;
unsigned long long result = 0;

unsigned long * p2 = p_list(200000, 2);
unsigned long * p5 = p_list(200000, 5);

#pragma omp parallel for private(k1,k2,k3,a) reduction(+ : result)
for (k1=0;k1<66667;k1++) {
// case 1: k1 = k2 < k3
k2 = k1;
k3 = 200000 - k1 - k2;
if (p2[k1]+p2[k2]+p2[k3]<199983 && p5[k1]+p5[k2]+p5[k3]<49987) {
result += 3;
}
// case 2: k1 < k2 = k3
if (k1 % 2 == 0) {
k2 = (200000 - k1)/2;
k3 = k2;
if (p2[k1]+p2[k2]+p2[k3]<199983 && p5[k1]+p5[k2]+p5[k3]<49987) {
result += 3;
}
}
// case 3: k1 < k2 < k3
for (a=1;2*a<(200000-3*k1);a++) {
k2 = k1 + a;
k3 = 200000 - k1 - k2;
if (p2[k1]+p2[k2]+p2[k3]<199983 && p5[k1]+p5[k2]+p5[k3]<49987) {
result += 6;
}
}
}

free(p2);
free(p5);

printf("result: %lld\n", result);

return 0;
}

This can be compiled and optimized using GCC as follows.

$gcc -O3 -fopenmp -o problem-154-omp problem-154-omp.c When executed on a 16-core machine, we get the following result.$ time ./problem-154-omp
result: 479742450

real    0m1.487s

This appears to be the fastest solution currently known, according to the forum of solutions on Project Euler. The CPUs on the 16-core machine are pretty weak compared to modern standards. When running on a single core on a new Intel Core i7, the result is returned in about 4.7 seconds.

### Motivation

Many interesting computational problems, such as those on Project Euler require that one find the sum of proper divisors of a given integer. I had a fairly crude brute-force method for doing this, and was subsequently emailed a comment by Bjarki Ágúst Guðmundsson who runs the site www.mathblog.dk. He pointed me in the direction of this page and provided some sample code illustrating how such an approach runs asymptotically faster than the approach I had been taking. Awesome! I’m going to expand on that a bit here, providing some mathematical proofs behind the claims and providing code for those who may want to take advantage of this.

### The Mathematics Behind It All

Let the function $\sigma(n)$ be the sum of divisors for a positive integer $n$. For example,
$\sigma(6)=1+2+3+6=12.$

It should seem obvious that for any prime $p$ we have $\sigma(p)=1+p$. What about powers of primes? Let $\alpha\in\mathbb{Z}_+$, and then
$\sigma(p^\alpha)=1+p+p^2+\cdots+p^\alpha.$

We’d like to write that in a closed form, i.e., without the “$+\cdots+$”. We use a standard series trick to do that.

\begin{align} \sigma(p^\alpha) &= 1+p+p^2+\cdots+p^\alpha\cr p\sigma(p^\alpha) &= p+p^2+p^3+\cdots+p^{\alpha+1}\cr p\sigma(p^\alpha)-\sigma(p^\alpha) &= (p+p^2+\cdots+p^{\alpha+1})-(1+p+\cdots+p^\alpha)\cr p\sigma(p^\alpha)-\sigma(p^\alpha) &= p^{\alpha+1}-1\cr (p-1)\sigma(p^\alpha) &= p^{\alpha+1}-1\cr \sigma(p^\alpha) &=\frac{p^{\alpha+1}-1}{p-1}.\end{align}

That solves the problem of finding the sum of divisors for powers of primes. It would be nice if we could show that $\sigma$ is multiplicative on powers of primes, i.e., that $\sigma(p_1^{\alpha_1}p_2^{\alpha_2})=\sigma(p_1^{\alpha_1})\sigma(p_2^{\alpha_2})$. We’ll prove that this is the case, and solve the problem in general along the way.

Proposition: The function $\sigma$ is multiplicative on powers of primes.

Proof: Let $n$ be a positive integer written (uniquely, by the fundamental theorem of arithmetic) as
$n=\prod_{i=1}^m p_i^{\alpha_i}$
for $m$ distinct primes $p_i$ with $\alpha_i\in\mathbb{Z}_+$. Any divisor $k$ of $n$ then has the form
$k=\prod_{i=1}^m p_i^{\beta_i}$
where each $\beta_i$ satisfies $0\le\beta_i\le\alpha_i$. Then
$\sigma(n)=\sigma\left(\prod_{i=1}^m p_i^{\alpha_i}\right)$
is the sum of all divisors $k$ of $n$ and can be written by summing over all possible combinations of the exponents $\beta_i$. There are $\prod_{i=1}^m \alpha_i$ combinations, and we can form their sum and simplify it as follows.
\begin{align}\sigma(n) &= \sum_{1\le i\le m,\ 0\le\beta_i\le\alpha_i}p_1^{\beta_i}p_2^{\beta_2}\cdots p_m^{\beta_m}\cr &= \sum_{\beta_1=0}^{\alpha_1}p_1^{\beta_1}\left(\sum_{2\le i\le m,\ 0\le\beta_i\le\alpha_i}p_2^{\beta_2}p_3^{\beta_3}\cdots p_m^{\beta_m}\right)\cr &= \sum_{\beta_1=0}^{\alpha_1}p_1^{\beta_1}\sum_{\beta_2=0}^{\alpha_2}p_2^{\beta_2}\left(\sum_{3\le i\le m,\ 0\le\beta_i\le\alpha_i}p_3^{\beta_3}p_4^{\beta_4}\cdots p_m^{\beta_m}\right) \cr &= \vdots\cr &=\sum_{\beta_1=0}^{\alpha_1}p_1^{\beta_1}\sum_{\beta_2=0}^{\alpha_2}p_2^{\beta_2}\sum_{\beta_3=0}^{\alpha_3}p_3^{\beta_3}\ \cdots\ \sum_{\beta_m=0}^{\alpha_m}p_m^{\beta_m}\cr &=\sigma(p_1^{\alpha_1})\sigma(p_2^{\alpha_2})\cdots\sigma(p_m^{\alpha_m}).\end{align}
This completes the proof. Q.E.D.

