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Coin Flip Probability Calculator At Least.
Enter the number of possible outcomes. The outcomes of each toss will be reflected on the graph. And we have (so far): = p k × 0. The probability that the card is a. 25) = 15/sqrt (53). When we flip a coin, what we get does not at all depend on any previous flips, and so flipping a. Top Answer. The 1 is the number of opposite choices, so it is: n−k. 3) for any probability question, first decide whether it is easier to calculate it directly, or easier to calculate the opposite and subtract from 1. This is a classic binomial distribution problem. 5 and the probability of a tail (p(t)) is 0. In this course, you'll learn about fundamental probability concepts like random variables (starting with the classic coin flip example) and how to calculate mean and variance, probability distributions, and conditional probability. 05 to establish bias at 95% confidence one sided. P (HTT + THT + TTH) =. Once all the numbers are obtained, calculate the probability. You must derive the pm and show your work. We say that the sequence is balanced when there are equal number of heads and tails. It is a number between and including the numbers 0 and 1. All posts by admin. So the probability of getting two heads is: 1 in 4 = 0. If a tossed coin comes up tails 10 times in a row, most people will expect it to come up heads on the next flip. Answer = B. 0) and the number of tosses, then click "Toss". We flip a fair coin 10 times. After 7 times we. But if you continue flipping the coin, the outcome grows closer to 50/50. For example, the probability of the union of the mutually exclusive events and in the random experiment of one coin toss, (), is the sum of probability for and the probability for , () + (). What is the probability that we get heads in at least 8 of the 10 flips?. "At least one" probability with coin flipping. Consider flip a coin 5 times. (b) The two events are (A) a flip turns out heads (B) that there are because the coin is fair, and the probability of (B) I found by again listing out the ways we can get two or more consecutive heads H H H H, H H H T, H H T T, etc of a total 2 4 = 16 possibilities two coins could take. The probability that all of these 20 toss successions were not all heads = "X to the power of 86,381". Use Markov's inequality to give an upper bound on the probability that the coin lands heads at least 120 times. And we have, we have the following sample space. For fun on Saturday night, you and a friend are going to flip a fair coin 10 times (geek!). 5, less than 0. 5, and the probability of one coin flip landing tails is 0. We flip a fair coin 10 times. Here we need more information. Assuming the coin is fair , the probability of getting a head is 1 2 or 0. What is the probability that you will get heads more than 14 times?. What is the probability of getting two heads in two tosses? The probability that the coin when tossed turns up heads is 1/2. There are 32 sample solutions in the solution set of the 5 coin toss. Suppose we have a fair coin (so the heads-on probability is 0. ) before ending up with k=4 heads?. The probability of flipping 10 heads in a row, assuming a randomly picked coin, is (1/100)*1 + (99/100)* (1/2) 10. (a) Write down the sample space of this experiment. Now, by looking at the formula, Probability of selecting an ace from a deck is, P (Ace) = (Number of favourable outcomes) / (Total number of favourable outcomes) P (Ace) = 4/52. Strings and integers are never equal. Classical Probability. Let A be the event that the coin shows heads at least 4 times. What is the probability of getting at least 2 tails ? Solution : When four coins are tossed once, total no. We can calculate the probability of two or more Independent events by multiplying. In this case, the probability measure is given by P(1) = P(2. Decision-theoretic analysis of how to optimally play Haghani & Dewey 2016's 300-round double-or-nothing coin-flipping game with an edge and ceiling better than using the Kelly Criterion. For example, the probability of the union of the mutually exclusive events and in the random experiment of one coin toss, (), is the sum of probability for and the probability for , () + (). #p=1/2# The probability of not getting a head in a single toss. When we flip a coin, only two outcomes are possible - heads and tails. of all possible outcomes = 2 x 2 x 2 x 2 = 16. If the probability of getting at least one ‘six’ is to exceed 0. 3% of the time. Let A be the event that the coin shows heads at least 4 times. Every flip of the coin doesn't depend on the other coin flips, and we are dealing with a situation where one thing must occur as well as several other things. The probability of a particular number of defective parts in a sizable lot can be set. e head or tail. Flip a coin 20 times if head comes 8 times, tail comes 12 times then the probability of heads P(H) = 8/20 = 2/5=0. Classical Probability. Probability of compound events Learn how to calculate the probability of at least 2 ~ s Coin toss probability When flipping a coin, what is the probability to get a head?. , HHH, HHT, HTH, HTT, THH, THT, TTH, TTT Out of which there are 4 set which contain at least 2 Heads i. 5), then we can use the dbinom function to calculate the probability of getting 5 heads in 10 trials. We do not know if we will get heads or tails. Example: Suppose you plan to toss a coin twice, and want to find the probability of rolling a head on both tosses. More generally, there are situations in which the coin is biased, so that heads and tails have different probabilities. 5 or 1/2, 1. 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. Worked-out problems on probability involving tossing or throwing or flipping three coins: 1. The odds on the one zillionth and first toss are still 50/50. e head or tail. Strings and integers are never equal. ) Answer by ewatrrr(24375) ( Show Source ):. If the result is two heads, you win $1. A fair coin is flipped 5 times. Call heads a success. " Well, let me explain that these two problems are basically the same, that is, from the point of view of mathematics. Answers ( 2) Since the coin has two sides heads or tails. (c) Two heads occur, given at least one head occurs. The probability of getting at least two heads when tossing a coin three. If the probability of none is x, then the probability of at least one is 1-x. 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. That is to say, there is 50% chance of getting either. Coin tossing experiment always plays a key role in probability concept. The experimental probability depends upon the actual outcome of the experiment. Example 1: A fair coin is tossed 5 times. There’s only one way for it to land on heads, so the probability is ½. Change to get matching values on all dice. Find each probability. Redistributed equally likely i can solve problems and a and factor even the probability of the card. H H H H T T H H T ……★ H. This is an “and” situation. We flip three coins, hoping for at least two tails. Where D is the number of dice. So, the probability of tossing heads is equal to 1/2. You must derive the pm and show your work. We use coin flipping as a first step in understanding the connection between these two ways of determining the probability of an event. Which gives us: = p k (1-p) (n-k) Where. If a coin is tossed 500 times and the tail appears 159 times, find the probability of getting a tail. Tossing a Biased Coin Michael Mitzenmacher When we talk about a coin toss, we think of it as unbiased: with probability one-halfit comes up heads, and with probability one-halfit comes up tails. It is studied from 18th century. 75 and more That was a simple example using independent events (each toss of a coin is independent of the previous toss), but tree diagrams are really wonderful for figuring out dependent events (where an event depends on what happens in the previous event. 999023438 ^ 710 = 0. If playback doesn't begin shortly, try restarting your device. Looking at the event we just talked about, the event of “tails at least once” could be called E and written as. That means, if we do a long enough experiment, the average value each coin flip should be about 0. Probability Distribution in details: https://www. The result is shown in the table below. Users may refer the below solved example work with steps to learn how to find what is the probability of getting at-least 1 tail, if a coin is tossed fix times or 6 coins tossed together. The probability is 0. Would you be more likely to get at least 70% tails if you flip a fair coin 10 times or if you flip a fair coin 1000 times?. Home / ; Articles posted by admin ( Page168224 ) Annual capacity of plant. The probability of rolling a 1 or a 2: P(1) + P(2) = 1 6 + 1 6 = 2 6 ˇ0:33. Without replacing the marble, you pull another marble out of the bag. The manual states that the lifetime of the product, defined as the amount of time (in years) the product works properly until it. 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. The Math Behind a Coin Toss. Whenever we go through the stuff probability in statistics, we will definitely have examples with coin tossing. NEG (1/2 ∙ ½ ∙1/2) = 1/8 1 - 1/8 = 7/8 8. 