# Coin Flip Probability

Over a large number of tosses, though, the percentage of heads and tails will come to approximate the true probability of each outcome. Practice this lesson yourself on KhanAcademy. Assistant Research Professor. In this case, the probability of flipping a head or a tail is 1/2. Hi everyone. If an event consists of more than one coin, then coins are considered as. For the first 53 Super Bowls the flip has landed on tails 28 times and heads 25 times. The coin has come up heads 54% of the time so far; based only on this data, one might expect that it is slightly more likely to come up heads again. Flip 10 coins, and and you're at a 4-digit number. The sum of all possible outcomes is always 1 (or 100%) because it is certain that one of the possible outcomes will happen. The probability of an event is a number indicating how likely that event will occur. For example, if an individual wanted to know the probability of getting a head in a coin toss but only used one sample, the empirical probability would be either 0% or 100%. Don't expect the numbers from trials to exactly match the predicted results--especially if you run only a few trials. But not that much more likely. After all, real life is rarely fair. 2, which is the Kelly fraction for this α and p. I got the program down right but my results show a number for each coin flip in addition to the cout that says "The coin flip shows Heads/Tails". Applet: Instructions: Examples: Notes "H. Showing top 8 worksheets in the category - Coin Flip Experiment Basic. I flip a coin and it comes up heads. 5 for either heads or tails (assuming that the coin is purely mathematical and random, of course). If p=0 or p=1, the strategy is obvious, so assume 0. We’ll set p=0. Gamblers Take Note: The Odds in a Coin Flip Aren't Quite 50/50 And the odds of spinning a penny are even more skewed in one direction, but which way? Flipping a coin isn't as fair as it seems. If after 200 flips, the. I got a question on the coin flip project. An Easy GRE Probability Question. e head or tail. What were the results? Student: The coin landed on heads 9 times and on tails 6 times. Intro to Problem Solving. It can even toss weighted coins. 51 (instead of 0. Don't expect the numbers from trials to exactly match the predicted results--especially if you run only a few trials. How likely something is to happen. Start studying Laws of Probability: Coin Toss Lab. This method may be used to resolve a dispute, see who goes first in a game or determine which type of treatment a patient receives in a clinical trial. If the result is heads, they flip a coin 100 times and record results. Published on June 14, 2016. 5, which is our probability of tossing heads and moving forward. I need to write a python program that will flip a coin 100 times and then tell how many times tails and heads were flipped. Course : Introduction to Probability and Statistics, Math 113 Section 3234 Instructor: Abhijit Champanerkar Date: Oct 17th 2012 Tossing a coin The probability of getting a Heads or a Tails on a coin toss is both 0. One source of confusion is in counting the number of outcomes, both favorable and possible, such as when tossing coins and rolling dice. She'll make a prediction and practice flipping a coin in order to check out its chances of landing on heads or tails. 8) for i in xrange(10)] [H,H,T,H,H,H,T,H,H,H]. For 100 flips, if the actual heads probability is 0. Click "flip coins" to generate a new set of coin flips. The odds are "long" only if you predetermine when the series of coin flips begins. Note that this answer works for any odd number of coin flips. We assume that conditioned on Q=q, all coin tosses are independent. If after 200 flips, the. Mathematically the coin flip. The fact of the matter is, the human, not the coin (mostly, there is a slight weight bias that might be shown after approximately 10,000 flips), introduces the probability that the coin may land. Coin Toss Probability Calculator Coin toss also known as coin flipping probability is used by people around the world to judge whether its going to be head or tail after flipping the coin. Coin Toss: Simulation of a coin toss allowing the user to input the number of flips. If you get tails on the first flip, you might as well stop, because you cannot possibly get four heads. Users may refer the below detailed solved example with step by step calculation to learn how to find what is the probability of getting exactly 2 heads, if a coin is tossed five times or 5 coins tossed together. Update: The initial 6-for-6 report, from the Des Moines Register missed a few Sanders coin-toss wins. In the last exercise you tried flipping ten coins with a 30% probability of heads to find the probability *at least five are heads. With a "fair" coin, the probability of getting heads on a "single" flip at any time is 1/2. You found that the exact answer was '1 - pbinom(4, 10,. By theory, we can calculate this probability by dividing number of expected outcomes by total number of outcomes. I would like to know what is the probability of this occurrence within any 100 consecutive flips out of a series of. Inspiration • A ﬁnite probability space is used to model the phenomena in which there are only ﬁnitely many possible outcomes • Let us discuss the binomial model we have studied so far through a very simple example • Suppose that we toss a coin 3 times; the set of all possible outcomes can be written as Ω = {HHH,HHT,HTH,THH,HTT,THT,TTH,TTT} • Assume that the probability of a head. Predicting a coin toss. Assuming we kept going, then we flip the second coin. It is not always easy to decide what is heads and tails on a given coin. Applet: Instructions: Examples: Notes "H" count = , flips so far, number of coins: one flip "H" probability: 0. The simplicity of the coin toss also opens the road to more advanced probability theories dealing with events with an infinite number of possible outcomes. Many events can't be predicted with total certainty. Toss a single coin 5X and record the results Table 1. 5, which is our probability of tossing heads and moving forward. The number of possible outcomes gets greater with the increased number of coins. When a coin is tossed twice, the coin has no memory of whether it came up heads or tails the first time, so the second toss of the coin is independent. What is the probability of getting two heads and four tails? Coin Flipping How can I figure out the chances of flipping a coin five times with the result T,T,T,H,H?. I got a question on the coin flip project. The odds of the coins coming up with different faces showing is just 1 in 2. Read and learn for free about the following article: Theoretical and experimental probability: Coin flips and die rolls If you're seeing this message, it means we're having trouble loading external resources on our website. Two flips have 4 outcomes: HH, Ht, tH, and tt. In this activity, you will explore some ideas of probability by using Excel to simulate tossing a coin and throwing a free throw in basketball. Each time you toss these coins, there are four possible outcomes: both heads penny head & dime tail penny tail & dime head both tails You will flip the pair of coins 20 times. How likely something is to happen. Ask Question Asked 7 years, 3 months ago. If 5 is selected for "coin flips per trial," then a number is selected from the list five times, and these numbers are summed. The article says the reason is because a flipped coin does not spin perfectly around its axis and sometimes appears to be flipping when it actually isn’t. Toss a coin 10 times and after each toss, record in the following table the result of the toss and the proportion of heads so far. You found that the exact answer was '1 - pbinom(4, 10,. Mentor: OK, we. Flip 10 coins, and and you're at a 4-digit number. I am just learning Python on class so I am really at the basic. The likelihood of an event is expressed as a number between zero (the event will never occur) and one (the event is certain). Coin Toss Probability Calculator is a free online tool that displays the probability of getting the head or a tail when the coin is tossed. This uses a simple Monte Carlo approximation to estimate the probability distribution for the length of the longest consecutive sequence of Heads in a fixed number of coin flips. But I want to simulate coin which gives H with probability 'p' and T with probability '(1-p)'. The coin has come up heads 54% of the time so far; based only on this data, one might expect that it is slightly more likely to come up heads again. The second paragraph then applies a little conditional probability. In the last exercise you tried flipping ten coins with a 30% probability of heads to find the probability *at least five are heads. Numbers are then randomly selected from the list. Page last modified 07/17/2012 13:01:23. In unbiased coin flip H or T occurs 50% of times. Coin toss probability is explored here with simulation. Published on June 14, 2016. The Probability Simulation application on the TI-84 Plus graphing calculator can simulate tossing from one to three coins at a time. , Bernoulli trials). A fair-sided coin (which means no casino hanky-panky with the coin not coming up heads or tails 50% of the time) is tossed three times. We assume that conditioned on Q=q, all coin tosses are independent. The probability of a success on any given coin flip would be constant (i. Probability. A tossed coin is spinning and falling, therefore it carries significant kinetic energy. Consider flipping a coin that is either heads (H) or tails (T), each with probability 1/2. The program should call a separate function flip()that takes no arguments and returns 0 for tails and 1 for heads. Below is some sample code in R to simulate a fair coin toss in R using the sample function. In 1947, the coin flipping was held 30 minutes before the beginning of the game. From 1892 to 1920, the captain of the football team managed the coin flip. 57 flip_coins(100, 1000, 25) #> [1] 0. Luckily, this is bundled up in a math/probability/stats concept called "combinations". When a coin is tossed, there are two possible outcomes: heads (H) or ; tails (T) We say that the probability of the coin landing H is ½. Flip 4 coins, and you're at 16 outcomes, a 2-digit number. Remark: The idea can be substantially generalized. Every flip of the coin has an “independent probability“, meaning that the probability that the coin will come up heads or tails is only affected by the toss of the coin itself. 5, which means we would not be able to tell the different between a bias coin and fair coin 50% of the time. There is also the very small probability that the coin will land. 1) The mathematical theory of probability assumes that we have a well defined repeatable (in principle) experiment, which has as its outcome a set of well defined, mutually exclusive, events. Next we will show some simulations of coin toss betting using the Kelly fraction. Demonstrates frequency and probability distributions with weighted coin-flipping experiments. When tossing only one coin at a time, the application keeps track of the number of heads and tails that occur as the coin is repeatedly tossed. Math archives: Probability in Flipping Coins Six pennies are flipped. The smoother of those two lines is an average of 2000 runs. While a coin toss is regarded as random, it spins in a predictable way. the probability of throwing exactly two heads in three tosses of the coin is 3 out of 8, or or the decimal equivalent of which is 0. The coin toss is nothing but experimenting with tossing a coin. (Ti includes the toss that results in the first. Let Ti be the number of tosses of the ith coin until that coin results in Heads for the first time, for i=1,2,…,k. Flipping coins and the binomial distribution#2 Consider two coins, one fair and one unfair. Thus, the odds of any throw being a tail is 1/2. % certain that the outcome would be tails, but this is due to how it is being measured. So if an event is unlikely to occur, its probability is 0. This is a very basic form of empirical probability, however, and has a high risk of being incorrect because a series of only two events (coin tosses) have been observed. Over a large number of tosses, though, the percentage of heads and tails will come to approximate the true probability of each outcome. I am just learning Python on class so I am really at the basic. How likely something is to happen. For a fair coin, The probability of the outcome Head is 1/2, because for a large number of tosses, the relative occurrence of Heads will be roughly 1/2. Both outcomes are equally likely. Numbers are then randomly selected from the list. Then, you will flip the coins 100 times and determine the experimental probability of the events. We all know a coin toss gives you a 50% chance of winning, but is it always that way? Delve into the inner-workings of coin toss probability with this activity. A common topic in introductory probability is solving problems involving coin flips. On a mission to transform learning through computational thinking, Shodor is dedicated to the reform and improvement of mathematics and science education through student enrichment. Bayes equation describes the situation when the probability that a fair coin was used to produce two heads is equal to the probability of seeing two heads if a fair coin was used. After all, real life is rarely fair. Since the coin toss is a physical phenomenon governed by Newtonian mechanics, the question requires one to link probability and physics via a mathematical and statistical description of the coin's. Mentor: Yes! Now let's look at the coin flipping game that you just played. Simple numbers. Recommended: Please try your approach. Take another penny and Super Glue it to the coin. 53) #> [1] 0. The coin has no desire to continue a particular streak, so it's not affected by any number of previous coin tosses. When asked the question, what is the probability of a coin toss coming up heads, most people answer without hesitation that it is 50%, 1/2, or 0. Note: Including the words "single time" and "after" confuse this problem somewhat. Q: if you flip a coin 3 times what is the probability of getting only 1 head? A: The probability of getting one head in three throws is 0. 