Thus, we now have a formula for the sum of divisors of an arbitrary positive integer $n$ using the factorization of $n$. Namely,
$\sigma(n)=\sigma\left(\prod_{i=1}^m p_i^{\alpha_i}\right)=\prod_{i=1}^m\left(\frac{p_i^{\alpha_i+1}-1}{p_i-1}\right).$

This is something I use quite a bit for various problems and programming exercises, so I figured I could post it here. It’s a basic post that isn’t advanced at all, but that doesn’t mean that the implementation given below won’t save work for others. The idea is to create a list of primes in C by malloc’ing a sieve, then malloc’ing a list of specific length based on that sieve. The resulting list contains all the primes below a given limit (defined in the code). The first member of the list is an integer representing the length of the list.

#include <stdio.h>
#include <stdlib.h>
#include <malloc.h>

#define                 bool    _Bool

static unsigned long    prime_limit = 1000000;

unsigned long sqrtld(unsigned long N) {
int                 b = 1;
unsigned long       res,s;
while(1<<b<N) b+= 1;
res = 1<<(b/2 + 1);
for(;;) {
s = (N/res + res)/2;
if(s>=res) return res;
res = s;
}
}

unsigned long * make_primes(unsigned long limit) {
unsigned long      *primes;
unsigned long       i,j;
unsigned long       s = sqrtld(prime_limit);
unsigned long       n = 0;
bool *sieve = malloc((prime_limit + 1) * sizeof(bool));
sieve[0] = 0;
sieve[1] = 0;
for(i=2; i<=prime_limit; i++) sieve[i] = 1;
j = 4;
while(j<=prime_limit) {
sieve[j] = 0;
j += 2;
}
for(i=3; i<=s; i+=2) {
if(sieve[i] == 1) {
j = i * 3;
while(j<=prime_limit) {
sieve[j] = 0;
j += 2 * i;
}
}
}
for(i=2;i<=prime_limit;i++) if(sieve[i]==1) n += 1;
primes = malloc((n + 1) * sizeof(unsigned long));
primes[0] = n;
j = 1;
for(i=2;i<=prime_limit;i++) if(sieve[i]==1) {
primes[j] = i;
j++;
}
free(sieve);
return primes;
}

int main(void) {

unsigned long * primes = make_primes(prime_limit);

printf("There are %ld primes <= %ld\n",primes[0],prime_limit);

free(primes);
return 0;
}

Say one wanted to form a list of all primes below 1,000,000. That’s what the above program does by default, since “prime_limit = 1000000.” If one compiles this and executes, you would get something like what follows. The timing is relatively respectable.

$gcc -O3 -o prime-sieve prime-sieve.c$ time ./prime-sieve
There are 78498 primes <= 1000000

real	0m0.008s
user	0m0.004s
sys	0m0.000s

The code is linked here: prime-sieve.c


# Math Help - Quickie #13

1. ## Quickie #13

This is a classic (very old).

An irrational number raised to an irrational power is always irrational.

Prove it or provide a counterexample.

2. Seen it. But I did not know it was a classic.

3. Originally Posted by Soroban
This is a classic (very old).

An irrational number raised to an irrational power is always irrational.

Prove it or provide a counterexample.

I'll give several counter examples;

For the most part,

[sqrt(x)^(sqrt(x))]^(sqrt(x)), where x is even, with the exception of when x is a square.

EX:

[sqrt(2)^(sqrt(2))]^(sqrt(2))

[sqrt(2)^(sqrt(2))] = 2^(sqrt(2)/2)

[2^(sqrt(2)/2)]^(sqrt(2)) = 2

4. Originally Posted by AfterShock
I'll give several counter examples;

For the most part,

[sqrt(x)^(sqrt(x))]^(sqrt(x)), where x is even, with the exception of when x is a square.

EX:

[sqrt(2)^(sqrt(2))]^(sqrt(2))

[sqrt(2)^(sqrt(2))] = 2^(sqrt(2)/2)

[2^(sqrt(2)/2)]^(sqrt(2)) = 2
Interesting, but you are assuming that $\sqrt{2}^{\sqrt{2}}$ is irrational, else this isn't a counterexample. I have little doubt that it is, but this isn't a good counterexample without proving this fact.

-Dan

5. Originally Posted by topsquark
Interesting, but you are assuming that $\sqrt{2}^{\sqrt{2}}$ is irrational, else this isn't a counterexample. I have little doubt that it is, but this isn't a good counterexample without proving this fact.

-Dan
Funny thing is you do not need to. The assumption is that irrational raised to irrational is always irrational. Thus, he is using this assumption. And arrives at a contradiction.

6. Originally Posted by ThePerfectHacker
Funny thing is you do not need to. The assumption is that irrational raised to irrational is always irrational. Thus, he is using this assumption. And arrives at a contradiction.
Aaaaah! I get it now.

-Dan

7. The "classic" solution goes like this:

Theorem: . $(\text{irrational})^{\text{irrational}}$ can be rational.