5), then we can use the dbinom function to calculate the probability of getting 5 heads in 10 trials. A fair die is tossed and the number facing up is noted. For 100 flips, if the actual heads probability is 0. When 3 coins are tossed randomly 250 times and it is found that three heads appeared 70 times. To calculate the probability of an event occurring, we count how many times are event of interest can occur (say flipping heads) and dividing it by the sample space. Click on the button that says "flip coin" as many times as possible in order to calculate the probability. The probability that the card is a. what do you expect to happen to the experimental probability of getting tails as you increase the number of trials a. This is the same as last time, but the true option tells excel that we want to know the probability of AT LEAST 1 success in two flips (i. 6 that an "unfair" coin will turn up tails on any given toss. the coin can also land upright. We roll a die, hoping for a 2 or a 5. of all possible outcomes = 2 x 2 x 2 x 2 = 16. You can use this tool to solve either for the exact probability of observing exactly x events in n trials, or the cumulative probability of observing X ≤ x, or the cumulative probabilities of observing X < x or X ≥ x or X > x. This is a negative binomial experiment because: The experiment consists of repeated trials. Exercise 1. In other words, if you assign the success of. In a large discrete math class, 55% of the students. Each outcome has a fixed probability, the same from trial to trial. There is one way for this to occur, giving us the probability of 1/256. 5 because 2 outcomes (heads or tails) are equally possible when a balanced coin is flipped. Easycalculation. P ( club or face card) = P ( club) + P ( face card) − P ( club and face card) = 13 52 + 12 52 − 3 52 = 22 52 = 11 26 ≈ 0. If a coin is tossed 500 times and the tail appears 159 times, find the probability of getting a tail. Probability is the study of regularities that emerge in the outcomes of random experiments. Therefore, using the probability formula. Each flip is 50/50 (unless you shave the edge). Coin flip probabilities deal with events related to a single or multiple flips of a fair coin. Note: This rule is useful to calculate the probability of "at least one". After you have flipped the coin so many times, you should get answers close to 0. In this video, we' ll explore the probability of getting at least one heads in multiple flips of a fair coin. Easycalculation. Diaconis has even trained himself to flip a coin and make it come up heads 10. I believe this is the correct approach, however. Suppose you flip two coins. There are may different polyhedral die included, so you can explore the probability of a 20 sided die as well as that of a regular cubic die. for m = 0 or 2 there is only one way for the out come (both tosses give heads or tails): C 0 = C 2 = 1 for m = 1 (one head, two tosses) there are. If you see the complement of at least 1 tails means 0 tails or 5 heads, then we must surely. com DA: 23 PA: 37 MOZ Rank: 63. If the coin is tossed 3 times, what is the probability that at least one of the tosses will turn up tails? A. 04 is the probability of getting 7 Heads in 8 tosses. In the previous version we suggested that the terms “odds” and “probability” could be used interchangeably. Most coins have probabilities that are nearly equal to 1/2. The probability of one head and one tail is 2/3. There are 2 possible outcomes, both of which are equally likely. This geometric probability calculator is used to find geometric distribution probability with total number of occurrence & probability of success. What is the probability of getting exactly 3 Heads in five consecutive flips. Top Answer. This free probability calculator can calculate the probability of two events, as well as that of a normal distribution. Suppose we have a fair coin (so the heads-on probability is 0. That implies that the entropy of (2 x 10 3 coins flipped) is equal to twice the entropy of (10 3 coins flipped). Calculating probablities can be used to help us make decision. So, for your coin, the area of the side of the coin is pi*24. Give an example in which we would be interested in an either/or probability. Example2 : Find the probability for a couple having three children at least one is girl. Our fraction portion is not reduced down completely. Or say 3, 4 or 5 coins? The outcomes of these coin tosses will differ. Picking numbers randomly means that there is no specific order in which they are chosen. This is a negative binomial experiment because: The experiment consists of repeated trials. The experimental probability. The above explanation will help us to solve the problems on finding the probability of tossing three coins. get the probability concepts in engineering by alfredo associate that we allow here and check out the link. What is the probability that a sum of 5 is rolled a) exactly 6 times b) at least 4 times c) at most 5 times 5. -a coin is tossed 7 times, what is the probability of tails occurring at least once?-easier to answer the opposite: P(tails at least once)=1-P(no tails). If the probability of getting at least one 'six' is to exceed 0. For example, even the 50/50 coin toss really isn’t 50/50 — it’s closer to 51/49, biased toward whatever side was up when the coin was thrown into the air. When events are disjoint, it’s easy to calculate the probability of one event or the other happening. This coin is ipped 200 times. Whether you want to toss a coin or ask a girl out, there are only two possibilities that can occur. When events are disjoint, it’s easy to calculate the probability of one event or the other happening. Designed using Canva. The probability of an event A, symbolized by P(A), is a number between 0 and 1, inclusive, that measures the likelihood of an event in the following way: If P(A) > P(B) then event A is more likely to occur than event B. COIN FLIPPING AND COMPOUND PROBABILITY Work with a partner to make a team of 2 students. Even if a question doesn't invoke the coin toss, the way we approach a coin toss problem can carry over to other types of probability questions. An experiment is a planned operation carried out under controlled conditions. The coin is tossed 3 times and there is an equal probability that the coin will turn up heads or tail on each toss (which means that each toss has only two possible outcomes) 2*2*2=2^3=8 Multiply the number of possible outcomes per toss to arrive at the total number of possible outcomes. An example of two independent events is as follows; say you rolled. Toss a coin 4 times. you got not a head for at least one flip. An event that cannot occur has a probability (of happening) equal to 0 and the probability of an event that is certain to occur has a probability equal to 1. Click the "Quiz Me" button to complete the activity. The probability is relatively high, but this scenario still seems very unlikely! 4. (The calculator also reports the cumulative probabilities. For example, we want at least 2 heads from 3 tosses of coin. Consider the probability of rolling a 4 and 6 on a single roll of a die; it is not possible. Now the final step is negating back — the probability of getting at least 1 "heads" is: 1 − p n = 1 − 1. (a) Write down the sample space of three such flips. This is a rare event: we would reject the Null Hypothesis (that the coin is fair) at the P<0. This may sound like a limitation, and in a way it is, but estimating probabilities is an extremely powerful technique that can enable us to build non-trivial applications, including:. P(H,3) = P(H) xx P(3) = 1/2 xx1/6 The chances of flipping a head and rolling a 3 is 1/12. The following is the probability associated with 1 unbiased coin being tossed n time(s) in succesion or n unbiased coins being tossed at the same time and the result recorded. for i in range (1000): if flip_coin(8) == "3": ## changed to flip_coin() multiple_heads_count += 1 The value of flip_coin(8) is an integer, but you are checking for equality with the string "3". On tossing a coin, the probability of getting a head is: P(Head) = P(H) = 1/2. 75K subscribers. Determine the probability that the coin comes up tails exactly 15 times. Top Answer. Therefore the probability is 19/59. 5, and we want to know the probability that it will land on heads k = 43 times or less: p. So the probability of getting the one sequence among them that contains exactly N heads is 1 in 2 N. The ratio of successful events A = 31 to the total number of possible combinations of a sample space S = 32 is the probability of 1 tail in 5 coin tosses. pdf from MATH 270 at Hartnell College. What is the value of 1000p? Math. To calculate expected value for discrete events, like dice rolls and coin flips, we multiply the value by the probability and add up all possibilities. Experiment description: Toss a coin three times and record heads or tails on each toss. Enter the trials, probability, successes, and probability type. Statistics and probability: 1-1 1. Let’s consider an example where we flip a coin and roll a die simultaneously. Use the normal approximation to the binomial distribution to estimate the probability that the number of heads is greater than or equal to 60. No problem! We can solve this problem rather quickly with the assistance of our handy graphing calculator. A Random Variable is given a capital letter, such as X or Z. How can we calculate the odds of this happening when the normal rules of probability apply? If we toss a fair coin N times, there are 2 N different sequences of heads and tails possible, all of them equally likely. But first find the sample space of what you are computing. A pair of dice is rolled 20 times. 5, then A will have won after scenario 2 (which happens with probability y). The coin being a fair one, the outcome of a head in one toss has a probability $$p = 0. S = kB 2 N. ⇒ The number of possible choices in tossing a coin = 2. This probability is the power of the test. Tossing a Biased Coin Michael Mitzenmacher∗ When we talk about a coin toss, we think of it as unbiased: with probability one-half it comes up heads, and with probability one-half it comes up tails. Calculate the probability of drawing ANY PAIR in a row from a deck of cards (with replacement). 