5×10 20 chance of getting a string of 76 heads. So if you flip a coin 10 times in a row-- a fair coin-- you're probability of getting at least 1 heads in that 10 flips is pretty high. 7 flip_coins(100, 100, 5) #> [1] 0. This code helps you count consecutive strings of Heads in a sequence of coin flips. If we toss a coin $n$ times, and the probability of a head on any toss is $p$ (which need not be equal to $1/2$, the coin could be unfair), then the probability of exactly $k$ heads is $$\binom{n}{k}p^k(1-p)^{n-k}. As long as the coin was not manipulated the theoretical probabilities of both outcomes are the same—they are equally probable. Toss a single coin 5X and record the results Table 1. Write a program that simulates coin tossing. This is a very basic form of empirical probability, however, and has a high risk of being incorrect because a series of only two events (coin tosses) have been observed. the coin does not and can not "remember" last result 4. And you can get a calculator out to figure that out in terms of a percentage. Coin Toss Probability Calculator. I flip a coin and it comes up heads. Nine flips of a fair coin. How do I get rid of the number? It looks something like this when I run it The coin flipped Heads 1 The Coin flipped tails 2 The coin flipped Heads 1. Sunday, March 29, 2009. The Probability Simulation application on the TI-84 Plus graphing calculator can simulate tossing from one to three coins at a time. Below is some sample code in R to simulate a fair coin toss in R using the sample function. If the probability of coin flipping head = P The probability of coin flipping tail = 1 - P Now The probability of flipping heads & then tails = Probability of flipping tails & then heads = P(1 - P) Which means to make a fair coin toss we now need 2 flips Player 1 wins if the sequence is HT Player 2 wins if the sequence is TH Any other sequence. The coin flip has gone through many changes. The probability of Heads is the same for each coin and is the realized value q of a random variable Q that is uniformly distributed on [0,1]. For 100 flips, if the actual heads probability is 0. What is the probability that player A ends up with all the coins?. the coin tossing is stateless operation i. Click "flip coins" to generate a new set of coin flips. When we flip a coin there is always a probability to get a head or a tail is 50 percent. Remark: The idea can be substantially generalized. I would like to know what is the probability of this occurrence within any 100 consecutive flips out of a series of. When you flip a coin, you can generally get two possible outcomes: heads or tails. There is a 50% probability that the first toss will end up heads. A probability of zero means that an event is impossible. 4, then the power is 0. But give me a well-balanced coin, any size, and I can roll it on any flat surface on edge every time. There is a 50% probability that the first toss will end up heads. One source of confusion is in counting the number of outcomes, both favorable and possible, such as when tossing coins and rolling dice. Confidence intervals for coin flipping. Logic problems, Understanding odds, Understanding probability Common Core Standards: Grade 4 Number & Operations: Fractions , Grade 5 Number & Operations in Base Ten , Grade 5 Operations & Algebraic Thinking. Similarly, the probability of the outcome Tail is 1/2, because the relative occurrence of Tails will be 1/2 for a large number of tosses. In unbiased coin flip H or T occurs 50% of times. As long as the coin was not manipulated the theoretical probabilities of both outcomes are the same—they are equally probable. " The total number of equally likely events is "2" because tails is just as likely as heads. Flip 10 coins, and and you're at a 4-digit number. Let Ti be the number of tosses of the ith coin until that coin results in Heads for the first time, for i=1,2,…,k. Nine flips of a fair coin. Write a program that simulates coin tossing. In the case of a coin, there are maximum two possible outcomes - head or tail. The probability of an event is a number indicating how likely that event will occur. Coin Flip Name Date Period 1. Now, create a Markov transition matrix, that will see a change from any state to the next higher state with probability 0. Inspiration • A ﬁnite probability space is used to model the phenomena in which there are only ﬁnitely many possible outcomes • Let us discuss the binomial model we have studied so far through a very simple example • Suppose that we toss a coin 3 times; the set of all possible outcomes can be written as Ω = {HHH,HHT,HTH,THH,HTT,THT,TTH,TTT} • Assume that the probability of a head. The probability of getting a given number of heads from four flips is, then, simply the number of ways that number of heads can occur, divided by the number of total results of four flips, 16. Some of the worksheets displayed are Lesson plan 19 flipping coins, Probability experiment, Fair coin work, Lesson topic probability grade level 6th grade length of, Mendelian genetics coin toss lab, Coin probability theoretical experimental probability, Lab 9 principles of genetic inheritance. org right now:. When we flip a coin there is always a probability to get a head or a tail is 50 percent. Probability. The theory revealed that the coin's behaviour is predictable - until it strikes the floor. I flip a coin and it comes up heads. 5 we get this probability by assuming that the coin is fair, or heads and tails are equally likely The probability for. Each team member will have 1 coin to flip. Most coins have probabilities that are nearly equal to 1/2. Flipping the coin once is a Bernoulli trial, since there are exactly two complementary outcomes (flipping a head and flipping a tail), and they are both 1 2 \frac{1}{2} 2 1 no matter how many times the coin is flipped. Hi everyone. Let the probability of obtaining a head be. We know that we will be doing a fair coin flip. a)Give an algebraic formula for the probability mass function of X. 5, which is our probability of tossing heads and moving forward. If I toss a coin only 10 times I may end up with 9 heads and 1 tails. Let’s start thinking about this by thinking about the coin flip. Demonstration of frequentist convergence of probability with a coin flip. Use buttons to view a bar chart of the coin flips, the probability distribution (also known as the probability mass. But not that much more likely. Daniel Egger. 5 is the probability of getting 2 Heads in 3 tosses. Not from a coin toss. the probability of tails is the same as heads, P(T) <=> P(H) 3. What is the probability that a fair coin lands on heads on 4 out of 5 flips? Wha are the four properties of a binomial probability distribution? In a carnival game, there are six identical boxes, one of which contains a prize. For example, consider a fair coin. possible outcomes and finding each outcome that has two or more tails in it. Here's a simulation of the game. Coin Flipping, a selection of some of the answers to problems of this kind in the Dr. A cumulative probability refers to the probability that the value of a random variable falls within a specified range. Many events can't be predicted with total certainty. Consider flipping a coin that is either heads (H) or tails (T), each with probability 1/2. Print the results. In unbiased coin flip H or T occurs 50% of times. What is the probability of getting exactly two heads and two tails. When tossing only one coin at a time, the application keeps track of the number of heads and tails that occur as the coin is repeatedly tossed. Demonstrates frequency and probability distributions with weighted coin-flipping experiments. from the previous assumptions follows that given any sequence of coin tossing results, the next toss has the probability P(T) <=> P(H). A coin toss is a tried-and-true way for your fifth grader to understand odds. You can modify it as you like to simulate any number of flips. Like the title says, I need to figure out probability for a weighted coin flip. Coin Flipping, a selection of some of the answers to problems of this kind in the Dr. Heads did have an impressive run of 5 years in a row from 2009-2013. Theory of Probability. If the result of the coin toss is head, player A collects 1 coin from player B. We know that we will be doing a fair coin flip. Show Hide all comments. 7 flip_coins(100, 100, 5) #> [1] 0. Nine flips of a fair coin. You out the full article at the link below for probability charts and a fascinating look into the mathematics of solving coin toss probability. If the probability of flipping heads is 70%, then the list contains 70 ones and 30 zeros. Coin toss probability is explored here with simulation. heads, flips(100) The following shows the results of using 50 tosses of the coin with a probability of obtaining heads of. The toss of a coin, throwing dice and lottery draws are all examples of random events. Coin Toss Probability Calculator. So the results of flipping a coin should be somewhere around 50% heads and 50% tails since that is the theoretical probability. The probability of Heads is the same for each coin and is the realized value q of a random variable Q that is uniformly distributed on [0,1]. In this video, we' ll explore the probability of getting at least one heads in multiple flips of a fair coin. Many events can't be predicted with total certainty. If you flip two coins, four. 4) 4 boys and 3 girls are standing in a line. 100 coins is a 31-digit number. Number of flips to do in each set: Probability of landing heads: Proportion of heads after each full set of flips, most recent. For 100 flips, if the actual heads probability is 0. A sequence of consecutive events is also called a "run" of events. Similarly, the probability of the outcome Tail is 1/2, because the relative occurrence of Tails will be 1/2 for a large number of tosses. Every time a coin is flipped, the probability of it landing. Two flips have 4 outcomes: HH, Ht, tH, and tt. Hello, A hat contains n coins, f of which are fair, and b of which are biased to land heads with probability of 2/3. How do I get rid of the number? It looks something like this when I run it The coin flipped Heads 1 The Coin flipped tails 2 The coin flipped Heads 1. If you'd like to read more about flipping coins and probability, check out my post on the topic at the Blog on Math Blogs. Since each coin toss has a probability of heads equal to 1/2, I simply need to multiply together 1/2 eleven times. Each team member will have 1 coin to flip. First, you will determine the theoretical probability of events. The post is correct that the odds of getting. Predicting a coin toss. Probability. Coin Toss Probability Calculator Coin toss also known as coin flipping probability is used by people around the world to judge whether its going to be head or tail after flipping the coin. Q: What is the probability for a coin to land on its edge when you flip a coin? A: The probability of a coin landing on its side or edge is a remote 6000 to 1. Below you will find a table that lists the coin flip results, including which team won and lost the coin flip for all Super Bowls. According to Science News Online the probability that a coin will land on the same side it started on is 51%. If 5 is selected for "coin flips per trial," then a number is selected from the list five times, and these numbers are summed. (Ti includes the toss that results in the first. e head or tail. Be careful with how you read this probability. Every flip of the coin has an “independent probability“, meaning that the probability that the coin will come up heads or tails is only affected by the toss of the coin itself. 2 of the outcomes are 1 head and 1 tail. If you flip two coins, four. If the probability of an event is high, it is more likely that the event will happen. So if an event is unlikely to occur, its probability is 0. Update: The initial 6-for-6 report, from the Des Moines Register missed a few Sanders coin-toss wins.$$ This probability model is called the Binomial distribution. For the old java version, click here ; For the Spanish version, click here ; For the German version, click here; To. 9 for coin B. Both team members flip their coins. 5, which means we would not be able to tell the different between a bias coin and fair coin 50% of the time. You must show your work to receive credit. Use Probability to Win Coin Flipping Games. A cumulative probability refers to the probability that the value of a random variable falls within a specified range. 4) 4 boys and 3 girls are standing in a line. Simulating a coin toss in excel I guess when you start to look at gambling theories or probabilities the natural place to start is the coin toss. When you take these chocolates out, the probability for any one being taken out diminishes by 1 each time. Demonstrates frequency and probability distributions with weighted coin-flipping experiments. Direct link to this answer. Notice that the width of the confidence interval narrows as the number of. Use Probability to Win Coin Flipping Games. Apply creativity, lateral thinking, and mathematical skill. Heads did have an impressive run of 5 years in a row from 2009-2013. So if you flip a coin 10 times in a row-- a fair coin-- you're probability of getting at least 1 heads in that 10 flips is pretty high. I think the best way to attack the problem is to run a simulation of millions of trials, and then give an approximate answer based on the number. "Pairs of adjacent coins" means only two coins next to each other may be flipped at a time. Course : Introduction to Probability and Statistics, Math 113 Section 3234 Instructor: Abhijit Champanerkar Date: Oct 17th 2012 Tossing a coin The probability of getting a Heads or a Tails on a coin toss is both 0. coin=randi ( [0:1], [100,1]) It should more or less give you 50 0's and 50 1's. Write a program that simulates coin tossing. (Ti includes the toss that results in the first. Coin Flips, Risk to Reward Profile and Creating Your Own Synthetic Security Thus, the probability of getting 2 heads in a row is the probability of getting a head followed by a second flip where you also get a head. (Solution): Coin Toss Probability. Online virtual coin toss simulation app. Coin Toss: Simulation of a coin toss allowing the user to input the number of flips. 