Proof

$\text{Consider }a \:=\:\sqrt{2}^{\sqrt{2}}$

There are only two possibilities:

. . (1) $a$ is rational.

. . (2) $a$ is irrational.

If (1) $a$ is rational, the theroem is verified.

If (2) $a$ is irrational, consider: . $a^{\sqrt{2}}$
. . an irrational number raised to an irrational power.

Then we have: . $a^{\sqrt{2}} \:=\:\left(\sqrt{2}^{\sqrt{2}}\right)^{\sqrt{2}} \:=\:\left(\sqrt{2}\right)^2\:=\:2$ . . . a rational number.

Either way, an irrational raised to an irrational power can be rational.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I've always found this proof amusing.

Note that we still don't know if $\sqrt{2}^{\sqrt{2}}$ is rational or irrational
. . but it doesn't matter . . .

8. I found a different counterexample.

$e$ is irrational.

$\ln 2$ is irrational (but do not know how to show it).

Then,
$e^{\ln 2}=2$


A simple non-planar graph with minimum number of vertices is the complete graph K 5. A complete graph is a graph in which every vertex has an edge to all other vertices is called a complete graph, In other words, each pair of graph vertices is connected by an edge. Two further examples are shown in Figure 1.14. A regular graph with vertices of degree k {\displaystyle k} is called a k {\displaystyle k} ‑regular graph or regular graph of degree k {\displaystyle k}. 3.A graph is k-regular if every vertex has degree k. How do 1-regular graphs look like? If every vertex of a simple graph has the same degree, then the graph is called a regular graph. A simple graph is called regular if every vertex of this graph has the same degree. Before you go through this article, make sure that you have gone through the previous article on various Types of Graphsin Graph Theory. An important property of graphs that is used frequently in graph theory is the degree of each vertex. C Tree. Let $G$ be a regular graph, that is there is some $r$ such that $|\delta_G(v)|=r$ for all $v\in V(G)$. 1.3 Find out whether the complete graph, the path and the cycle of order n 1 are bipartite and/or regular. 1.6.Show that if a k-regular bipartite graph with k>0 has a bipartition (X;Y), then jXj= jYj. I think you wanted to ask about a spanning 1-regular graph, also known as a perfect matching or 1-factor. This means that (assuming this is not a multigraph, no self-edges, etc) if you have n vertices, then each vertex has n-1 edges. Regular Graph c) Simple Graph d) Complete Graph … View Answer ... B Regular graph. Another plural is vertexes. hence, The edge defined as a connection between the two vertices of a graph. That is, if a graph is k-regular, every vertex has degree k. Exercises: Draw all 0-regular graphs with 1 vertex; 2 vertices; 3 vertices. If every vertex in a regular graph has degree k,then the graph is called k-regular. Conjecture 8 : Let G be a 3-regular cyclically 4-edge-connected graph of order n.Then G contains a cycle of length at least cn where c is a positive num- ber. G is said to be regular of degree r (or r-regular) if deg(v) = r for all vertices v in G. Complete graphs of order n are regular of degree n − 1, and empty graphs are regular of degree 0. Explanation: In a regular graph, degrees of all the vertices are equal. Some sources claim that the letter K in this notation stands for the German word komplett, but the German name for a complete graph, vollständiger Graph, does not contain the letter K, and other sources state that the notation honors the contributions of Kazimierz Kuratowski to graph theory. The problem seems similar to Hamiltonian Path which is NP complete problem for a general graph. How to create a program and program development cycle? yes No Not enough information to decide If Ris the equivalence relation defined by the panition {{1. A complete graph Km is a graph with m vertices, any two of which are adjacent. D n2. {5}. Explanation of Complete Graph with Diagram and Example, Explanation of Abstract Data Types with Diagram and Example, What is One Dimensional Array in Data Structure with Example, What is Singly Linked List? The first example is an example of a complete graph. What is Polynomials Addition using Linked lists With Example. Vertex Cover (VC): A vertex cover in an undirected graph G = (V;E) is a subset of vertices V0 V such that every edge in G has at least one endpoint in V0. therefore, In a directed graph, an edge goes from one vertex, the source, to another, the target, and hence makes the connection in only one direction. I'm not sure about my anwser. In the first, there is a direct path from every single house to every single other house. Every strongly regular graph is symmetric, but not vice versa. B n*n. C nn. Kn has n(nâ1)/2 edges and is a regular graph of degree nâ1. (a) every induced subgraph of a complete graph is complete; (b) every subgraph of a bipartite graph is bipartite. In the given graph the degree of every vertex is 3. The vertex is defined as an item in a graph, sometimes referred to as a node, The plural is vertices. definition. The complete graph on n vertices is denoted by Kn. 1.7.Show that, in any group of two or more people, there are always two with exactly the same number of friends inside the group. Theorem 9 : Let G be a 3-connected 3-regular graph , and let S be a set of nine vertices of G.Then G has a cycle which includes every vertex of S. (Aolton et al., 1982; Kelmans and Lomonosov, 1982) A graph of this kind is sometimes said to be an srg(v, k, λ, μ).Strongly regular graphs were introduced by Raj Chandra Bose in 1963.. A complete graph is a graph in which every vertex has an edge to all other vertices is called a complete graph, In other words, each pair of graph vertices is connected by an edge. 4. DEFINITION : Complete graph: In a graph, if there exist an edge between every pair of vertices,then such a graph is called complete graph. A connected graph may not be (and often is not) complete. In the graph, a vertex should have edges with all other vertices, then it called a complete graph. 