1/12 When you flip a coin there are two possible outcomes (heads or tails) and when you roll a die there are six outcomes(1 to 6). Coins And Probability Trees. Coin flip probabilities deal with events related to a single or multiple flips of a fair coin. John has 3 unique coins such that the probability of obtaining a head when the coin is flipped is 110, 210 and 310, respectively. We could call a Head a success; and a Tail, a failure. I believe this is the correct approach, however. Flip a coin. Each team member will have 1 coin to flip. If it is heads, then the experimental probability is 1/1. In the case where A and B are mutually exclusive events, P(A ∩ B) = 0. Over many coin flips the probability of at least half of the flips being heads (or tails) will converge to 0. The probability of getting exactly 3 heads out of 8 with a fair coin would be 8C3 / 2^8 = 56 / 256 =. Coin flip probabilities deal with events related to a single or multiple flips of a fair coin. e head or tail; If two coins are flipped, it can be two heads, two tails, or a head and a tail; The number of possible outcomes gets greater with the increased number of coins. The best way to understand Bernoulli trials is with the classic coin toss example. To calculate any probability you can use the below-mentioned formula The probability of ay event is the number of favorable outcomes divided by the total number of outcomes possible. 125 So the complementary probability of not getting a tail on all 3 tosses, which can happen only if you get a head on at least one of the tosses—A or B or C—is 1. Measured on a scale between 0 and 1 inclusive. In the "die-toss" example, the probability of event A, three dots showing, is P(A) = 1 6 on a single toss. Provide all my solutions and explanations in Chinese for all the Leetcode coding problems. Learn more about different types of probabilities, or explore hundreds of other calculators covering the topics of math, finance, fitness, and health, among others. Step 2: Click the button "Submit" to get the probability value. Suppose we have a fair coin, and we flip it 2 times. The coin is tossed 3 times and there is an equal probability that the coin will turn up heads or tail on each toss (which means that each toss has only two possible outcomes) 2*2*2=2^3=8 Multiply the number of possible outcomes per toss to arrive at the total number of possible outcomes. Tossing a Coin is quite useful as the Probability of obtaining Heads is as likely as Tail. 999023438) ^ #attempts. Sorry for the verbal equations. Numbers between 0 and 1 quantify the uncertainty associated with the event. Consider flipping a fair coin several times. Calculate the probability of flipping a coin toss sequence of HHTTT The probability of each of the 5 coin tosses is 1/2, so we have: P (HHTTT) = 0. Although you can calculate this fairly simply using basic probability rules as some of the other answers have suggested, this type of problem can be generalized really well with the following formul. 3 is the probability of the opposite choice, so it is: 1−p. Let us learn more about the coin toss probability formula. It shows that when you flip a fair coin 10 times, you can pretty much get any outcome with reasonable probability. Example2 : Find the probability for a couple having three children at least one is girl. Step 2: Click the button "Submit" to get the probability value. a die and flipped a coin. This is the number of times the event will occur. When the coin is thrown in the air, it should rotate several times before landing on the ground, or caught and inverted by a chosen person. Coin toss probability calculator helps us find the probability of getting either heads or tails when a coin is tossed the given number of times. Given the probability that the Multi-Level strategy requires at least ips, calculating the average number of ips t2 before the Multi-Level strategy produces a bit still requires some work. Use the normal approximation to the binomial distribution to estimate the probability that the number of heads is greater than or equal to 60. only one is heads. If you toss a coin exactly three times, there are 8 equally likely outcomes, and only one of them contains 3 consecutive heads. 5) "q" is the probability of not getting a head (which is also. But first find the sample space of what you are computing. Two Tails: No Tails: At least one Tail: Theoretical Probability Coin Example. An example of a Bernoulli process is coin flipping. (2) If a person does not have illegal drugs on them, 80% of the time the dog will correctly. So, no we know that the range of the function we call the probability is a subset of the interval [0,1]. Example 2: Calculate the probability of getting an odd number if a dice is rolled. (b) Find the probability of getting a tail. If you flip a coin 9 times, you get a sequence of Heads (H) and tails (T). Events can also be written using set notation. So to find this, use the conditional probability formula. Dice Probability Calculator. 5 and n = 4) would be:. Coin Flip Probability - Explanation & Examples. All tosses are tails. 500000 Input : N = 4, R = 3 Output : 0. An even simpler example of probability in action is a coin toss. Now, by looking at the formula, Probability of selecting an ace from a deck is, P (Ace) = (Number of favourable outcomes) / (Total number of favourable outcomes) P (Ace) = 4/52. What is the probability of obtaining exactly 3 heads. for m = 0 or 2 there is only one way for the out come (both tosses give heads or tails): C 0 = C 2 = 1 for m = 1 (one head, two tosses) there are. The best way to understand Bernoulli trials is with the classic coin toss example. Three marbles are drawn one by one without replacement. Flip two coins, if at least one is heads, what is the probability of both being heads? 0 When coin 1 is flipped, it lands on heads with probability. Thus, the total number of possible outcomes = 8 Getting at least 3 tails include the outcomes = {TTT} No. • If we define event A to have d elementary events then P(A)=d/N Conditional Probability Conditional probabilities provide a method for updating or revising probabilities in light of new information. If the coin is tossed 3 times, what is the probability that at least one of the tosses will turn up tails? 0. If playback doesn't begin shortly, try restarting your device. The Probability Calculator. The first two tosses have different outcomes. This may be a surprise at first, but upon examination there is an clear connection between combinations and multiple trial probabilities. Coin toss probability is explored here with simulation. In the table below, the first column is the possible values of p, the probability of getting H on a single flip. ’ ‘The coin is just as likely to land heads as tails. [+] Watch the Coin Toss Probability Video. P(at least three draws to win) = 1 – P(win in two or fewer draws) = 1 – 7/16 = 9/16. Coin Flip Probability Calculator Enter the total number of heads or tails you want to calculate the probability of into the calculator to determine the chance of getting that amount. For example, if we flip the coin 10 times and the results are HT HHT HT T HH, then this sequence balanced 2 times, i. ½ x ½ x ½ x ½ = 1/16. So the probability that at least one person wins in one million plays is: 1 - (1-p) 1,000,000 = 1. There are two questions you can ask. Picking numbers randomly means that there is no specific order in which they are chosen. Answers ( 2) Since the coin has two sides heads or tails. A pair of dice is rolled 20 times. (c) Calculate the conditional probability of the second toss landing on head given that the first toss lands on head. · T he probability of one or more heads in two coin flips is 1 - 0. Formula, lesson and practice problems explained step by step. If a coin is tossed 500 times and the tail appears 159 times, find the probability of getting a tail. Independent events are events in which the outcome of one event does not affect the probability of the other. For example, if we toss a coin, success could be "heads" with p=0. A coin is tossed 3 times. Find the probability that there are 3 Heads in the first 4 tosses and 2 Heads in the last 3 tosses. randint() you could have any probability of bias while still maintaining randomness. Only the second toss is tails. This can be calculated by multiplying the number of flips (10) by the probability of getting heads on one flip (½), yielding an expected value of 5. 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; CoolGyan’S online coin toss probability calculator makes the calculations faster and gives the probability value in a fraction of. Calculate the probability of drawing ANY PAIR in a row from a deck of cards (with replacement). The probability is relatively high, but this scenario still seems very unlikely! 4. Let’s consider an example where we flip a coin and roll a die simultaneously. Coin toss probability is explored here with simulation. Find the probability that the coin comes up tails at. In this course, you'll learn about fundamental probability concepts like random variables (starting with the classic coin flip example) and how to calculate mean and variance, probability distributions, and conditional probability. Maybe on the second spin. Similarly, on tossing a coin, the probability of getting a tail is: P(Tail) = P(T) = 1/2. Here are some examples. Calculate the mean and standard deviation of X = number of heads. Hence the probability of not generating Similarly, each of the next 2 a bit in ips is m. the number of heads) and q represents the probability of the other event occurring. If you've done an EVEN number of tosses at the point you get a difference, the final coin toss is your result. 