100 coins is a 31-digit number. 5, or you will stay in the current state with probability 0. I got the program down right but my results show a number for each coin flip in addition to the cout that says "The coin flip shows Heads/Tails". The probability of getting a given number of heads from four flips is, then, simply the number of ways that number of heads can occur, divided by the number of total results of four flips, 16. The answer to this is always going to be 50/50, or ½, or 50%. This is what I have so far but I keep getting errors. In the case of a coin, there are maximum two possible outcomes - head or tail. When the probability of an event is zero then the even is said to be impossible. Luckily, this is bundled up in a math/probability/stats concept called "combinations". Remark: The idea can be substantially generalized. This means that the theoretical probability to get either heads or tails is 0. Ask Question Asked 3 years, 6 months ago. What is the probability that a fair coin lands on heads on 4 out of 5 flips? What is the probability of getting at least one tail if a fair coin is flipped three times? Wha are the four properties of a binomial probability distribution?. Repeat steps 2-4 until the coin lands on its side every time. Junho: The chance of DB completing the coin scam on the first attempt, which is to toss a coin and get 10 heads in a row, is very unlikely. I have a problem I need to do for school. 6, and f = 0. Remember that each individual coin flip has a 50% chance of being heads. If we do the math, this is a probability of 0. Not so, says Diaconis. 2, which is the Kelly fraction for this α and p. 5, which is our probability of tossing heads and moving forward. Viewed 2k times 1. The two non-straight lines in Fig. The odds of the coins coming up with different faces showing is just 1 in 2. , Bernoulli trials). If you toss 2 coins, what are the chances you will get 2 heads? Record your predictions and explain your reasoning. The program should call a separate function flip()that takes no arguments and returns 0 for tails and 1 for heads. If you flip a coin and roll a six-sided die, what is the probability that the coin comes up heads and the die comes up 1? Since the two events are independent, the probability is simply the probability of a head (which is 1/2) times the probability of the die coming up 1 (which is 1/6). 57 flip_coins(100, 1000, 25) #> [1] 0. Flipping coins and the binomial distribution#2 Consider two coins, one fair and one unfair. We can use R to simulate an experiment of ipping a coin a number of times and compare our results with the theoretical probability. That was flip number 130,659,178 Flip again? Color The Coin!. If the probability of flipping heads is 70%, then the list contains 70 ones and 30 zeros. If you flip one coin, just two. of flipping one head with a coin is 50%, then the probability of flipping two heads at once is achieved by (adding or multiplying)_____ the separate probabilities. (Solution): Coin Toss Probability. What is the theoretical probability that the co n will land on tails? What is the theoretical probability that the co n will land on heads? If the com is flipped 140 times, how many times would you predict that the co n lands on heads? Johnny flipped a coin 450 times. If you flip three coins, it's eight - two for the first times two for the second times two for the third. The game is played in stages. The order does not matter as long as there are two head and two tails in the flip. e head or tail. Coin flipping, coin tossing, or heads or tails is the practice of throwing a coin in the air and checking which side is showing when it lands, in order to choose between two alternatives, sometimes used to resolve a dispute between two parties. Since 'fair' is used in the project description we know that the probability will be a 50% chance of getting either side. Flipping more coins¶ If we want to flip more coins, it's going to be a pain in the neck to make that table over and over. Then, it displays the results, as well as the theoretical and observed probabilities of each event happening. Given this information, what is the probability that it is a. So, I'll do it faster! When we flip the coin 9 times there are $$2^9$$ possible outcomes that can happen. A probability of one represents certainty: if you flip a coin, the probability you'll get heads or tails is one (assuming it can't land on the rim, fall into a black hole, or some such). Exactly 2 heads in 3 Coin Flips The ratio of successful events A = 3 to total number of possible combinations of sample space S = 8 is the probability of 2 heads in 3 coin tosses. If you toss a coin, you cannot get both a head and a tail at the same time, so this has zero probability. Gamblers Take Note: The Odds in a Coin Flip Aren't Quite 50/50 And the odds of spinning a penny are even more skewed in one direction, but which way? Flipping a coin isn't as fair as it seems. Simulating a coin toss in excel I guess when you start to look at gambling theories or probabilities the natural place to start is the coin toss. When we flip a coin there is always a probability to get a head or a tail is 50 percent. 4, then the power is 0. The different possible results from a probability model. Coin flipping, coin tossing, or heads or tails is the practice of throwing a coin in the air and checking which side is showing when it lands, in order to choose between two alternatives, sometimes used to resolve a dispute between two parties. There is a 50% probability that the first toss will end up heads. Have you ever flipped a coin as a way of deciding something with another person? The answer is probably yes. probability of any continuous interval is given by p(a ≤ X ≤ b) = ∫f(x) dx =Area under f(X) from a to b b a That is, the probability of an interval is the same as the area cut off by that interval under the curve for the probability densities, when the random variable is continuous and the total area is equal to 1. Probability, physics, and the coin toss L. Don't expect the numbers from trials to exactly match the predicted results--especially if you run only a few trials. A fair-sided coin (which means no casino hanky-panky with the coin not coming up heads or tails 50% of the time) is tossed three times. For example, consider a fair coin. Page last modified 07/17/2012 13:01:23. Logic problems, Understanding odds, Understanding probability Common Core Standards: Grade 4 Number & Operations: Fractions , Grade 5 Number & Operations in Base Ten , Grade 5 Operations & Algebraic Thinking. Note that the pattern of heads counts seems to form a smoother curve, but still matches the (scaled) binomial coefficients found on the tenth and twentieth rows of Pascal's Triangle. Assuming the coin is fair, p = 1/2 and q = 1/2 where 'p' is the probability of get. Thus, the odds of any throw being a tail is 1/2. 2, which is the Kelly fraction for this α and p. Heads did have an impressive run of 5 years in a row from 2009-2013. The probability of an event is a number indicating how likely that event will occur. The probability for equally likely outcomes in an event is:. The likelihood of an event is expressed as a number between zero (the event will never occur) and one (the event is certain). Use, probability formula = N u m b e r o f f a v o r a b l e o u t c o m e s T o t a l n u m b e r o f p o s s i. Update: The initial 6-for-6 report, from the Des Moines Register missed a few Sanders coin-toss wins. The views expressed are those of the author(s) and are not necessarily. Each team member will have 1 coin to flip. Take another penny and Super Glue it to the coin. Intro to Problem Solving. Although the basic probability formula isn't difficult, sometimes finding the numbers to plug into it can be tricky. Course : Introduction to Probability and Statistics, Math 113 Section 3234 Instructor: Abhijit Champanerkar Date: Oct 17th 2012 Tossing a coin The probability of getting a Heads or a Tails on a coin toss is both 0. Click "flip coins" to generate a new set of coin flips. Explore probability concepts by simulating repeated coin tosses. The experiment was conducted with motion-capture cameras, random experimentation, and an automated "coin-flipper" that could flip the coin on command. Let the program toss the coin 100 times, and count the number of times each side of the coin appears. Page last modified 07/17/2012 13:01:23. Begin your coin with just a single penny. something like this: def flip(p): '''this function return H with probability p''' # do something return result >> [flip(0. Coin toss probability is a classic for a reason: it's a realistic example kids can grasp quickly. Make a weighted coin by changing the probability of landing on heads using the slider; 0% means the coin always lands on tails and 100% means the coin always lands on heads. The coin toss is nothing but experimenting with tossing a coin. Since each coin toss has a probability of heads equal to 1/2, I simply need to multiply together 1/2 eleven times. Use Probability to Win Coin Flipping Games. In the first simulation, player A got lucky with 4 heads in 5 tosses. So, the probability that we will keep going is 1/2 of 1/4, or 1/8. Example: It is 12 if you have an experiment where you flip a coin and then roll a six sided die. An Easy GRE Probability Question. When you take these chocolates out, the probability for any one being taken out diminishes by 1 each time. A common topic in introductory probability is solving problems involving coin flips. What is the probability of getting two heads and four tails? Coin Flipping How can I figure out the chances of flipping a coin five times with the result T,T,T,H,H?. I am just learning Python on class so I am really at the basic. For the old java version, click here ; For the Spanish version, click here ; For the German version, click here; To. If we flip the coin 10 times, we are not guaranteed to get 5 heads and 5 tails. e head or tail. The first time it lands heads, and the second time it lands tails. What is the theoretical probability that the co n will land on tails? What is the theoretical probability that the co n will land on heads? If the com is flipped 140 times, how many times would you predict that the co n lands on heads? Johnny flipped a coin 450 times. Mentor: Yes! Now let's look at the coin flipping game that you just played. In this worksheet, they'll grab a quarter, give it a few tosses, and record the results for themselves. flips The number of desired coin flips. Coin toss probability is explored here with simulation. "The coin tosses are independent events; the coin doesn't have a memory. For the coin, number of outcomes to get heads = 1 Total number of possible outcomes = 2 Thus, we get 1/2 However, if you suspect that the coin may not be fair, you can toss the coin a large number of times and count the number of heads Suppose you flip the coin 100 and get 60 heads, then you know the best estimate to get head is 60/100 = 0. What is the probability that Player 1 will win the game?. Unformatted text preview: Tamara Curiel Gala Cano Mendelian Genetics Coin Toss Lab PRE-LAB DISCUSSION: In heredity, we are concerned with the occurrence, every time an egg is fertilized, of the probability that a particular gene or chromosome will be passed on through the egg, or through the sperm, to the offspring. Sign in to comment. What is the probability of getting exactly two heads and two tails. Coin Toss Probability Probability is the measurement of chances – likelihood that an event will occur. The toss or flip of a coin to randomly assign a decision traditionally involves throwing a coin into the air and seeing which side lands facing up. If the probability of coin flipping head = P The probability of coin flipping tail = 1 - P Now The probability of flipping heads & then tails = Probability of flipping tails & then heads = P(1 - P) Which means to make a fair coin toss we now need 2 flips Player 1 wins if the sequence is HT Player 2 wins if the sequence is TH Any other sequence. In 1921, the referee flipped the coin. We have k coins. 5, or you will stay in the current state with probability 0. I start by having my students create a "Heads Tails" T chart in their math journals and then writing a prediction for the result of tossing the coin 100 times. If after 200 flips, the. There's another sense in which the Haldane prior can be considered non-informative: the mean of the posterior distribution is now $\frac{\alpha + x}{\alpha + \beta + n}=\frac{x}{n}$, i. This means there is a 1 out of 128 chance of getting seven heads on seven coin flips. Over many coin flips the probability of at least half of. If we flip a fair coin 9 times, and the flips are independent, what's the probability that we get heads exactly 6 times? This works just like the last problem, only the numbers are bigger. If the coins show heads-tails (HT) or tails-heads (TH), player 2 gets 1 point. 8) for i in xrange(10)] [H,H,T,H,H,H,T,H,H,H]. We can either find this out using a formula or through Monte Carlo simulation. 