4)A star graph of order 7. A simple graph with ‘n’ mutual vertices is called a complete graph and it is denoted by ‘K n ’. The complete graph on n vertices is denoted by Kn. Regular Graphs A graph G is regular if every vertex has the same degree. As the above graph n=7 Ans - Statement p is true. The complete bipartite graph K m, n is planar if and only if m ≤ 2 or n ≤ 2. A complete graph is a graph that has an edge between every single vertex in the graph; we represent a complete graph … Q.1. Theorem 2.4 If G is a k-regular bipartite graph with k > 0 and the bipartition of G Privacy therefore, The total number of edges of complete graph = 21 = (7)*(7-1)/2. Solution: A 1-regular graph is just a disjoint union of edges (soon to be called a matching). Complete graphs correspond to cliques. regular graph : a regular graph is a graph in which every node has the same degree • connected graph : a graph is connected if any two points can be joined by a path (a sequence of edges that are pairwise adjacent) MATH3301 EXTREMAL GRAPH THEORY Deﬂnition: A near regular complete multipartite graph is a complete multipartite graph with orders of its partite sets diﬁering by at most 1. {6} {7}} which of the graphs betov/represents the quotient graph G^R of the graph G represented below. 45 The complete graph K, has... different spanning trees? Some authors exclude graphs which satisfy the definition trivially, namely those graphs which are the disjoint union of one or more equal-sized complete graphs, and their complements, the complete multipartite graphs with equal-sized independent sets. Advantage and Disadvantages. 1 2 3 4 QUESTION 3 Is this graph regular? therefore, in an undirected graph pair of vertices (A, B) and (B, A) represent the same edge. Statement Q Is True. In graph theory, a regular graph is a graph where each vertex has the same number of neighbors; i.e. 1.8.1. A regular graph of degree r is strongly regular if there exist nonnegative integers e, d such that for all vertices u, v the number of vertices … The complete graph with n graph vertices is denoted mn. A regular directed graph must also satisfy the stronger condition that the indegree and outdegree of each vertex are equal to each other. Regular, Complete and Complete Bipartite. What is the Classification of Data Structure with Diagram, Explanation array data structure and types with diagram, Abstract Data Type algorithm brief Description with example, What is Algorithm Programming? 4.How many (labelled) graphs exist on a given set of nvertices? Important graphs and graph classes De nition. Could you please help me on Discrete-mathematical-structures. A graph and its complement. Shelly has narrowed it down to two different layouts of how she wants the houses to be connected. The set of vertices V(G) = {1, 2, 3, 4, 5} for n 3, the cycle C A regular graph is called n-regular if every vertex in this graph has degree n. Match the values of n (in the right column) for which the graphs (in the left column) are regular? 2)A bipartite graph of order 6. A 2-regular graph is a disjoint union of cycles. A complete graph is connected. Acomplete graphhas an edge between every pair of vertices. What is Data Structures and Algorithms with Explanation? Question: Let Statements P And Q Be As Follows P = "Every Complete Graph Is Regular." Any graph with 8 or less edges is planar. $\begingroup$ @Igor: I think there's some terminological confusion here - an induced subgraph of a complete graph is a complete graph... $\endgroup$ – ndkrempel Jan 17 '11 at 17:25 $\begingroup$ @ndkrempel: yes, confusion reigns. A symmetric graph is one in which there is a symmetry (graph automorphism) taking any ordered pair of adjacent vertices to any other ordered pair; the Foster census lists all small symmetric 3-regular graphs. Definition: Regular. The complete graph with n graph vertices is denoted mn. In a complete graph, for every two vertices in a graph, there is an edge that directly connects the two. the complete graph with n vertices has calculated by formulas as edges. The vertex cover problem (VC) is: given an undirected graph G and an integer k, does G have a vertex cover of size k? What are the basic data structure operations and Explanation? Every non-empty graph contains such a graph. We have discussed- 1. A single edge connecting two vertices, or in other words the complete graph K 2 on two vertices, is a 1-regular graph. Then, we have $|\delta_\bar{G}(v)|=n-r-1$, where $\bar{G}$ is the complement of $G$ and $n=|V(G)|$. $\endgroup$ – Igor Rivin Jan 17 '11 at 17:40 Any graph with 4 or less vertices is planar. Complete Graph. View Answer Answer: Tree ... Answer: The number of edges in walk W 49 If for some positive integer k, degree of vertex d(v)=k for every vertex v of the graph G, then G is called... ? A graph is called Eulerian if it has an Eulerian Cycle and called Semi-Eulerian if it has an Eulerian Path. Fortunately, we can find whether a given graph has a … graph when it is clear from the context) to mean an isomorphism class of graphs. Regular Graph - A graph in which all the vertices are of equal degree is called a regular graph. Suppose a contractor, Shelly, is creating a neighborhood of six houses that are arranged in such a way that they enclose a forested area. The line graph H of a graph G is a graph the vertices of which correspond to the edges of G, any two vertices of H being adjacent if and…. 1.4 Give the size: 1)of an r-regular graph of order n; 2)of the complete bipartite graph K r;s. For all natural numbers nwe de ne: the complete graph complete graph, K n K n on nvertices as the (unlabeled) graph isomorphic to [n]; [n] 2. They are called 2-Regular Graphs. ... A k-regular graph G is one such that deg(v) = k for all v ∈G. If all the vertices in a graph are of degree ‘k’, then it is called as a “ k-regular graph “. Aregular graphis agraphwhereevery vertex has the same degree.Therefore, every compl, Let statements p and q be as follows p = "Every complete graph is regular." A nn-2. In this article, we will discuss about Bipartite Graphs. Which of the following statements for a simple graph is correct? In a weighted graph, every edge has a number, it’s called “weight”. Output Result In both the graphs, all the vertices have degree 2. View desktop site. Properties of Regular Graphs: A complete graph N vertices is (N-1) regular. 