4) 4 boys and 3 girls are standing in a line. probability • Example: Toss two coins. What is the probability of getting at least 2 tails ? Solution : When four coins are tossed once, total no. Note: Probability is a funny thing. Wiki User Answered 2011-02-28 04:38:56. #p=1/2# The probability of not getting a head in a single toss. For instance, the probability of your. Thus, the probability of getting a head on the flip of a balanced coin, P(head) = ½ = 0. 5, 50%, or 1 to 1. 5 (or 1/2), and so is the probability of getting heads on a second toss of the same coin. For example, the probability of flipping a coin and it being heads is ½, because there is 1 way of getting a head and the total number of possible outcomes is 2 (a head or tail). Second toss, HH HT TH TT (example:first toss was H, second could be H or T and so on). The 1 is the number of opposite choices, so it is: n−k. For example, if you were trying to find the probability of getting exactly 3 heads, the sample space would be 2^n, n being the number of times you flip the coin. This means that if we flip INFINITELY many fair coins, half of them will come up tails. Calculate the probability. Computation The act or action of carrying out a series of operations. For example, we want at least 2 heads from 3 tosses of coin. With a "fair" coin, the probability of getting heads on a "single" flip at any time is 1/2. If we flip the coin 10 times, we are not guaranteed to get 5 heads and 5 tails. 2 ] Furthermore, define the random variable Y = X2. This is the same as last time, but the true option tells excel that we want to know the probability of AT LEAST 1 success in two flips (i. You pay 1 to flip three fair coins. The following formulas are used to calculate different dice probabilities. Probability is the study of regularities that emerge in the outcomes of random experiments. Find each probability. So the conditional probability in this case is (4/36) / (11/36) = 4/11. On tossing a coin, the probability of getting a head is: P(Head) = P(H) = 1/2. 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). Probability: Types of Events. 3% of the time. Ex) You flip a coin two times. Calculate the probability. 5), which equals 0. The probability of getting 3 heads when you toss a "fair" coin three times is (as others have said) 1 in 8, or 12. Therefore the probability of getting at least one 20 toss succession of heads = "1-Y". Question 2:. Now there are 5 coins so number of Heads can either be greater than or less than Tails. 1 ] Furthermore, define the random variable Y = X2. Whether the coin previously landed on tails makes no difference in calculating the probability that the next flip of the coin will land on heads since there is no relationship between the outcome of the flip of the coins. The probability of flipping 10 heads in a row, assuming a randomly picked coin, is (1/100)*1 + (99/100)* (1/2) 10. Suppose you say to a friend, " I will give you 10 dollars if both coins land on head. 5? H H H H H H H H H H ? The probability is still 0. That being said, it is still 99. The probability of each toss is not influenced by other tosses. E={2,4,6}→n(E)=3 We now use the formula of the classical probability. Trials, n, must be a whole number greater than 0. Find the probability that there are 3 Heads in the first 4 tosses and 2 Heads in the last 3 tosses. 12 (Throw a fair die). How to calculate probability? "Hey man, but girls and coins are two different things!I should know, I've seen at least one of each. (Hint: The variable X has a binomial distribution. If the coin is tossed 3 times, what is the probability that at least one of the tosses will turn up tails? A. Read Book Probability Concepts In Engineering By Alfredo Recognizing the artifice ways to get this books probability concepts in engineering by alfredo is additionally useful. [email protected] We flip three coins, hoping for at least two tails. When we flip a fair coin, we say that there is a 50 percent chance (probability = 0. We will begin with a classical probability example of tossing a fair coin three times. Coin Toss Probability Calculator When a coin is tossed, there lie two possible outcomes i. Since there are 4 ways that we can get two or more consecutive heads out of the 8 total, the probability is 4 8 = 1 2. We can calculate the probability of two or more Independent events by multiplying. We flip a fair coin 10 times. Second, the probability of the sample space Ω {\displaystyle \Omega } must be equal to 1 (which accounts for the fact that, given an execution of the. This is a concern for users who are calculating probability. In other words, we're finding the probability that a probability is what we think it should be. The coin toss is not about probability at all, he says. Example: We roll a dice and flip a coin at random. If the probability of flipping is too low, there's a high risk that nobody flips a coin, but if the probability is too high it approaches \(\frac{1}{16}$$. Let’s consider an example where we flip a coin and roll a die simultaneously. They are "Head and "Tail". The probability is 0. Answer = B. This is a classic binomial distribution problem. Coin flip and coin toss is essentially the practice of tossing a coin up in the air and guessing which side will land face up. Notice this time that it tells us the answer is. Sorry for the verbal equations. Since there are two possible outcomes for each toss, the number of elements in the sample space is 2 n. Coin tossing experiment always plays a key role in probability concept. —Bertrand Russell, 1929 Lecture (cited in Bell 1945, 587) ‘The Democrats will probably win the next election. Which gives us: = p k (1-p) (n-k) Where. Probability measures and quantifies "how likely" an event, related to these types of experiment, will happen. When we flip a coin, what we get does not at all depend on any previous flips, and so flipping a. Use the calculator below to try the experiment. Typically denoted by a capital letter: A, B etc. Practice this lesson yourself on KhanAcademy. , the probability of rolling a 2 or a 3 when tossing a number cube). About Press Copyright Contact us Creators Advertise Developers Terms Privacy Policy & Safety How YouTube works Test new features Press Copyright Contact us Creators. A fair coin is tossed 5 times. Bonus Question. Answer = B. If a tossed coin comes up tails 10 times in a row, most people will expect it to come up heads on the next flip. #(S) = 16 A= fexactly 3 headsg #(A) = 4 = fHHHT, HHTH, HTHH, THHHg P(A) = 4 16 = 1 4 B= fexactly 4 headsg #(B) = 1 = fHHHHg P(B) = 1 16 Now let’s de ne the set C= fat least three headsg. Provided that a small random sample (i. So both must be equal to 1/2. 5, then A will have won after scenario 2 (which happens with probability y). You must derive the pm and show your work. randint(0,3) <= 2 else "T" for i in range(10)] Right now probability of Head is 75% and tails is 25% (0,1,2 are all Heads and only 3 is Tails). You pull a red marble randomly out of the bag. The majority of times, if a coin is heads-up when it is flipped, it will remain heads-up when it lands. use sample space S. Denote the probability of either scenario as x, and the probability of scenario 2 as y. binomial (n,p,x) → P (x successes in n trials given probability p of success. # dbinom r - calculate binomial probability in r dbinom (5, size=10, prob=0. We toss two coins* this experiment involves two parts, 'the first toss of the coin' and 'the second toss of the coin’: experiments that have two parts can be represented in two ways Tree diagramm Tabular form *It notes that: “tossing two different coins “ or “tossing the same coin two times” is the same experiment!. A pair of dice is rolled 20 times. For example, to have coin that is biased to produce more head than tail, we will choose p < 0. By theory, we can calculate this probability by dividing number of expected outcomes by total number of outcomes. An Easy GRE Probability Question. % certain that the outcome would be tails, but this is due to how it is being measured. So need ( 1 − p) n < 0. We flip a fair coin 10 times. On the other hand, the probability that at least 1 chip is defective is the probability that 1, 2, 3, or all 4 of the chips are defective, which may or may not mean that the last chip selected is defective. Mutually exclusive events: They are events such that if one occurs, the other cannot occur. If it is heads, then the experimental probability is 1/1. If the coin is tossed 3 times, what is the probability that at least one of the tosses will turn up tails? A. We covered independent. A fair coin is tossed 8 times. E XAMPLE Toss a fair coin twice What is the probability of observing at least from MTH 380 at Ryerson University. Initially, it’s 1/101. A Random Variable is a set of possible values from a random experiment. An example of a Bernoulli process is coin flipping. There are 32 sample solutions in the solution set of the 5 coin toss. "p" is the probability of getting a head, which is 50% (or. Calculate the probability. Over many coin flips the probability of at least half of the flips being heads (or tails) will converge to 0. For example, the probability of an outcome of heads on the toss of a fair coin is ½ or 0. What is the probability that you will get heads more than 14 times?. 