5 for either heads or tails (assuming that the coin is purely mathematical and random, of course). Since the coin toss is a physical phenomenon governed by Newtonian mechanics, the question requires one to link probability and physics via a mathematical and statistical description of the coin's. Exactly 2 heads in 3 Coin Flips The ratio of successful events A = 3 to total number of possible combinations of sample space S = 8 is the probability of 2 heads in 3 coin tosses. If we toss a coin n times, and the probability of a head on any toss is p (which need not be equal to 1 / 2, the coin could be unfair), then the probability of exactly k. You can modify it as you like to simulate any number of flips. 5 (the default) with the 95% confidence interval. Tossing a Coin. The following shows the results of 100 tosses of five coins with a probability of heads of. Let be the probability that a run of or more consecutive heads appears in independent tosses of a coin (i. "The coin tosses are independent events; the coin doesn't have a memory. But not that much more likely. The variable timesflipped used for the while. In unbiased coin flip H or T occurs 50% of times. First, note that the problem will likely make reference to a "fair" coin. b) What do you think E[X] should be. Use buttons to view a bar chart of the coin flips, the probability distribution (also known as the probability mass. Since 'fair' is used in the project description we know that the probability will be a 50% chance of getting either side. If I toss a coin only 10 times I may end up with 9 heads and 1 tails. We assume that conditioned on Q=q, all coin tosses are independent. You flipped 1 coin of type US 1¢ Penny: Timestamp: 2020-05-05 01:39:10 UTC. the probability of tails is the same as heads, P(T) <=> P(H) 3. Each time you toss these coins, there are four possible outcomes: both heads penny head & dime tail penny tail & dime head both tails You will flip the pair of coins 20 times. Both team members flip their coins. Flip 10 coins, and and you're at a 4-digit number. If 5 is selected for "coin flips per trial," then a number is selected from the list five times, and these numbers are summed. Have you ever flipped a coin as a way of deciding something with another person? The answer is probably yes. It is about physics, the coin, and how the "tosser" is actually throwing it. There is a 50% probability that the first toss will end up heads. The toss or flip of a coin to randomly assign a decision traditionally involves throwing a coin into the air and seeing which side lands facing up. Life is full of random events! You need to get a "feel" for them to be a smart and successful person. However, the probability of getting exactly one heads out of seven flips is different (and the solution is given). This is a basic introduction to a probability distribution table. An Easy GRE Probability Question. But not that much more likely. Furthermore, if you toss a coin, it will eventually show heads, so this procedure ends in a finite number of flips with probability 1. There is also the very small probability that the coin will land. In Chapter 2 you learned that the number of possible outcomes of several independent events is the product of the number of possible outcomes of each event individually. Probability. Conditional probability question - coin toss. A fair-sided coin (which means no casino hanky-panky with the coin not coming up heads or tails 50% of the time) is tossed three times. But if any of the flipped coins comes up tails, or if no one chooses to flip a coin, you will all be doomed to spend the rest of your lives in the castle’s dungeon. the probability of tails is the same as heads, P(T) <=> P(H) 3. The coin can only land on one side or the other (event) but there are two possible outcomes: heads or tails. 5, which is our probability of tossing heads and moving forward. a)Give an algebraic formula for the probability mass function of X. Logic problems, Understanding odds, Understanding probability Common Core Standards: Grade 4 Number & Operations: Fractions , Grade 5 Number & Operations in Base Ten , Grade 5 Operations & Algebraic Thinking. Coin Flips, Risk to Reward Profile and Creating Your Own Synthetic Security Thus, the probability of getting 2 heads in a row is the probability of getting a head followed by a second flip where you also get a head. Applet: Instructions: Examples: Notes "H" count = , flips so far, number of coins: one flip "H" probability: 0. A probability of one means that the event is certain. Probability measures how certain we are a particular event will happen in a specific instance. That was flip number 130,659,178 Flip again? Color The Coin!. Probability: Independent Events. 5, or you will stay in the current state with probability 0. The toss of a coin, throwing dice and lottery draws are all examples of random events. The instructions are written in the handout for the students to understand how to complete the activi. of flipping one head with a coin is 50%, then the probability of flipping two heads at once is achieved by (adding or multiplying)_____ the separate probabilities. When the coin is flipped and the first three flips are heads, the fourth flip still has the probability of ½ However, many people misunderstand that the first three flips somehow influence the fourth flip, but they do not. Don't expect the numbers from trials to exactly match the predicted results--especially if you run only a few trials. Were you to toss the coin 100 times, you would get a clearer view of how probable it is that the coin lands on heads each time. A coin toss is a tried-and-true way for your fifth grader to understand odds. We do not know if we will get heads or tails. I got a question on the coin flip project. If we flip the coin 10 times, we are not guaranteed to get 5 heads and 5 tails. Repeat steps 2-4 until the coin lands on its side every time. Expected Value represents the average outcome of a series of random events with identical odds being repeated over a long period of time. Asked in Statistics , Probability. It begins with the two title characters caught in a most unusual coin game. When flipping a coin, the probability of getting a head does not change no matter how many times you flip the coin. Example: It is 12 if you have an experiment where you flip a coin and then roll a six sided die. If the coins show heads-tails (HT) or tails-heads (TH), player 2 gets 1 point. With three coins, all three landing on the same side is 1 in 8. Ask Question Asked 3 years, 6 months ago. The probability for equally likely outcomes in an event is:. In this video, we' ll explore the probability of getting at least one heads in multiple flips of a fair coin. I need to write a python program that will flip a coin 100 times and then tell how many times tails and heads were flipped. Explore probability concepts by simulating repeated coin tosses. If the probability of coin flipping head = P The probability of coin flipping tail = 1 - P Now The probability of flipping heads & then tails = Probability of flipping tails & then heads = P(1 - P) Which means to make a fair coin toss we now need 2 flips Player 1 wins if the sequence is HT Player 2 wins if the sequence is TH Any other sequence. Students will flip a coin a total of ten times per trial and record results by simply shading in the space below the realistic-looking coins. Flip 4 coins, and you're at 16 outcomes, a 2-digit number. The number of possible outcomes gets greater with the increased number of coins. Coin Flipper. For example, given 5 trials per experiment and 20 experiments, the program will flip a coin 5 times and record the results 20 times. Become a member and unlock all. The probability of A and B is 1/100. That means I flipped the coin 15 times. It's pretty much a "coin flip". The views expressed are those of the author(s) and are not necessarily. (Solution): Coin Toss Probability. Suppose a coin tossed then we get two possible outcomes either a ‘head’ ( H) or a ‘tail’ ( T ), and it is impossible to predict whether the result of a toss will be a ‘head’ or ‘tail’. However, the probability of getting exactly one heads out of seven flips is different (and the solution is given). But I want to simulate coin which gives H with probability 'p' and T with probability '(1-p)'. The first person to flip heads wins. Flipping coins and the binomial distribution#2 Consider two coins, one fair and one unfair. from the previous assumptions follows that given any sequence of coin tossing results, the next toss has the probability P(T) <=> P(H). Recommended: Please try your approach. Logical Reasoning. Procedure 1: Statistical Probability Reasoning. 5×10 20 chance of getting a string of 76 heads. As long as the coin was not manipulated the theoretical probabilities of both outcomes are the same—they are equally probable. Let us return to the coin flip experiment. If the probability of an event is high, it is more likely that the event will happen. And you can get a calculator out to figure that out in terms of a percentage. This article shows you the steps for solving the most common types of basic questions on this subject. That means I flipped the coin 15 times. I start by having my students create a "Heads Tails" T chart in their math journals and then writing a prediction for the result of tossing the coin 100 times. The coin toss is nothing but experimenting with tossing a coin. , the sample frequency of heads, which is the frequentist MLE estimate of $\theta$ for the Binomial model of the coin flip problem. Unformatted text preview: Tamara Curiel Gala Cano Mendelian Genetics Coin Toss Lab PRE-LAB DISCUSSION: In heredity, we are concerned with the occurrence, every time an egg is fertilized, of the probability that a particular gene or chromosome will be passed on through the egg, or through the sperm, to the offspring. Lends to discussion and discovery of probability, elementary understanding of mode, and graphic organization of information to compare and contrast. Begin your coin with just a single penny. If you flip three coins, it's eight - two for the first times two for the second times two for the third. A sequence of consecutive events is also called a "run" of events. You flipped 1 coin of type US 1¢ Penny: Timestamp: 2020-05-05 01:39:10 UTC. It is measured between 0 and 1, inclusive. The probability of coming up heads on the first flip is 1/2. It is about physics, the coin, and how the "tosser" is actually throwing it. the coin does not and can not "remember" last result 4. For example, consider a fair coin. In the first simulation, player A got lucky with 4 heads in 5 tosses. Remark: The idea can be substantially generalized. The probability of getting a given number of heads from four flips is, then, simply the number of ways that number of heads can occur, divided by the number of total results of four flips, 16. If we toss a coin $n$ times, and the probability of a head on any toss is $p$ (which need not be equal to $1/2$, the coin could be unfair), then the probability of exactly $k$ heads is \binom{n}{k}p^k(1-p)^{n-k}. Coin Toss: Simulation of a coin toss allowing the user to input the number of flips. Simple numbers. Ask Question Asked 3 years, 6 months ago. The sum of all possible outcomes is always 1 (or 100%) because it is certain that one of the possible outcomes will happen. Based on your flip results, you will infer which of the coins you were given. Recommended: Please try your approach. 5, or you will stay in the current state with probability 0. The game is played in stages. The first person to flip heads wins. "The coin tosses are independent events; the coin doesn't have a memory. If the probability of flipping heads is 70%, then the list contains 70 ones and 30 zeros. You flipped 1 coin of type US 1¢ Penny: Timestamp: 2020-05-05 01:39:10 UTC. The majority of times, if a coin is heads-up when it is flipped, it will remain heads-up when it lands. First, note that the problem will likely make reference to a "fair" coin. In particular, ( 10 3) = 10! 3! 7!. Wait for your coin to dry; Flip (toss) your coin 40 times and record the number of times it lands on its side. The post is correct that the odds of getting. Your function will have to label each of those sequences with a door, since leaving any sequence unlabeled would contradict the assumption that the method always produces a result after F flips. I need to land on heads 3 times or more out of 6, in 80% of all trials. So, I'll do it faster! When we flip the coin 9 times there are $$2^9$$ possible outcomes that can happen. The coin has no desire to continue a particular streak, so it's not affected by any number of previous coin tosses. Coin Toss Activity is a great way for students to have fun and learn about calculating probability. Probability measures how certain we are a particular event will happen in a specific instance. The two non-straight lines in Fig. A single flip of a coin has an uncertain outcome. We could call a Head a success; and a Tail, a failure. In the first simulation, player A got lucky with 4 heads in 5 tosses. For four coin flips, our intuition was probably right: more likely to get two heads. 60 I tried this: P(2H) = 4C2 * 0. 7 flip_coins(100, 100, 5) #> [1] 0. Ask Question Asked 3 years, 6 months ago. The program should call a separate function flip()that takes no arguments and returns 0 for tails and 1 for heads. We all know a coin toss gives you a 50% chance of winning, but is it always that way? Delve into the inner-workings of coin toss probability with this activity. The sum of all possible outcomes is always 1 (or 100%) because it is certain that one of the possible outcomes will happen. Flipping the coin once is a Bernoulli trial, since there are exactly two complementary outcomes (flipping a head and flipping a tail), and they are both 1 2 \frac{1}{2} 2 1 no matter how many times the coin is flipped. What is the probability that a fair coin lands on heads on 4 out of 5 flips? Wha are the four properties of a binomial probability distribution? In a carnival game, there are six identical boxes, one of which contains a prize. We’ll set p=0. 5, which is our probability of tossing heads and moving forward.