1.8. Terms complete. In the second, there is a way to get from each of the houses to each of the other houses, but it's not necessarily … Complete Graph defined as An undirected graph with an edge between every pair of vertices. A graph G is said to be complete if every vertex in G is connected to every other vertex in G. Thus a complete graph G must be connected. therefore, the complete digraph is a directed graph in which every pair of distinct vertices is connected by a pair of unique edges (one in each direction). (Thomassen et al., 1986, et al.) …the graph is called a complete graph (Figure 13B). A K graph. 3)A complete bipartite graph of order 7. The set of edges E(G) = {(1, 2), (1, 4), (1, 5), (2, 3), (3, 4), (3, 5), (1, 3)} To calculate total number of edges with N vertices used formula such as = ( n * ( n â 1 ) ) / 2. Note: An undirected graph represented as a directed graph with two directed edges, one “to” and one “from,” for every undirected edge. A graph in which degree of all the vertices is same is called as a regular graph. Regular Graph: A graph is said to be regular or K-regular if all its vertices have the same degree K. A graph whose all vertices have degree 2 is known as a 2-regular … A graph is a collection of vertices connected to each other through a set of edges. © 2003-2021 Chegg Inc. All rights reserved. | The complete graph with n vertices is denoted by K n. The Figure shows the graphs K 1 through K 6. D Not a graph. A complete graph K n is planar if and only if n ≤ 4. 2} {3 4}. Statement P Is True. Defined Another way you can say, A complete graph is a simple undirected graph in which every pair of distinct vertices is connected by a unique edge. every vertex has the same degree or valency. Definition, Example, Explain the algorithm characteristics in data structure, Divide and Conquer Algorithm | Introduction. In this article, we will show that every bipartite graph is 2 chromatic ( chromatic number is 2 ).. A simple graph G is called a Bipartite Graph if the vertices of graph G can be divided into two disjoint sets – V1 and V2 such that every edge in G connects a vertex in V1 and a vertex in V2. q = "Every regular graph Is complete" Select the option below that BEST applies to these statements. In simple words, no edge connects two vertices belonging to the same set. Hence, the complement of $G$ is also regular. An undirected graph is defined as a graph containing an unordered pair of vertices is Know an undirected graph. 1)A 3-regular graph of order at least 5. The study of graphs is known as Graph Theory. And 2-regular graphs? The graphs in the chapter are always regular of degree r, that is, every vertex in the graph is incident to r edges in the graph. Q = "Every Regular Graph Is Complete" Select The Option Below That BEST Applies To These Statements. Let Statements P And Q Be As Follows P = "Every Complete Graph Is Regular." Kn For all n … Both statments are true Neither statement is true QUESTION 2 Find the degree of vertex 5. Statement p is true. 2. Every graph has certain properties that can be used to describe it. therefore, A graph is said to complete or fully connected if there is a path from every vertex to every other vertex. A simple graph }G ={V,E is said to be regular of degree k, or simply k-regular if for each v∈V, δ(v) =k. Statement q is true. the complete graph with n vertices has calculated by formulas as edges. & Is planar is same is called a regular graph has the same degree, then jYj! =  every complete graph K n ’ BEST Applies to These Statements each vertex are equal to each.. Between the two vertices, then the graph is defined as a connection between two!, 1986, et al. soon to be called a regular graph is complete '' the... Less vertices is planar if and only if n ≤ 4 plural is vertices has... different trees... Operations and explanation to create a program and program development cycle by Kn statments are true Neither statement true. ) = K for all n … 45 the complete graph n vertices is Know an undirected graph pair vertices., is a 1-regular graph is bipartite satisfy the stronger condition that the indegree and of... ) every subgraph of a graph G is one such that deg ( v ) = K for n... Between every pair of vertices K 2 on two vertices, is graph! Soon to be called a regular directed graph must also satisfy the condition! Vertices is denoted by Kn the path and the cycle of order 1! ) every subgraph of a bipartite graph is symmetric, but not vice versa the shows... Two of which are adjacent vertices belonging to the same degree, then graph., then the graph G is regular if every vertex is defined as perfect. The first, there is a disjoint union of edges pair of vertices the graph! Simple graph with 8 or less edges is planar if and only if n ≤ or. Many ( labelled ) graphs exist on a given set of edges a single edge connecting two vertices, in! Layouts of how she wants the houses to be connected jXj= jYj on! K > 0 has a number, it ’ s called “ weight ” graph n vertices is denoted Kn... ‘ K ’, then the graph is complete ; ( B ) and B! The vertex is defined as a connection between the two vertices belonging the! P =  every complete graph with n vertices is denoted by Kn... a bipartite! ) = K for all v ∈G ( nâ1 ) /2 edges and is a collection of vertices a! 6 } { 7 } } which of the graphs, all the vertices are of equal is... And only if n ≤ 4 every vertex has the same degree v ∈G this graph regular regular. Has an Eulerian cycle and called Semi-Eulerian if it has an Eulerian path an... N graph vertices is denoted mn G represented below not vice versa to Hamiltonian which... By K n. the Figure shows the graphs K 1 through K 6 other words complete. 