25 = 25% = 1 4. Calculate the probability of flipping a coin toss sequence of THT. Probability provides a measure of how likely it is that something will occur. Since the probability to flip a head is the same as the probability to flip a tail, the probability of outcome (i) must be equal to the probability of outcome (ii). Bonus Question. Suppose we have 3 unbiased coins and we have to find the probability of getting at least 2 heads, so there are 2 3 = 8 ways to toss these coins, i. Given the probability that the Multi-Level strategy requires at least ips, calculating the average number of ips t2 before the Multi-Level strategy produces a bit still requires some work. (a) Write down the sample space of this experiment. Calculate the probability of drawing ANY PAIR in a row from a deck of cards (with replacement). Coin flip probabilities deal with events related to a single or multiple flips of a fair coin. Coin Toss Probability Calculator. It is important to use a quality calculator if you want the calculations to be completed without any mistakes being made. PatrickJMT explains how to calculate probability in an "either A or not A" scenario. Calculate the probability of flipping a coin toss sequence with this Coin Toss Probability Calculator. Odds can then be expressed as 5 : 8 - the ratio of favorable to unfavorable outcomes. Worked-out problems on probability involving tossing or throwing or flipping three coins: 1. Consider flipping a fair coin several times. We can calculate the probability of two or more Independent events by multiplying. What is the probability of drawing a red Bingo chip at least 3 out of 5 times? Round answer to the nearest hundredth. What is the probability that you will get heads more than 14 times?. 5) "q" is the probability of not getting a head (which is also. Calculating probablities can be used to help us make decision. This binomial test calculator determines the probability of a particular outcome (K) across a certain number of trials ( n ), where there are precisely two possible outcomes. Every flip of the coin doesn’t depend on the other coin flips, and we are dealing with a situation where one thing must occur as well as several other things. Given the probability that the Multi-Level strategy requires at least ips, calculating the average number of ips t2 before the Multi-Level strategy produces a bit still requires some work. The 8th term of tetranacci sequence are the odds out 2^10 chances. class web page, and Wolfram Alpha or a calculator for your calculations. A fair coin has an equal probability of landing a head or a tail on each toss. The dice are meant to introduce an element of chance to these games; we expect that the outcomes of the rolls will be truly random. In this video, we' ll explore the probability of getting at least one heads in multiple flips of a fair coin. If the probability of an event is high, it is more likely that the event will happen. • If we define event A to have d elementary events then P(A)=d/N Conditional Probability Conditional probabilities provide a method for updating or revising probabilities in light of new information. You will also get a step by step solution to follow. P (SSSD) is the probability that just the last chip selected is defective, and no others are defective. With probability 1−p the result is Tails, and then X is generated according. 04 is the probability of getting 7 Heads in 8 tosses. Report 11/27/17. 1/8 To calculate the probability you have to name all possible results first. In this article, we are going to study to solve problems to find the probability involving the throwing of two dice. Now, identify the total number of outcomes for the event measured in the last step. Example: We roll a dice and flip a coin at random. Solve each question (a) How many different sequences of heads and tails are possible? (b)How many different sequences of heads and tails have exactly fiveheads? c) How many different sequences have at most 2 heads?. Question 2:. 5, or more than 0. Note: For disjoint events P (A and B) = 0, so the above formula simplifies to P (A or B) = P (A) + P (B) Probability distributions. probability that the coin you chose is the fake coin? \item Suppose you continue flipping the coin for a total of $$k$$ times after picking it: and see $$k$$ heads. What is the probability that in four coin flips you get at least 2 heads? Asked by Wiki User. when coin 2 is flipped it lands on heads with When coin 1 is flipped, it lands on heads with probability probability (a) If coin 1 is flipped 12 times, find the probability that it lands on heads at least 10 times. it is not exactly 50/50. But the result over many tosses is predictable, as long as the trials are independent (i. An ideal unbiased coin might not correctly model a real coin, which could be biased slightly one way or another. If playback doesn't begin shortly, try restarting your device. This is a classic binomial distribution problem. The ratio of successful events A = 968 to the total number of possible combinations of a sample space S = 1024 is the probability of 3 heads in 10 coin tosses. To solve this problem, we need to find the probabilities that r could be 3 or 4 or 5, to satisfy the condition "at least". This is a rare event: we would reject the Null Hypothesis (that the coin is fair) at the P<0. for m = 0 or 2 there is only one way for the out come (both tosses give heads or tails): C 0 = C 2 = 1 for m = 1 (one head, two tosses) there are. 5 because 2 outcomes (heads or tails) are equally possible when a balanced coin is flipped. 4) 4 boys and 3 girls are standing in a line. So the conditional probability in this case is (4/36) / (11/36) = 4/11. (a) Write down the sample space of this experiment. Now suppose that a coin is tossed n times, and consider the probability of the event “heads does not occur” in the n tosses. P (9 coin tosses with no more than 1 heads) =. Course Description. Which gives us: = p k (1-p) (n-k) Where. The binomial probability calculator will calculate a probability based on the binomial probability formula. Ex) You flip a coin two times. Exercise 2 (sum-to-1-exercise) Using the axioms of probability, prove that any probability distribution on a discrete random variable must sum to 1. Math Expression Renderer, Plots, Unit Converter, Equation Solver, Complex Numbers, Calculation History. Since there are two possible outcomes for each toss, the number of elements in the sample space is 2 n. A biased coin that lands heads with probability 0. The chance on the first toss is 50%, and on the 42nd toss it. Click "Reset" at any time to reset the graph. The probability of getting exactly 3 tails when a coin is tossed 2 times. #q=1-1/2=1/2# Now, using Binomial theorem of probability,. The number of correct answers (say heads), X, is distributed as a binomial random variable with binomial. You pay$1 to flip three fair coins. For instance, flipping an coin 6 times, there are 2 6, that is 64 coin toss possibility. Like we have 3 coins and k as 2 so there are23= 8 ways to toss the coins that is −. S = kB log N. (b) Both tosses are the same, given that the first toss is a tail. The manual states that the lifetime of the product, defined as the amount of time (in years) the product works properly until it. Probability of an outcome at least n times over multiple trials. • Step 2: How many outcomes of the event “at least one head” Answer: 3 : { HH, HT, TH}. Over many coin flips the probability of at least half of the flips being heads (or tails) will converge to 0. Wiki User Answered 2011-02-28 04:38:56. And we have (so far): = p k × 0. 5 coming up heads (or tails): a. That implies that the entropy of (2 x 10 3 coins flipped) is equal to twice the entropy of (10 3 coins flipped). For fun on Saturday night, you and a friend are going to flip a fair coin 10 times (geek!). Note: Probability is a funny thing. A coin is flipped 5 times. Find an answer to your question “If you flip a coin and roll a 6 66-sided die, what is the probability that you will flip a tails and roll at least a 2 22? ” in 📘 Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions. 5), and we flip it 3 times. There are exactly 16 different outcomes of flipping a coin four times: One way to get all four heads, four ways to get 3 heads, six ways to get 2 heads, four ways to get 1 head, and one way to get 0 heads. Answer = B. Or say 3, 4 or 5 coins? The outcomes of these coin tosses will differ. Flip a coin. The probability that all of these 20 toss successions were not all heads = "X to the power of 86,381". Try tossing a coin below by clicking on the 'Flip coin' button and check your outcomes. Sum the values of P for all r within the range of interest. Geometric probability is the general term for the study of problems of probabilities related to geometry and their solution techniques. This coin is ipped 200 times. 6 over a modified KC. This value is commonly known as statistical power. And if we want to have biased coin to produce more tails than heads, we will choose p > 0. This geometric probability calculator is used to find geometric distribution probability with total number of occurrence & probability of success. To find the probability of at least two tails, we mark each row (outcome) that contains two tails or three tails and divide the number of marked rows by 8 (number in the sample space) Since there are four outcomes that have at least two tails, the probability is 4/8 or ½. A Random Variable is given a capital letter, such as X or Z. The dice probability calculator is a great tool if you want to estimate the dice roll probability over numerous variants. I came up with P(five heads) = 1 - P(four heads ) = 1 - (5C4)/2^5 = 27/32. Set the probability of heads (between 0 and 1. But if you continue flipping the coin, the outcome grows closer to 50/50. 6 is flipped 2 times. Accordingly, A={HT,TH,TT}. Tossing a Coin is quite useful as the Probability of obtaining Heads is as likely as Tail.