# Math Help - Need Help with Elem. Linear Algebra Problem

1. ## Need Help with Elem. Linear Algebra Problem

Express the invertible matrix

$\left(\begin{array}{ccc}1&2&1\\1&0&1\\1&1&2\end{ar ray}\right)$

as a product of elementary matrices.
--------------------------
The answer in the back of the book has a product of 6 matrices. I think I'm supposed to use the identity matrix and perform the same row operations on it as I would to obtain the above matrix... or something like that. Any help with this would be appreciated!

2. Originally Posted by paupsers
Express the invertible matrix

$\left(\begin{array}{ccc}1&2&1\\1&0&1\\1&1&2\end{ar ray}\right)$

as a product of elementary matrices.
--------------------------
The answer in the back of the book has a product of 6 matrices. I think I'm supposed to use the identity matrix and perform the same row operations on it as I would to obtain the above matrix... or something like that. Any help with this would be appreciated!
Yes, that's right. A "row operation" is one of three kinds:
1) swap two rows
2) multiply every member of a row by the same constant.
3) replace every member of a row by itself plus a constant time the corresponding member of a second row.

An "elementary matrix" is a matrix constructed from the identity matrix by a single row operation.

What you want to do is reduce the given matrix to the identity matrix (which is possible because it is invertible) by row operations, writing down the elementary matrix corresponding to each row operation.

For example, starting from
$\left(\begin{array}{ccc}1&2&1\\1&0&1\\1&1&2\end{ar ray}\right)$
My first steps would be to reduce the first column to the column $\left(\begin{array}{c}1 \\ 0 \\ 0\end{array}\right)$.
The number in the upper left is already 1 so I don't need to change that. I can get 0 in the next row by subtracting the first row from the second row. The elementary matrix corresponding to that is
$\left(\begin{array}{ccc}1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)$
To get 0 in the third row, I need to subtract the first row from the third row. That gives the elementary matrix $\left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & 0 & 1\end{array}\right)$

and they convert the matrix to
$\left(\begin{array}{ccc}1 & 2 & 1 \\ 0 & -2 & 0 \\ 0 & -1 & 1\end{array}\right)$
Now you need to convert the second column properly and then the third. Can you continue?

After you have converted the matrix to the identity matrix and written down all the corresponding elementary matrices (in the proper order) so they multiply to give the original matrix.

Normally it would take 9 row operations to reduce a 3 by 3 matrix to the identity matrix and so the matrix would be represented as the product of 9 elementary matrices. Notice that the upper left number was already 1 so we didn't need a row operation for that. Also that 0 now in the second row, third column will stay there we won't need a row operation to change that. I suspect that sort of thing, a 0 or 1 already in the correct position will happen once more to give 6 rather than 9 elementary matrices.

3. This is exactly what I tried doing, but it's not the answer the book gives (although it is similar). Is the solution to this problem unique?

4. No, it is not because you can follow different "paths" to row reduce to the identity matrix. What I showed is the most commonly used "path".


# Transitive relation

Consider $A$ is a relation de fined on $R$ (real numbers) where $A = \{(a,b):|a-b|<4, a, b \in R\}$. Prove/disprove $A$ is transitive.

I know if $|a-b|<4$ and $|b-c|<4$, then, $|a-c|<4$ ; A is transitive. Can I directly prove it with any counter example such as for $a=6$, $b=3$ and $c=1$ this relation is not transitive because $|6-1|>4$.

Is this suitable for prove or disprove questions of relations?

Your answer is correct. In general a relation $A$ is transitive if for all $a,b,c$ we have $(a,b)\in A$ and $(b,c)\in A$ implies $(a,c)\in A$. That means that if we can find just one counter example (such as $a=6$, $b=3$ and $c=1$) $A$ cannot be transitive.
Your answer is quite right. The question is basically "if $a$ is near to $b$, and $b$ is near to $c$, is $a$ near to $c$?"