4 QUESTION 3 is this graph regular with minimum number of vertices denoted! Algorithm characteristics in data structure operations and explanation directed graph must also satisfy the stronger condition the! With K > 0 has a bipartition ( X ; Y ), then it a... Follows P =  every complete graph ( Figure 13B ) of every vertex in a graph..., then it is denoted by ‘ K ’, then the graph G is regular ''! Has certain properties that can be used to describe it bipartite graphs, referred... | Introduction of graphs is known as a graph is called as a connection the... Less edges is planar 2-regular graph is called a regular directed graph must also satisfy the stronger that... Or 1-factor many ( labelled ) graphs exist on a given set of nvertices the graphs, all the are. “ weight ” a complete graph with 4 or less vertices is Know an undirected graph is a., is a graph G is regular. a node, the edge defined as a node the... Or fully connected if there is a graph with n graph vertices is denoted by K n. Figure! And ( B, a vertex should have edges with all other vertices or. Select the Option below that BEST Applies to These Statements every regular graph is complete graph regular ) a complete graph m. A single edge connecting two vertices of a bipartite graph of degree ‘ K n is planar given graph degree... To every other vertex spanning trees explanation: in a regular graph induced subgraph of a complete graph n! Then jXj= jYj is also regular. be called a matching ) we will discuss bipartite! K n. the Figure shows the graphs betov/represents the quotient graph G^R of the graphs betov/represents the quotient G^R. N is planar layouts of how she wants the houses to be called a matching ) graph it. 0 has a number, it ’ s called “ weight ” indegree and outdegree of each.. Shows the graphs, all the vertices is denoted by Kn with ‘ n ’ mutual vertices is the graph. Other vertices, any two of which are adjacent operations and explanation matching or 1-factor the to! Of Graphsin graph Theory that BEST Applies to These Statements regular graphs: a complete bipartite graph called... The stronger condition that the indegree and outdegree of each vertex which of the graph, also as... A graph are of degree ‘ K n is planar if and only if n ≤ 4 the! By K n. the Figure shows the graphs K 1 through K 6 G represented.... Called a regular graph Follows P =  every regular graph a weighted graph, also known as Theory. G^R of the graph is called k-regular Divide and Conquer algorithm | Introduction both the K., there is a 1-regular graph is symmetric, but not vice versa path and the cycle of 7., it ’ s called “ weight ” B, a ) induced... Has n ( nâ1 ) /2 edges and is a direct path from every vertex has the same.! V ) = K for all v ∈G plural is vertices the stronger condition that the and. Pair of vertices a regular graph what are the basic data structure, and. Edge connecting two vertices, any two of which are adjacent every regular graph is complete graph an isomorphism of... K 5 { { 1 ’, then jXj= jYj al. out whether complete... Edge defined as a graph defined as an item in a weighted,! N is planar minimum number of vertices the quotient graph G^R of the graph said... Or 1-factor if a k-regular graph G is regular if every vertex is 3 ) complete path. Calculated by formulas as edges a given set of edges ( soon to be connected exist on a given of! Have edges with all other vertices, then the graph G is one such that deg ( v =. P and Q be as Follows P =  every complete graph vertex is defined as item... Used frequently in graph Theory is a direct path from every single other house Graphsin graph.... Direct path from every single other house with all other vertices, is a direct path every! Has calculated by formulas as edges for n 3, the path and cycle... A 1-regular graph has... different spanning trees Ris the equivalence relation defined the. K 5 an important property of graphs that is used frequently in graph Theory equal is... \$ is also regular. said to complete or fully connected if there a! The first example is an example of a graph, a graph G is one such that deg ( )... Graph, every edge has a bipartition ( X ; Y ) then. ) /2 edges and is a direct path from every vertex to every single other house it called regular... To every single house to every single house to every single house to every other vertex other house graph! K > 0 has a bipartition ( X ; Y ), it! As edges seems similar to Hamiltonian path which is NP complete problem for a general graph vertex has the degree. Has certain properties that can be used to describe it but not vice versa graph! A weighted graph, a ) represent the same set has an Eulerian cycle and called Semi-Eulerian if has! Called as a node, the edge defined as an item in regular., et al. Addition using Linked lists with example a direct path from every single other house whether. Symmetric, but not vice versa the quotient graph G^R of the graph is ;... As a “ k-regular graph “ all other vertices, is a from... Has degree K, then it is called k-regular cycle C a graph is to! Vertex are equal QUESTION: Let Statements P and Q be as Follows =. The study of graphs has narrowed it down to two different layouts of how she wants the houses be... Q be as Follows P =  every regular graph graphhas an edge between every pair vertices. On n vertices is denoted by Kn a vertex should have edges with other... Complete ; ( B ) every induced subgraph of a bipartite graph of degree.! 1-Regular graph is regular. of vertices connected to each other through a set of edges ( to... Denoted by Kn the graphs K 1 through K 6 N-1 ).. Is every regular graph is complete graph, example, Explain the algorithm characteristics in data structure operations and?. S called “ weight ” { 6 } { 7 } } which of the graph is bipartite the graph!, it ’ s called “ weight ” graph of order 7 statments! Property of graphs is known as graph Theory a vertex should have edges with all other vertices is...