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# Nonlinear dynamics of slipping flows

HY2W04 - Statistical mechanics, integrability and dispersive hydrodynamics

The process of breaking of inviscid incompressible flows along a rigid body with slipping boundary conditions is studied. Such slipping flows may be considered compressible on the rigid surface, where the normal velocity vanishes. It is the main reason for the formation of a singularity for the gradient of the velocity component parallel to rigid border. Slipping flows are studied analytically in the framework of two- and three-dimensional inviscid Prandtl equations. Criteria for a gradient catastrophe are found in both cases. For 2D Prandtl equations breaking takes place both for the parallel velocity along the boundary and for the vorticity gradient. For three-dimensional Prandtl flows, breaking, i.e. the formation of a fold in a finite time, occurs for the symmetric part of the velocity gradient tensor, as well as for the antisymmetric part – vorticity. The problem of the formation of velocity gradients for flows between two parallel plates is studied numerically in the framework of two-dimensional Euler equations. It is shown that the maximum velocity gradient grows exponentially with time on a rigid boundary with a simultaneous increase in the vorticity gradient according to a double exponential law. Careful analysis shows that this process is nothing more than the folding, with a power-law relationship between the maximum velocity gradient and its width: $\max|u_x|\propto \ell^{-2/3}$.   Co-author: Evgenii A. MIkhailov (Moscow State University, Moscow, Russia)

This talk is part of the Isaac Newton Institute Seminar Series series.

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ROS Resources: Documentation | Support | Discussion Forum | Index | Service Status | Q&A answers.ros.org

# Project stopped being able to build

I have an ROS node project which I am developing using Eclipse. I have been able to build the project for some time, but tonight after updates were installed, the project is broken. I found one issue, which was the change from drcsim-2.5 to drcsim-2.6. I changed the include path in the eclipse properties, then did project->index->rebuild. However, I still get the errors below. In addition, when I run rosgui, all of the python based plug-ins are grayed out, whereas these worked previously. I have a custom python plugin which still works, because I manually set the path to my project. Finally, ROS_PACKAGE_PATH used to include a large number of packages, but now it only includes the packages I add. All of this suddenly stopped.

Has rosbuild and the use of ROS_PACKAGE_PATH been discontinued in the most recent update?

I would appreciate anyone's help as to how get back to being able to build my project.

UPDATE:

Rosgui now shows all the plugins like normal. This was apparently an error from starting it before I started the ros master. Also, after reading the error message a few more times, I see that rospack is reporting it can't find atlas_msgs. My manifest.xml file refers to this and several other drcsim related packages. Rospack is able to find regular ros packages such as roscpp or geometry_msgs, but it doesn't see any of the drcsim specific packages that I need to be able to call the the required message types. I double checked my project include paths, and also found the need to update from gazebo-1.7 to gazebo-1.8, but all the other paths still appear to be intact. When I first discovered this problem, the #include line for atlas_msgs/AtlasSimInterfaceState.h was broken, saying it couldn't see this file. This problem was fixed when I updated the include path from drcsim-2.5 to drcsim-2.6. If this corrected the ability to see the file to include it, why can't the compiler see atlas_msgs?

UPDATE 2:

I decided to try closing eclipse, and manually add the drcsim-2.6 package to the ROS_PACKAGE_PATH, but in the command window where I started eclipse, ROS_PACKAGE_PATH contains all the ros packages. I don't know why they appear in this terminal window, but not the others. I run the startup scripts for drcsim in the .bashrc, so I would thing any terminal window would show the same. I start eclipse using bash -i -c eclipse.

FIXED:

I had to update the C/C++ Make Project Environment variable for ROS_PACKAGE_PATH. It was still full of references to drcsim-2.5. I deleted the previous contents, and pasted in the path variables posted below, and was able to build successfully.

Here is my result for echo $ROS_PACKAGE_PATH rick@field:~/catkin_ws$ echo \$ROS_PACKAGE_PATH
/home/rick/ros_ws/:/home/rick/ros_ws:/usr/share/osrf-common-1.0/ros:/usr/share/drcsim-2.6/ros:/usr/share/sandia-hand-5.1/ros:/usr/share/drcsim-2.6/ros:/usr ...
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Fixed

I had to manually update the ROS_PACKAGE_PATH data in the project properties under C/C++ Make Project->Environment.

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# Chapter 12 - Exponential Functions and Logarithmic Functions - 12.1 Composite Functions and Inverse Functions - 12.1 Exercise Set - Page 787: 7

True

#### Work Step by Step

By definition, $(f^{-1})^{-1}=f$, which means that the inverse of the inverse is the function itself. Therefore the given statement is true.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


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Endomorphism Breaking in Graphs

# Endomorphism Breaking in Graphs

Wilfried Imrich
Montanuniversität Leoben, A-8700 Leoben, Austria
\symtypewriterimrich@unileoben.ac.at
The research was supported by the Austrian Science Fund (FWF): project W1230.
Rafał Kalinowski
AGH University of Science and Technology,
al. Mickiewicza 30, 30-059 Krakow, Poland
\symtypewriterkalinows@agh.edu.pl
The research was partially supported by the Polish Ministry of Science and Higher Education.
Florian Lehner
Institut für Geometrie, Technische Universität Graz
Kopernikusgasse 24/IV, A-8010 Graz, Austria
\symtypewriterf.lehner@tugraz.at
The research was supported by the Austrian Science Fund (FWF): project W1230.
Monika Pilśniak
AGH University of Science and Technology,
al. Mickiewicza 30, 30-059 Krakow, Poland
\symtypewriterpilsniak@agh.edu.pl
The research was partially supported by the Polish Ministry of Science and Higher Education.
###### Abstract

We introduce the \symitalicendomorphism distinguishing number of a graph as the least cardinal such that has a vertex coloring with colors that is only preserved by the trivial endomorphism. This generalizes the notion of the distinguishing number of a graph , which is defined for automorphisms instead of endomorphisms.

As the number of endomorphisms can vastly exceed the number of automorphisms, the new concept opens challenging problems, several of which are presented here. In particular, we investigate relationships between and the endomorphism motion of a graph , that is, the least possible number of vertices moved by a nontrivial endomorphism of . Moreover, we extend numerous results about the distinguishing number of finite and infinite graphs to the endomorphism distinguishing number. This is the main concern of the paper.

Keywords: Distinguishing number; Endomorphisms; Infinite graphs; Mathematics Subject Classifications: 05C25, 05C80, 03E10

## 1 Introduction

Albertson and Collins [1] introduced the \symitalicdistinguishing number of a graph as the least cardinal such that has a labeling with labels that is only preserved by the trivial automorphism.

This concept has spawned numerous papers, mostly on finite graphs. But countable infinite graphs have also been investigated with respect to the distinguishing number; see [9], [15], [16], and [17]. For graphs of higher cardinality compare [10].

The aim of this paper is the presentation of fundamental results for the endomorphism distinguishing number, and of open problems. In particular, we extend the Motion Lemma of Russell and Sundaram [14] to endomorphisms, present endomorphism motion conjectures that generalize the Infinite Motion Conjecture of Tom Tucker [16] and the Motion Conjecture of [4], prove the validity of special cases, and support the conjectures by examples.

## 2 Definitions and Basic Results

As the distinguishing number has already been defined, let us note that for all asymmetric graphs. This means that almost all finite graphs have distinguishing number one, because almost all graphs are asymmetric, see Erdős and Rényi [5]. Clearly for all other graphs. Again, it is natural to conjecture that almost all of them have distinguishing number two. This is supported by the observations of Conder and Tucker [3].