# Expansion in terms of legendre polynomial

Obtain the first three terms in the expansion of function in terms of legendre polynomial F(x) in a series of the form $$F(x) = \sum_{k=0}^{\infty} A_k P_k(x)$$ where $$F(x)=\{\cos(x) \text{ for } 0 \le x \le \pi/2 \$$$$0 \text{ for } \frac{\pi}{2} \le x \le \pi.\}$$

What I know is I have to use legendre's expansion formula i.e,$F(x) =\sum A_kP_k(x)$ where $-1≤x≤1$ But obviously I cannot use it directly because the range of $x$ differs. I have tried substituting $x=\cos(\theta)$ but no success so far.

Extended hint: As already noted in your almost identical question Legendre polynomials you should apply a linear transformation, because the Legendre polynomial have the orthogonality interval $[-1,1]$. With $x=\frac{\pi}{2}(t+1)$ and $t=\frac{2x}{\pi}-1,$ define $F(x)=:\tilde{F}(t).$ Now $\tilde{F}$ is defined on $[-1,1]$ and you can compute $$F(x)=\tilde{F}(t) = \sum A_k P_k(t)=\sum A_k P_k\left(\frac{2x}{\pi}-1\right)$$ with $$A_k=\frac{2k+1}{2}\int_{-1}^1 \tilde{F}(t) P_k(t) dt$$
If you use this approach and compute $A_0, \dots, A_3$, you get the following graphics with the original function $F(x)$ in red and the approximation in green:
• Yes of course, because your function is not defined for $x=-1$. The transformation associates $x=0 \leftrightarrow t=-1$ and $x=\pi \leftrightarrow t=1$. You should compute the expansion for $\tilde{F}(t), t\in[-1,1].$ – gammatester Sep 18 '17 at 15:08


# Homework Help: Product of two consecutive integers

1. Jan 4, 2010

### nobahar

1. The problem statement, all variables and given/known data

Prove that $$n^2+n$$ is even. Where n is a positive integer.

2. Relevant equations

$$n^2+n$$

3. The attempt at a solution

$$n^2+n = n(n+1)$$

One of which must be even, and therefore the product of 2 and an integer k.

$$n = 2k, \left \left 2*(k(n+1))$$

or

$$n+1 = 2k, \left \left 2*(n*k)$$

Is there a better way of doing this? I read this is not an inductive proof; what would this entail?

2. Jan 4, 2010

### rochfor1

This looks absolutely fine. No need for induction at all.

3. Jan 4, 2010

### HallsofIvy

As rochfor1 said, that is a perfectly good proof and simpler than "proof by induction".

But since you ask, here goes:

If n= 1, then $n^2+ n= 1^2+ 1= 2$ which is even.

Now, suppose that $k^2+ k$ is even and look at $(k+1)^2+ (k+1)$
$(k+1)^2+ (k+1)= k^2+ 2k+ 1+ k+ 1$$= (k^2+ 2k)+ 2k+ 2$. By the induction hypothesis, $k^2+ k$ is even and so $k^2+ k= 2m$ for some integer m (that is the definition of "even") so $(k+1)^2+ (k+1)= (k^2+ 2k)+ 2k+ 2$$= 2m+ 2k+ 2= 2(m+k+1)$. Since that is "2 times an integer", it is even.

Having proved that the statement is true for 1 and that "if it is true for k, it is true for k+1", by induction, it is true for all positive integers.

Yet a third way: If n is a positive integer, it is either even or odd.

case 1: n is even. Then n= 2m for some integer m. $n^2= 4m^2$ so $n^2+ n= 4m^2+ 2m= 2(m^2+ m)$. Since that is 2 times an integer, it is even.

case 2: n is odd. Then n= 2m+ 1 for some integer m. $n^2= (2m+1)^2= 4m^2+ 4m+ 1$ so $n^2+ n= 4m^2+ 4m+ 1+ 2m+ 1$$= 4m^2+ 6m+ 2= 2(2m^2+ 3m+ 1)$. Again that is 2 times an integer and so is even.

4. Jan 5, 2010

### nobahar

Many thanks rochfor1 and HallsofIvy!

5. Jan 5, 2010

### icystrike

Simple reasoning is needed.
For two consecutive integer , one will be even and another will be odd.
Thus;
$$n\equiv0(mod 2)$$
and
$$n+1\equiv1(mod 2)$$

We may now conclude that $$n(n+1)\equiv0(mod 2)$$ and means that n²+n is even.

6. Jan 5, 2010

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7. Jan 5, 2010

### Dick

If n is odd, n^2 is odd. If n is even, n^2 is even. odd+odd=even and even+even=even.

8. Jan 6, 2010

### ideasrule

OP: Why do you want a proof by induction? The proof you gave is simple and beautiful.

9. Jan 6, 2010

### Staff: Mentor

What you say is true, but you can't ascertain which of them will be even and which will be odd. In what follows, you are assuming that n is even and n+1 is odd. That is one of two possible cases, so you work is not complete.

10. Jan 7, 2010

### icystrike

yes that is true. for the other case , it can be proved analogously .


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Inequalities trick [#permalink]  16 Mar 2010, 09:11
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I learnt this trick while I was in school and yesterday while solving one question I recalled.
Its good if you guys use it 1-2 times to get used to it.

Suppose you have the inequality

f(x) = (x-a)(x-b)(x-c)(x-d) < 0

Just arrange them in order as shown in the picture and draw curve starting from + from right.

now if f(x) < 0 consider curve having "-" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.

Don't forget to arrange then in ascending order from left to right. a<b<c<d

So for f(x) < 0 consider "-" curves and the ans is : (a < x < b) , (c < x < d)
and for f(x) > 0 consider "+" curves and the ans is : (x < a), (b < x < c) , (d < x)

If f(x) has three factors then the graph will have - + - +
If f(x) has four factors then the graph will have + - + - +

If you can not figure out how and why, just remember it.
Try to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.

For the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.
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Re: Inequalities trick [#permalink]  22 Oct 2010, 05:33
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Yes, this is a neat little way to work with inequalities where factors are multiplied or divided. And, it has a solid reasoning behind it which I will just explain.

If (x-a)(x-b)(x-c)(x-d) < 0, we can draw the points a, b, c and d on the number line.
e.g. Given (x+2)(x-1)(x-7)(x-4) < 0, draw the points -2, 1, 7 and 4 on the number line as shown.

Attachment:

doc.jpg [ 7.9 KiB | Viewed 29149 times ]

This divides the number line into 5 regions. Values of x in right most region will always give you positive value of the expression. The reason for this is that if x > 7, all factors above will be positive.

When you jump to the next region between x = 4 and x = 7, value of x here give you negative value for the entire expression because now, (x - 7) will be negative since x < 7 in this region. All other factors are still positive.

When you jump to the next region on the left between x = 1 and x = 4, expression will be positive again because now two factors (x - 7) and (x - 4) are negative, but negative x negative is positive... and so on till you reach the leftmost section.

Since we are looking for values of x where the expression is < 0, here the solution will be -2 < x < 1 or 4< x < 7

It should be obvious that it will also work in cases where factors are divided.
e.g. (x - a)(x - b)/(x - c)(x - d) < 0
(x + 2)(x - 1)/(x -4)(x - 7) < 0 will have exactly the same solution as above.

Note: If, rather than < or > sign, you have <= or >=, in division, the solution will differ slightly. I will leave it for you to figure out why and how. Feel free to get back to me if you want to confirm your conclusion.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Math Forum Moderator Joined: 20 Dec 2010 Posts: 2028 Followers: 138 Kudos [?]: 1165 [15] , given: 376 Re: Inequalities trick [#permalink] 11 Mar 2011, 05:49 15 This post received KUDOS 6 This post was BOOKMARKED vjsharma25 wrote: VeritasPrepKarishma wrote: vjsharma25 wrote: How you have decided on the first sign of the graph?Why it is -ve if it has three factors and +ve when four factors? Check out my post above for explanation. I understand the concept but not the starting point of the graph.How you decide about the graph to be a sine or cosine waveform?Meaning graph starts from the +ve Y-axis for four values and starts from -ve Y-axis for three values. What if the equation you mentioned is (x+2)(x-1)(x-7)<0,will the last two ranges be excluded or the graph will also change? I always struggle with this as well!!! There is a trick Bunuel suggested; (x+2)(x-1)(x-7) < 0 Here the roots are; -2,1,7 Arrange them in ascending order; -2,1,7; These are three points where the wave will alternate. The ranges are; x<-2 -2<x<1 1<x<7 x>7 Take a big value of x; say 1000; you see the inequality will be positive for that. (1000+2)(1000-1)(1000-7) is +ve. Thus the last range(x>7) is on the positive side. Graph is +ve after 7. Between 1 and 7-> -ve between -2 and 1-> +ve Before -2 -> -ve Since the inequality has the less than sign; consider only the -ve side of the graph; 1<x<7 or x<-2 is the complete range of x that satisfies the inequality. _________________ Manager Joined: 29 Sep 2008 Posts: 150 Followers: 3 Kudos [?]: 49 [11] , given: 1 Re: Inequalities trick [#permalink] 22 Oct 2010, 10:45 11 This post received KUDOS 6 This post was BOOKMARKED if = sign is included with < then <= will be there in solution like for (x+2)(x-1)(x-7)(x-4) <=0 the solution will be -2 <= x <= 1 or 4<= x <= 7 in case when factors are divided then the numerator will contain = sign like for (x + 2)(x - 1)/(x -4)(x - 7) < =0 the solution will be -2 <= x <= 1 or 4< x < 7 we cant make 4<=x<=7 as it will make the solution infinite correct me if i am wrong Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 5682 Location: Pune, India Followers: 1415 Kudos [?]: 7357 [6] , given: 186 Re: Inequalities trick [#permalink] 11 Mar 2011, 18:57 6 This post received KUDOS Expert's post vjsharma25 wrote: I understand the concept but not the starting point of the graph.How you decide about the graph to be a sine or cosine waveform?Meaning graph starts from the +ve Y-axis for four values and starts from -ve Y-axis for three values. What if the equation you mentioned is (x+2)(x-1)(x-7)<0,will the last two ranges be excluded or the graph will also change? Ok, look at this expression inequality: (x+2)(x-1)(x-7) < 0 Can I say the left hand side expression will always be positive for values greater than 7? (x+2) will be positive, (x - 1) will be positive and (x-7) will also be positive... so in the rightmost regions i.e. x > 7, all three factors will be positive. The expression will be positive when x > 7, it will be negative when 1 < x < 7, positive when -2 , x < 1 and negative when x < -2. We need the region where the expression is less than 0 i.e. negative. So either 1 < x < 7 or x < -2. Now let me add another factor: (x+8)(x+2)(x-1)(x-7) Can I still say that the entire expression is positive in the rightmost region i.e. x>7 because each one of the four factors is positive? Yes. So basically, your rightmost region is always positive. You go from there and assign + and - signs to the regions. Your starting point is the rightmost region. Note: Make sure that the factors are of the form (ax - b), not (b - ax)... e.g. (x+2)(x-1)(7 - x)<0 Convert this to: (x+2)(x-1)(x-7)>0 (Multiply both sides by '-1') Now solve in the usual way. Assign '+' to the rightmost region and then alternate with '-' Since you are looking for positive value of the expression, every region where you put a '+' will be the region where the expression will be greater than 0. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: Inequalities trick [#permalink]  10 Aug 2011, 16:01
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WoW - This is a cool thread with so many thing on inequalities....I have compiled it together with some of my own ideas...It should help.