# Numerically integrate a plotted function

I used Plot[NIntegrate[...]...] to plot a function of 5 different variables. It took really long. Right now I need to integrate this function one more time over the 5th variable and plot the result. Can someone tell me if there is an option to Nintegrate the plot to speed things up? Maybe there's a way to convert the plot to a list of $x$ and $y$ and then take a sum over $(y2-y1)\,(x2-x1)$ and list plot it or something similar?

I guess it's not gonna work for me. I have the following function:

f[ρ_, ϕ_, ζ_, ξ_, r_] := 2.*100.*20.*0.02*0.2*ρ*Cos[ϕ]*Sech[20*(ρ - 1.)]^2.*ζ*Exp[-ζ]*
BesselJ[0, 100.*ζ*Sqrt[r^2. + ρ^2. - 2.*r*ρ*Cos[ϕ]]]*(Sqrt[ζ^2. + 0.02^2.]*
Cosh[10.*Sqrt[ζ^2. + 0.02^2.]*(ξ + 1.)] + ζ*Sinh[10.*Sqrt[ζ^2. + 0.02^2.]
*(ξ + 1.)])/((2.*ζ^2. + 0.02^2.)*Sinh[10.*Sqrt[ζ^2. + 0.02^2.]] +
2.*ζ*Sqrt[ζ^2. + 0.02^2.]*Cosh[10.*Sqrt[ζ^2. + 0.02^2.]])

Here the function $f$ have 5 variables: $\rho,\ \phi,\ \zeta,\ \xi,\ r.$. First, I need to integrate over $\rho,\ \phi,\ \zeta,\ \xi.$ and plot $f=f(r).$ It takes about 1 hour. I use as I said before

Plot[NIntegrate[f, {ξ, -1., 0.}, {ζ, 0., ∞}, {ϕ, -3.1415, +3.1415}, {ρ, 0., ∞}], {r, 0, 10}]

The next step is I need to integrate one more time over $r$, {r,0,r1}and plot it over {r1,0,7}. I tried what you've said, but I failed. Is your method with NDsolve going to work with this kind of function?

• See the accepted answer here. – Jason B. Jan 21 '16 at 15:32
• It seems I misunderstood your question. Can you clarify why you want to tie together the plotting w.r.t r and the integration w.r.t r? Is your motivation performance? I would expect that integrating over all 5 variables in one step with NIntegrate should be much faster than this plot was. You can also speed up plotting by limiting the number of plot points. Set MaxRecursion -> 0 and manually set the desired number of PlotPoints (these symbols can be looked up in the documentation). – Szabolcs Jan 21 '16 at 21:35
• But it would be even better to not use Plot but build a Table of the values, which you can store for later, and the ListPlot them. If the resolution of the plot is not good enough, you can compute additional points without throwing away the already computed ones. This is not (easily) possible with Plot which will always recompute everything from the start and it's not possible to interrupt it then continue if necessary (at least not without advanced hacks ...) – Szabolcs Jan 21 '16 at 21:37

The trick is to use NDSolve instead of NIntegrate and thus in effect obtain a numerical antiderivative that can be evaluated fast at different points. NIntegrate will only do definite integration, so it needs to be run each time the integration bounds are changed. This is very slow, as you noticed. NDSolve will only need to be run once.

### Slow way (what you used)

We are going to integrate f:

f[x_] := Sin[x^2]

F[y_?NumericQ] := NIntegrate[f[x], {x, 0, y}]

Plot[F[x], {x, 0, 5}] // AbsoluteTiming

It took 5 seconds on my machine.

NIntegrate[F[x], {x, 0, 5}] // AbsoluteTiming
(* {0.686052, 2.63519} *)

Both integration and plotting are slow with this method. For complex functions, they can be prohibitively slow.

### Fast way (NDSolve)

iF = NDSolveValue[{F2'[x] == f[x], F2[0] == 0}, F2, {x, 0, 5}]

iF is an interpolating function.

Plot[iF[x], {x, 0, 5}] // AbsoluteTiming

Here's a not so widely known trick: the antiderivative of an interpolating function can be computed exactly using Integrate (not NIntegrate). It is returned by Mathematica as another InterpolatingFunction object. This is possible because interpolating functions are really just piecewise polynomials. Derivatives can be taken too.

So we can compute the second integral directly (and very quickly) as

Integrate[iF[x], {x, 0, 5}]

(* 2.63519 *)

Another trick: To get the antiderivative as a function object directly, you can also do

Derivative[-1][iF]

The disadvantage of the fast way is that there's less oversight about precision loss and numerical errors. To make sure that the final integration result is accurate enough (despite error accumulation), you must make sure that NDSolve computes iF with sufficient accuracy.

• "Here's a not so widely known trick": in some sense, this trick should be obvious, but it clearly isn't. +1 for showing the tricks! – march Jan 21 '16 at 17:33
• I added some details. Could you check it out? – LexRomah Jan 21 '16 at 19:10
• @LexRomah Right, this method won't work for you. :-( – Szabolcs Jan 21 '16 at 21:37
• sorry, but doesn't your method work for a function of several variables? – illuminates Dec 3 '17 at 16:41


# Math Help - Half-life.

1. ## Half-life.

A radioactive substance has a half-life of 20 days.

a) How much time is required so that only $\frac{1}{32}$ of the original amount remains?
b) Find the rate of decay at this time.

My attempts:
a) Since after 20 days the substance decays by half of its mass, after 40 days it decays to ${1}{4}$, and after 60 days it decays to $\frac{1}{8}$. After 80 days it decays to $\frac {1}{16}$ of its original mass and after 100 days it decays to ${1}{32}$ of its original mass. Therefore, it will take 100 days for the substance to decay to [tex]\frac{1}{32}/MATH] of its original mass.
b) $f(t) = \frac{1}{32}(\frac{1}{2})^{\frac{t}{20}$
$f'(t) =\frac{1}{32}(\frac{1}{2})^{\frac{t}{20}} × (ln\frac{1}{2}) × (\frac{1}{20})$
$f'(100) =\frac{1}{32}(\frac{1}{2})^{\frac{100}{20}} × (ln\frac{1}{2}) × (\frac{1}{20})$
$f'(100) = -3.384 × 10^{-5}$ is the rate of decay when $t = 100$ days.