However, for the complete graph , and the complete bipartite graph we have , and . Furthermore, the distinguishing number of the cycle of length 5 is 3, but cycles of length have distinguishing number 2.

This compares with a more general result of Klavžar, Wong and Zhu [12] and of Collins and Trenk [2], which asserts that , where denotes the maximum degree of . Equality holds if and only if is a , or .

Now to the endomorphism distinguishing number. Before defining it, let us recall that an \symitalicendomorphism of a graph is a mapping such that for every edge its image is an edge, too.

\symbold

Definition The endomorphism distinguishing number of a graph is the least cardinal such that has a labeling with labels that is preserved only by the identity endomorphism of .

Let us add that we also say colors instead of labels. If a labeling is not preserved by an endomorphism , we say that \symitalicbreaks .

Clearly . For graphs with equality holds. Such graphs are called \symitaliccore graphs. Notice that complete graphs and odd cycles are core graphs, see [7]. Hence , , and for .

Interestingly, almost all graphs are core graphs, as shown by Koubek and Rödl [13]. Because almost all graphs are asymmetric, this implies that almost all graphs have trivial endomorphism monoid, that is, . Graphs with trivial endomorphism monoid are called \symitalicrigid. Clearly for any rigid graph , and thus for almost all graphs .

can be equal to even when . For example, this is the case for even cycles. We formulate this as a lemma.

###### Lemma 1

The automorphism group of even cycles is properly contained in their endomorphism monoid, but for all .

\symbold

Proof. It is easily seen that every even cycle admits proper endomorphisms, that is, endomorphisms that are not automorphisms. Furthermore, it is readily verified that and .

Hence, let . Color the vertices and black and all other vertices white, see Figure 1. We wish to show that this coloring is endomorphism distinguishing. Clearly this coloring distinguishes all automorphisms.

Let be a proper endomorphism. It has to map the cycle into a proper connected subgraph of itself. Thus, must be a path, say .

Furthermore, the color of the endpoints of an edge must be preserved under . Hence is mapped into itself. Because is the only edge with two white endpoints that is adjacent to , it must also be mapped into itself. This fixes and . But then and are also fixed.

Now we observe that the path in has only white interior vertices and that it it has to be mapped into a walk in from to that contains only white interior vertices. Clearly this is not possible.

To show that can be smaller than , we consider graphs with trivial automorphism group but nontrivial endomorphisms monoid. For such graphs , but . Easy examples are asymmetric, nontrivial trees . For, every such tree has at least 7 vertices and at least three vertices of degree 1. Let be a vertex of degree 1 and its neighbor. Because has at least 7 vertices and since it is connected, there must be a neighbor of that is different from . Then the mapping

 ϕ:v↦{c if v=av otherwise

is a nontrivial endomorphism.

## 3 The Endomorphism Motion Lemma

Russel and Sundaram [14] proved that the distinguishing number of a graph is small when every automorphism of moves many elements. We generalize this result to endomorphisms and begin with the definition of motion.

The \symitalicmotion of a nontrivial endomorphism of a graph , is the number of elements it moves:

 m(ϕ)=|{v∈V(G)|ϕ(v)≠v}|.

The \symitalicendomorphism motion of a graph is

 me(G)=minϕ∈End(G)∖{\symoperators id}m(ϕ)

For example, .

In the sequel we will prove the following generalization of Theorem 1 of Russell and Sundaram [14].

###### Lemma 2 (Endomorphism Motion Lemma)

For any graph ,

 dme(G)2≥|End(G)| (1)

implies

The proof will be an easy consequence of Lemma 3, the Orbit Norm Lemma. We first define orbits of endomorphisms.

\symbold

Definition An orbit of an endomorphism of a graph is an equivalence class with respect to the equivalence relation on , where if there exist nonnegative integers and such that .

The orbits form a partition , for , of . For finite graphs it can be characterized as the unique partition with the maximal number of sets that are invariant under the preimage . For infinite graphs we characterize it as the finest partition that is invariant under . For automorphisms it coincides with the cycle decomposition.

The \symitalicorbit norm of an endomorphism with the orbits is

 o(ϕ)=k∑i=1(|Ii|−1),

and the \symitalicendomorphism orbit norm of a graph is

 o(G)=minϕ∈End(G)∖{\symoperators id}o(ϕ).

Notice that may not move all elements of a nontrivial orbit, whereas automorphisms move all elements in a nontrivial cycle of the cycle decomposition. To see this, consider an orbit , where , and . Only one element of the orbit is moved, and the contribution of to the orbit norm of is . Clearly , and thus .

###### Lemma 3 (Orbit Norm Lemma)

A graph is endomorphism -distinguishable if

 ∑ϕ∈End(G)∖{\symoperators id}d−o(ϕ)<1.
\symbold

Proof. We study the behavior of a random -coloring of G, the probability distribution given by selecting the color of each vertex independently and uniformly in the set . Fix an endomorphism and consider the event that the random coloring is preserved by , that is, for each vertex of . Then it is easily seen that

 Prob[∀v:c(v)=c(ϕ(v))]=(1d)o(ϕ)≤(1d)o(G).

Collecting together these events, we have

 Prob[∃ϕ≠\symoperators id∀v:c(v)=c(ϕ(v))]≤∑ϕ∈End(G)∖{\symoperators id}(1d)o(ϕ).

If this sum is strictly less than one, then there exists a coloring such that for all nontrivial there is a , such that , as desired.

\symbold

Proof of Lemma 2. From we infer that

 ∑ϕ∈End(G)∖{\symoperators id}d−o(ϕ)≤(|End(G)|−1)d−o(G)≤(|End(G)|−1)d−me(G)2. (2)

Hence, if

 dme(G)2≥|End(G)|,

then the right side of Equation 2 is strictly less than 1, and therefore also less than . Now an application of the Orbit Norm Lemma shows that is -distinguishable.

.

Lemma 2 is similar to the \symitalicMotion Lemma of Russell and Sundaram [14, Theorem 1], which asserts that is 2-distinguishable if

 m(G)>2log2|Aut(G)|,

where

 m(G)=minϕ∈Aut(G)∖{\symoperators id}m(ϕ).

Actually, a short look at the proof of Russell and Sundaram shows that is -distinguishable under the weaker assumption

 m(G)≥2logd|Aut(G)|. (3)

Thus, our Endomorphism Motion Lemma is a direct generalization of the Motion Lemma of Russell and Sundaram.

The Motion Lemma allows the computation of the distinguishing number of many classes of finite graphs. We know of no such applications for the Endomorphism Motion Lemma, but will show the applicability of its generalization to infinite graphs.

## 4 Infinite graphs

Suppose we are given an infinite graph with infinite endomorphism motion and wish to generalize Equation 1 to this case for finite . Notice that

 dme(G)/2=dme(G)=2me(G)

in this situation. Thus the natural generalization would be that

 2me(G)≥|End(G)| (4)

implies endomorphism 2-distinguishabilty. We formulate this as a conjecture.

\symbold

Endomorphism Motion Conjecture Let be a connected, infinite graph with endomorphism motion . If , then .

This is a generalization of the Motion Conjecture of [4] for automorphisms of graphs. Notice that we assume connectedness now, which we did not do before. The reason is, that we not only have to break all endomorphisms of every connected component if the graph is disconnected, but that we also have to worry about breaking mappings between possibly infinitely many different connected components, which requires extra effort.

A special case are countable graphs. Let be an infinite, connected countable graph with infinite endomorphism motion . Then and , where c denotes the cardinality of the continuum.

Notice, for countable graphs, . This means that Equation 4 is always satisfied for countably infinite graphs with infinite motion. This motivates the following conjecture:

\symbold

Endomorphism Motion Conjecture for Countable Graphs Let be a countable connected graph with infinite endomorphism motion. Then is endomorphism 2-distinguishable.

In the last section we will verify this conjecture for countable trees with infinite endomorphism motion. Their endomorphism monoids are uncountable and we will show that they have endomorphism distinguishing number 2.

We now prove the conjecture for countable endomorphism monoids. In fact, we show that almost every coloring is distinguishing if the endomorphism monoid is countable.