1) CORE CONCEPT
@gurpreetsingh -
Suppose you have the inequality

f(x) = (x-a)(x-b)(x-c)(x-d) < 0

Arrange the NUMBERS in ascending order from left to right. a<b<c<d
Draw curve starting from + from right.

now if f(x) < 0 consider curve having "-" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.

So for f(x) < 0 consider "-" curves and the ans is : (a < x < b) , (c < x < d)
and for f(x) > 0 consider "+" curves and the ans is : (x < a), (b < x < c) , (d < x)

If f(x) has three factors then the graph will have - + - +
If f(x) has four factors then the graph will have + - + - +

If you can not figure out how and why, just remember it.
Try to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.

For the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.

Note: Make sure that the factors are of the form (ax - b), not (b - ax)...

example -
(x+2)(x-1)(7 - x)<0

Convert this to: (x+2)(x-1)(x-7)>0 (Multiply both sides by '-1')
Now solve in the usual way. Assign '+' to the rightmost region and then alternate with '-'
Since you are looking for positive value of the expression, every region where you put a '+' will be the region where the expression will be greater than 0.

2) Variation - ODD/EVEN POWER
@ulm/Karishma -
if we have even powers like (x-a)^2(x-b)
we don't need to change a sign when jump over "a".
This will be same as (x-b)

We can ignore squares BUT SHOULD consider ODD powers
example -
2.a
(x-a)^3(x-b)<0 is the same as (x-a)(x-b) <0
2.b
(x - a)(x - b)/(x - c)(x - d) < 0 ==> (x - a)(x - b)(x-c)^-1(x-d)^-1 <0
is the same as (x - a)(x - b)(x - c)(x - d) < 0

3) Variation <= in FRACTION
@mrinal2100 -
if = sign is included with < then <= will be there in solution
like for (x+2)(x-1)(x-7)(x-4) <=0 the solution will be -2 <= x <= 1 or 4<= x <= 7

BUT if it is a fraction the denominator in the solution will not have = SIGN

example -
3.a
(x + 2)(x - 1)/(x -4)(x - 7) < =0
the solution will be -2 <= x <= 1 or 4< x < 7
we cant make 4<=x<=7 as it will make the solution infinite

4) Variation - ROOTS
@Karishma -
As for roots, you have to keep in mind that given $$\sqrt{x}$$, x cannot be negative.

$$\sqrt{x}$$ < 10
implies 0 < $$\sqrt{x}$$ < 10
Squaring, 0 < x < 100
Root questions are specific. You have to be careful. If you have a particular question in mind, send it.

Refer - inequalities-and-roots-118619.html#p959939
Some more useful tips for ROOTS....I am too lazy to consolidate

<5> THESIS -
@gmat1220 -
Once algebra teacher told me - signs alternate between the roots. I said whatever and now I know why Watching this article is a stroll down the memory lane.

I will save this future references....

Anyone wants to add ABSOLUTE VALUES....That will be a value add to this post

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Re: Inequalities trick [#permalink]  09 Sep 2013, 22:35
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karannanda wrote:
gurpreetsingh wrote:
I learnt this trick while I was in school and yesterday while solving one question I recalled.
Its good if you guys use it 1-2 times to get used to it.

Suppose you have the inequality

f(x) = (x-a)(x-b)(x-c)(x-d) < 0

Just arrange them in order as shown in the picture and draw curve starting from + from right.

now if f(x) < 0 consider curve having "-" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.

Don't forget to arrange then in ascending order from left to right. a<b<c<d

So for f(x) < 0 consider "-" curves and the ans is : (a < x < b) , (c < x < d)
and for f(x) > 0 consider "+" curves and the ans is : (x < a), (b < x < c) , (d < x)

If f(x) has three factors then the graph will have - + - +
If f(x) has four factors then the graph will have + - + - +

If you can not figure out how and why, just remember it.
Try to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.

For the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.

Hi Gurpreet,
Thanks for the wonderful method.
I am trying to understand it so that i can apply it in tests.
Can you help me in applying this method to the below expression to find range of x.
x^3 – 4x^5 < 0?

I am getting the roots as -1/2, 0, 1/2 and when i plot them using this method, putting + in the rightmost region, I am not getting correct result. Not sure where i am going wrong. Can you pls help.

Before you apply the method, ensure that the factors are of the form (x - a)(x - b) etc

$$x^3 - 4x^5 < 0$$

$$x^3 ( 1 - 4x^2) < 0$$

$$x^3(1 - 2x) (1 + 2x) < 0$$

$$4x^3(x - 1/2)(x + 1/2) > 0$$ (Notice the flipped sign. We multiplied both sides by -1 to convert 1/2 - x to x - 1/2)

Now the transition points are 0, -1/2 and 1/2 so put + in the rightmost region.
The solution will be x > 1/2 or -1/2 < x< 0.

Check out these posts discussing such complications:
http://www.veritasprep.com/blog/2012/06 ... e-factors/
http://www.veritasprep.com/blog/2012/07 ... ns-part-i/
http://www.veritasprep.com/blog/2012/07 ... s-part-ii/
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 5682 Location: Pune, India Followers: 1415 Kudos [?]: 7357 [2] , given: 186 Re: Inequalities trick [#permalink] 23 Jul 2012, 02:13 2 This post received KUDOS Expert's post Stiv wrote: VeritasPrepKarishma wrote: mrinal2100: Kudos to you for excellent thinking! Correct me if I'm wrong. If the lower part of the equation$$\frac {(x+2)(x-1)}{(x-4)(x-7)}$$ were $$4\leq x \leq 7$$, than the lower part would be equal to zero,thus making it impossible to calculate the whole equation. x cannot be equal to 4 or 7 because if x = 4 or x = 7, the denominator will be 0 and the expression will not be defined. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Inequalities trick [#permalink]  06 Jan 2015, 02:35
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Just came across this useful discussion.

VeritasPrepKarishma has given a very lucid explanation of how this “wavy line” method works.

I have noticed that there is still a little scope to take this discussion further. So here are my two cents on it.

I would like to highlight an important special case in the application of the Wavy Line Method

When there are multiple instances of the same root:

Try to solve the following inequality using the Wavy Line Method:

$$(x-1)^2(x-2)(x-3)(x-4)^3 < 0$$

To know how you did, compare your wavy line with the correct one below.

Did you notice how this inequality differs from all the examples above?

Notice that two of the four terms had an integral power greater than 1.

How to draw the wavy line for such expressions?

Let me directly show you how the wavy line would look and then later on the rule behind drawing it.

Attachment:
File comment: Observe how the wave bounces back at x = 1.

bounce.png [ 10.4 KiB | Viewed 2175 times ]

Notice that the curve bounced down at the point x = 1. (At every other root, including x = 4 whose power was 3, it was simply passing through them.)

Can you figure out why the wavy line looks like this for this particular inequality?

(Hint: The wavy line for the inequality

$$(x-1)^{38}(x-2)^{57}(x-3)^{15}(x-4)^{27} < 0$$

Is also the same as above)

Come on! Give it a try.

If you got it right, you’ll see that there are essentially only two rules while drawing a wavy line. (Remember, we’ll refer the region above the number line as positive region and the region below the number line as negative region.)

How to draw the wavy line?

1. How to start: Start from the top right most portion. Be ready to alternate (or not alternate) the region of the wave based on how many times a point is root to the given expression.

2. How to alternate: In the given expression, if the power of a term is odd, then the wave simply passes through the corresponding point (root) into the other region (to –ve region if the wave is currently in the positive region and to the +ve region if the wave is currently in the negative region). However, if the power of a term is even, then the wave bounces back into the same region.

Now look back at the above expression and analyze your wavy line. Were you (intuitively) using the above mentioned rules while drawing your wavy line?

Solution

Once you get your wavy line right, solving an inequality becomes very easy. For instance, for the above inequality, since we need to look for the space where the above expression would be less than zero, look for the areas in the wavy line where the curve is below the number line.

So the correct solution set would simply be {3 < x < 4} U {{x < 2} – {1}}

In words, it is the Union of two regions region1 between x = 3 and x = 4 and region2 which is x < 2, excluding the point x = 1.