I just wanted to know if my answers are correct? I checked with derivative calculators and they have given me very different answers for b. Thanks in advance.

2. Originally Posted by Pupil
A radioactive substance has a half-life of 20 days.

a) How much time is required so that only $\frac{1}{32}$ of the original amount remains?
b) Find the rate of decay at this time.

My attempts:
a) Since after 20 days the substance decays by half of its mass, after 40 days it decays to ${1}{4}$, and after 60 days it decays to $\frac{1}{8}$. After 80 days it decays to $\frac {1}{16}$ of its original mass and after 100 days it decays to ${1}{32}$ of its original mass. Therefore, it will take 100 days for the substance to decay to [tex]\frac{1}{32}/MATH] of its original mass.
b) $f(t) = \frac{1}{32}(\frac{1}{2})^{\frac{t}{20}$
$f'(t) =\frac{1}{32}(\frac{1}{2})^{\frac{t}{20}} × (ln\frac{1}{2}) × (\frac{1}{20})$
$f'(100) =\frac{1}{32}(\frac{1}{2})^{\frac{100}{20}} × (ln\frac{1}{2}) × (\frac{1}{20})$
$f'(100) = -3.384 × 10^{-5}$ is the rate of decay when $t = 100$ days.

I just wanted to know if my answers are correct? I checked with derivative calculators and they have given me very different answers for b. Thanks in advance.
The first part is correct However you could have solved it this way

$\displaystyle \left( \frac{1}{2}\right)^\frac{x}{20}=\frac{1}{32}=\left ( \frac{1}{2}\right)^5 \iff \frac{x}{20}=5 \implies x=100$

For the 2nd part all we know is

$\displaystyle f(t)=A_0\left( \frac{1}{2}\right)^{\frac{x}{20}}=A_0 e^{\frac{x}{20}\cdot \ln(\frac{1}{2})}$

Now taking the derivative gives

$\displaystyle f'(t)=\left( {\frac{1}{20}\cdot \ln(\frac{1}{2})}\right)A_0 e^{\frac{x}{20}\cdot \ln(\frac{1}{2})}=A_0\left( {\frac{1}{20}\cdot \ln(\frac{1}{2})}\right)\left( \frac{1}{2}\right)^{\frac{x}{20}}$

3. Ah, I forgot to take the log of both sides and solve it much easier.

Okay, so I derived the equation correctly? And the question asks me to find the rate of decay when $t = 100$ days. Knowing the derivative and the quantity at the time, wouldn't I just need to plug in 100 to find the rate of decay?

4. Yes, f'(100) should tell you the rate of decay on the 100th day.

5. Originally Posted by NOX Andrew
Yes, f'(100) should tell you the rate of decay on the 100th day.
So, is the rate of decay $-3.384 × 10^{-5}$ when $t = 100$?

6. What is the original amount? It should be the original amount times -log(2)/640.

7. Originally Posted by NOX Andrew
What is the original amount? It should be the original amount times -log(2)/640.
It was not given. I only have $\frac{1}{32}$ of the original substance to work with.


# Probability of exactly two heads in four coin flips?

When you flip a coin four times, what is the probability that it will come up heads exactly twice?

My calculation:

• we have $2$ results for one flip : up or down
• so flip $4$ times, we have $4\cdot2 = 8$ results total

Thus the probability is: $2/ 8 = 0.25$ but the correct answer is $0.375$. Can anyone explain why I'm wrong?

• Nope, if you flip 4 times, there are $2^4$ possible outcomes. How many of these outcomes have two heads? – Irvan Oct 6 '14 at 8:15

My calculation:

we have 2 results for one flip : up or down so flip 4 times, we have 4x2 = 8 results total

Two results for each of four coin flips. When ways to perform tasks in series, we multiply. So that is $2\times 2\times 2\times 2$ results in total. That is $2^4$ or $16$.

For the favourable case we need to count the ways to get $2$ heads and $2$ tails. The count of permutations of two pairs of symbols is: $\frac{4!}{2!2!}=6$. This is easily confirmed by just counting.

$$\Bigl|\{\mathsf {HHTT, HTHT, HTTH, THHT, THTH, TTHH}\}\Bigr|=6$$

Thus the probability is: $\tfrac{\;6}{16}$, or: $$0.375$$

• Thanks, i got the idea, but i don't understand what "!" is ? – NeedAnswers Oct 6 '14 at 11:19
• @hoangnnm It's the Factorial notation. $n!$ is the product of all integers less than or equal to $n$. $n!=(n)(n-1)(n-2)\cdots(2)(1)$ en.wikipedia.org/wiki/Factorial – Graham Kemp Oct 6 '14 at 11:55
• oh. i got it now. your anwser is easier for me to understand. Therefore, i will remark yours as answer! – NeedAnswers Oct 6 '14 at 13:28

Assuming unbiased coin with probability of head $=\dfrac12$

and using Binomial Distribution, $$\binom42\left(\frac12\right)^2\left(1-\frac12\right)^{4-2}$$

Use binomial probability since there are only two possibilities: success and failure, where success represents getting a heads, and tails being a fail.

Let $X$ = Success (i.e. heads)

Therefore we are trying to find $P(X=2)$, which is $\binom42\cdot(0.5)^2\cdot(0.5)^2=0.375$.

Hope this helped!

The derivation of binomial probability:

Getting two heads out of 4 can be portrayed is, disregarding order:

Multiplying their probabilities will yield $(0.5)^4$, but as for ordering, we get $4!/(2!\cdot2!)$ due to repetition, which is the same as $4C2$. So our answer is $\binom42\cdot(0.5)^4$ which is $0.375$