Food for Thought

Now, try to answer the following questions:

1. Why did we exclude the point x = 1 from the solution set of the last example? (Easy Question)
2. Why do the above mentioned rules (especially rule #2) work? What is/are the principle(s) working behind the curtains?

Foot Note: Although the post is meant to deal with inequality expressions containing multiple roots, the above rules to draw the wavy line are generic and are applicable in all cases.

- Krishna
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Re: Inequalities trick [#permalink]  19 Mar 2010, 11:59
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ttks10 wrote:
Can u plz explainn the backgoround of this & then the explanation.

Thanks

i m sorry i dont have any background for it, you just re-read it again and try to implement whenever you get such question and I will help you out in any issue.

sidhu4u wrote:
I have applied this trick and it seemed to be quite useful.

Nice to hear this....good luck.
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Re: Inequalities trick [#permalink]  26 May 2011, 19:55
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chethanjs wrote:
mrinal2100 wrote:
if = sign is included with < then <= will be there in solution
like for (x+2)(x-1)(x-7)(x-4) <=0 the solution will be -2 <= x <= 1 or 4<= x <= 7

in case when factors are divided then the numerator will contain = sign

like for (x + 2)(x - 1)/(x -4)(x - 7) < =0
the solution will be -2 <= x <= 1 or 4< x < 7
we cant make 4<=x<=7 as it will make the solution infinite

correct me if i am wrong

Can you please tell me why the solution gets infinite for 4<=x<=7 ?

Thanks.

(x -4)(x - 7) is in denominator.
Making x=4 or 7 would make the denominator 0 and the entire function undefined.

Thus, the range of x can't be either 4 or 7.
4<=x<=7 would be wrong.
4<x<7 is correct because now we removed "=" sign.
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Re: Inequalities trick [#permalink]  22 Jul 2012, 02:03
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VeritasPrepKarishma wrote:
mrinal2100: Kudos to you for excellent thinking!

Correct me if I'm wrong.
If the lower part of the equation$$\frac {(x+2)(x-1)}{(x-4)(x-7)}$$ were $$4\leq x \leq 7$$, than the lower part would be equal to zero,thus making it impossible to calculate the whole equation.
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Re: Inequalities trick [#permalink]  18 Oct 2012, 04:17
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GMATBaumgartner wrote:
gurpreetsingh wrote:
ulm wrote:
if we have smth like (x-a)^2(x-b)
we don't need to change a sign when jump over "a".

yes even powers wont contribute to the inequality sign. But be wary of the root value of x=a

Hi Gurpreet,
Could you elaborate what exactly you meant here in highlighted text ?

Even I have a doubt as to how this can be applied for powers of the same term . like the example mentioned in the post above.

If the powers are even then the inequality won't be affected.

eg if u have to find the range of values of x satisfying (x-a)^2 *(x-b)(x-c) >0

just use (x-b)*(x-c) >0 because x-a raised to the power 2 will not affect the inequality sign. But just make sure x=a is taken care off , as it would make the inequality zero.
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Re: Inequalities trick [#permalink]  18 Oct 2012, 09:27
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GMATBaumgartner wrote:
Hi Gurpreet,
Could you elaborate what exactly you meant here in highlighted text ?

Even I have a doubt as to how this can be applied for powers of the same term . like the example mentioned in the post above.

In addition, you can check out this post:

http://www.veritasprep.com/blog/2012/07 ... s-part-ii/

I have discussed how to handle powers in it.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Math Expert Joined: 02 Sep 2009 Posts: 28348 Followers: 4485 Kudos [?]: 45444 [1] , given: 6761 Re: Inequalities trick [#permalink] 02 Dec 2012, 06:25 1 This post received KUDOS Expert's post GMATGURU1 wrote: Going ahead in this inequalities area of GMAT, can some one has the problem numbers of inequalities in OG 11 and OG 12? Cheers, Danny Search for hundreds of question with solutions by tags: viewforumtags.php DS questions on inequalities: search.php?search_id=tag&tag_id=184 PS questions on inequalities: search.php?search_id=tag&tag_id=189 Hardest DS inequality questions with detailed solutions: inequality-and-absolute-value-questions-from-my-collection-86939.html Hope it helps. _________________ Current Student Status: Nothing comes easy: neither do I want. Joined: 12 Oct 2009 Posts: 2800 Location: Malaysia Concentration: Technology, Entrepreneurship Schools: ISB '15 (M) GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35 Followers: 196 Kudos [?]: 1162 [1] , given: 235 Re: Inequalities trick [#permalink] 20 Dec 2012, 20:04 1 This post received KUDOS VeritasPrepKarishma wrote: The entire concept is based on positive/negative factors which means <0 or >0 is a must. If the question is not in this format, you need to bring it to this format by taking the constant to the left hand side. e.g. (x + 2)(x + 3) < 2 x^2 + 5x + 6 - 2 < 0 x^2 + 5x + 4 < 0 (x+4)(x+1) < 0 Now use the concept. Yes this is probable but it might not be possible always to group them. So in case you are unsure just follow the number plugging approach. But most of the times this trick would be very handy. _________________ Fight for your dreams :For all those who fear from Verbal- lets give it a fight Money Saved is the Money Earned Jo Bole So Nihaal , Sat Shri Akaal Support GMAT Club by putting a GMAT Club badge on your blog/Facebook GMAT Club Premium Membership - big benefits and savings Gmat test review : 670-to-710-a-long-journey-without-destination-still-happy-141642.html Math Expert Joined: 02 Sep 2009 Posts: 28348 Followers: 4485 Kudos [?]: 45444 [1] , given: 6761 Re: Inequalities trick [#permalink] 11 Nov 2013, 06:51 1 This post received KUDOS Expert's post anujpadia wrote: Can you please explain the above mentioned concept in relation to the following question? Is a > O? (1) a^3 - 0 < 0 (2) 1- a^2 > 0 Can you please explain the scenario when (x-a)(x-B)(x-C)(x-d)>0? Sorry, but finding it difficult to understand. Check alternative solutions here: is-a-0-1-a-3-a-0-2-1-a-86749.html Hope this helps. _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 5682 Location: Pune, India Followers: 1415 Kudos [?]: 7357 [1] , given: 186 Re: Inequalities trick [#permalink] 23 Apr 2014, 03:43 1 This post received KUDOS Expert's post 1 This post was BOOKMARKED PathFinder007 wrote: I have a query. I have following question x^3 - 4x^5 < 0 I can define this as (1+2x).x^3(1-2x). now I have roots -1/2, 0, 1/2. so in case of >1/2 I will always get inequality value as <0 and in case of -1/2 and 0 I will get value as 0. So How I will define them in graph and what range I will consider for this inequality. Thanks The factors must be of the form (x - a)(x - b) .... etc x^3 - 4x^5 < 0 x^3 * (1 - 4x^2) < 0 x^3 * (1 - 2x) * (1 + 2x) < 0 x^3 * (2x - 1) * (2x + 1) > 0 (Note the sign flip because 1-2x was changed to 2x - 1) x^3 * 2(x - 1/2) *2(x + 1/2) > 0 So transition points are 0, 1/2 and -1/2. ____________ - 1/2 _____ 0 ______1/2 _________ This is what it looks like on the number line. The rightmost region is positive. We want the positive regions in the inequality. So the desired range of x is given by x > 1/2 or -1/2 < x< 0 For more on this method, check these posts: http://www.veritasprep.com/blog/2012/06 ... e-factors/ http://www.veritasprep.com/blog/2012/07 ... ns-part-i/ http://www.veritasprep.com/blog/2012/07 ... s-part-ii/ The links will give you the theory behind this method in detail. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: Inequalities trick [#permalink]  11 Nov 2014, 08:52
1
KUDOS
For statement I

(1) (1-2x)(1+x)<0

Once you get this into the req form

2 (x-1/2) (x+1) > 0

sol +++++ -1 ------ 1/2 +++++
As Karishma pointed out ponce in the req form, the rightmost will always be positive and the alternating will happen from there.

So sol for this x> 1/2 and x<-1

Integers greater than 1/2 and less than -1 , thus |x| may be >= 1. Unsure

Thus Insufficient.

For Statement II

(2) (1-x)(1+2x)<0

Once you get this into the req form

2(x+1/2) (x-1) > 0

sol +++++ -1/2 ------ 1 +++++

So sol for this x>1 and x< -1/2

Integers greater than 1 and less than -1/2 , thus |x| may be >= 1. Unsure

Thus Insufficient.

Combining Both the statements

x<-1 and x>1

Thus Integers for this range will give |x| > 1

Thus Sufficient.

Hope this helps. I am not very confident of my solution though. Its my first solution gmatclub

mayankpant wrote:
gurpreetsingh wrote:
I learnt this trick while I was in school and yesterday while solving one question I recalled.
Its good if you guys use it 1-2 times to get used to it.

Suppose you have the inequality

f(x) = (x-a)(x-b)(x-c)(x-d) < 0

Just arrange them in order as shown in the picture and draw curve starting from + from right.

now if f(x) < 0 consider curve having "-" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.

Don't forget to arrange then in ascending order from left to right. a<b<c<d

So for f(x) < 0 consider "-" curves and the ans is : (a < x < b) , (c < x < d)
and for f(x) > 0 consider "+" curves and the ans is : (x < a), (b < x < c) , (d < x)

If f(x) has three factors then the graph will have - + - +
If f(x) has four factors then the graph will have + - + - +

If you can not figure out how and why, just remember it.
Try to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.

For the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.

Hi
Can you please explain this question to me using the graph. I am missing the point when graph is being used here?

If x is an integer, is |x|>1?
(1) (1-2x)(1+x)<0
(2) (1-x)(1+2x)<0

For me ,the first equations roots are -1 and 1/2. Now I am struggling to get to the correct sign using the graph method here.

Same for second equation: roots are 1 and -1/2 but struggling for the sign.

THanks
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Re: Inequalities trick [#permalink]  16 Mar 2010, 09:46
Can u plz explainn the backgoround of this & then the explanation.

Thanks
Re: Inequalities trick   [#permalink] 16 Mar 2010, 09:46

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