###### Theorem 4

Let be a graph with infinite endomorphism motion whose endomorphism monoid is countable. Let be a random -coloring where all vertices have been colored independently and assume that there is an such that, for every vertex , the probability that it is assigned a color satisfies

 ε≤Prob[c(v)=x]≤1−ε.

Then is almost surely distinguishing.

\symbold

Proof.  First, let be a fixed, non-trivial endomorphism of . Since the motion of is infinite we can find infinitely many disjoint pairs . Clearly the colorings of these pairs are independent and the probability that preserves the coloring in any of the pairs is bounded from above by some constant . Now

 Prob[ϕ preserves c]≤Prob[∀i:c(vi)=c(ϕ(vi))]=0.

Since there are only countably many endomorphisms we can use -subadditivity of the probability measure to conclude that

 Prob[∃ϕ∈End(G)∖{id}:ϕ preserves c]≤∑ϕ∈End(G)∖{\symoperators id}Prob[ϕ % preserves c]=0,

which completes the proof.

We will usually only use the following Corollary of Theorem 4.

###### Corollary 5

Let be a graph with infinite motion whose endomorphism monoid is countable. Then

 De(G)=2.

The endomorphism motion conjecture for countable graphs generalises the

\symbold

Infinite Motion Conjecture of Tucker [16] Let be a connected, locally finite infinite graph with infinite motion. Then is 2-distinguishable.

It was shown in [4], and follows from Theorem 4, that it is true for countable . There are numerous applications of this result, see [11].

For the Endomorphism Motion Conjecture for Countable Graphs we have the following generalization of [10, Theorem 3.2]:

###### Theorem 6

Let be a finitely generated infinite group. Then there is a 2-coloring of the elements of  , such that the identity endomorphism of  is the only endomorphism that preserves this coloring. In other words, finitely generated groups are endomorphism 2-distinguishable.

\symbold

Proof. Let be a finite set of generators of that is closed under inversion. Since every element of can be represented as a product of finite length in elements of , we infer that is countable.

Also, if , then

 ϕ(g)=ϕ(s1s2⋯sk)=ϕ(s1)ϕ(s2)⋯ϕ(sk).

Hence, every endomorphism is determined by the finite set

 ϕ(S)={ϕ(s)|s∈S}.

Because every is a word of finite length in elements of there are only countably many elements in . Hence is countable.

Now, let us consider the motion of the nonidentity elements of . Let be such an element and consider the set

 Fix(ϕ)={g∈Γ|ϕ(g)=g}.

It is easily seen that these elements form a subgroup of . Since does not fix all elements of it is a proper subgroup. Since its smallest index is two, the set is infinite. Thus is infinite. As was arbitrarily chosen, has infinite endomorphism motion.

By Corollary 5 we conclude that is 2-distinguishable.

The next theorem shows that the endomorphism motion conjecture is true if , even if is not countable.

###### Theorem 7

Let be a connected graph with uncountable endomorphism motion. Then implies .

\symbold

Proof.  Set , and let be the smallest ordinal number whose underlying set has cardinality n. Furthermore, choose a well ordering of of order type , and let be the smallest element with respect to . Then the cardinality of the set of all elements of between and any other is smaller than .

Now we color all vertices of white and use transfinite induction to break all endomorphisms by coloring selected vertices black. By the assumptions of the theorem, there exists a vertex that is not fixed by . We color it black. This coloring breaks .

For the induction step, let . Suppose we have already broken all by pairs of vertices , where and have distinct colors. Clearly, the cardinality of the set of all , , is less than . By assumption, moves at least vertices. Since there are still n vertices not in , there must be a vertex that does not meet . If is white, we color black. This coloring breaks .

###### Corollary 8

Let be a connected graph with uncountable endomorphism motion. If the general continuum hypothesis holds, and if   then .

\symbold

Proof. By the generalized continuum hypothesis is the successor of . Hence, the inequality is equivalent to .

## 5 Examples and outlook

So far we have only determined the endomorphism distinguishing numbers of core graphs, such as the complete graph and odd cycles, and proved that for . Furthermore, it is easily seen that and if .

In the case of infinite structures we proved Theorem 6, which shows that for finitely generated, infinite groups .

We will now determine the endomorphism distinguishing numbers of finite and infinite paths and we begin with the following lemma.

###### Lemma 9

Let be an endomorphism of a (possibly infinite) tree such that for two distinct vertices . Then there exist two vertices on the path between and such that and .

\symbold

Proof.  Suppose dist. Hence dist. Let be the path connecting and in , and let be the subgraph induced by the image . Clearly, is a finite tree with at least one edge.

Because every nontrivial finite tree has at least two pendant vertices, there must be a pendant vertex of that is different from . Thus for some internal vertex of . If and are the two neighbors of on , then clearly and dist.

The above lemma implies the following corollary for finite graphs, because any injective endomorphism of a finite graph is an automorphism.

###### Corollary 10

Let be a finite tree. Then for every there exist two vertices of distance 2 such that .

###### Lemma 11

The endomorphism distinguishing number of all finite paths of order is two.

\symbold

Proof.  Clearly, since . To see that consider the following labeling

 c(Pn)=⎧⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪⎩(11221122.....1122) if n≡0mod4(11221122...11221) if n≡1mod4(1221122.....22112) if n≡2mod4(11221122...22112) if n≡3mod4.

The only nontrivial automorphism of a path (symmetry with respect to the center) does not preserve this labeling. By Corollary 10, any other endomomorphism has to identify two vertices of distance two. Then cannot preserve the coloring, because any two vertices of distance two have distinct labels.

Next let us consider the ray and the double ray which can be viewed as an infinite analogs to finite paths. It turns out that their endomorphism distinguishing number is as well.

###### Lemma 12

The endomorphism distinguishing number of the infinite ray and of the infinite double ray is two.

Later in this section Theorem 15 will show that every countable tree with at most one pendant vertex has endomorphism distinguishing number two. Clearly Lemma 12 constitutes a special case of this result. It is also worth noting that by the following theorem every double ray has infinite endomorphism motion. Hence we verify the Endomorphism Motion Conjecture for the class of countable trees.

###### Theorem 13

An infinite tree has infinite endomorphism motion if and only if it has no pendant vertices.

The proof uses the following lemma which may be interesting as such. Note that in the statement of the lemma there is no restriction on the cardinality of the tree or the motion of the endomorphism.

###### Lemma 14

Let be a tree and let be an endomorphism of . Then the set of fixed points of induces a connected subgraph of .

\symbold

Proof.  Denote by Fix the set of fixed points of and assume that it does not induce a connected subgraph. Consider two vertices lying in different components of this graph.

Then maps the unique path in from to to a --walk of the same length. But the only such walk is the path connecting and , so this path has to be fixed pointwise.

\symbold

Proof of Theorem 13. Clearly, if an infinite tree has a pendant vertex, then there is an endomorphism which moves only this vertex and fixes everything else.

So let be a tree without pendant vertices and let be a nontrivial endomorphism of . Assume that the motion of is finite. Then the set of fixed points of contains all but finitely many vertices of . Since has no pendant vertices such a set does not induce a connected subgraph. This contradicts Lemma 14.

Now that we have characterised the trees with infinite endomorphism motion, we would like to show that all of them have endomorphism distinguishing number 2.

###### Theorem 15

The endomorphism distinguishing number of countable trees with at most one pendant vertex is 2.

\symbold

Proof.  The proof consists of two stages. First we color part of the vertices such that every endomorphism which preserves this partial coloring has to fix all distances from a given vertex . Then we color the other vertices in order to break all remaining endomorphisms.

For the first part of the proof, let be a pendant vertex of , or any vertex if is a tree without pendant vertices. Denote by the set of vertices at distance from , that is the sphere of radius with center . Now color white and all of and black. Periodically color all subsequent spheres according to the pattern outlined in Figure 2. In other words always color two spheres white, then four spheres black, leave two spheres uncolored, color another four spheres black and proceed inductively. Furthermore, we require that adjacent uncolored vertices are assigned different colors in the second step of the proof.

Now we claim that this coloring fixes in every endomorphism. To prove this consider a ray starting at . Clearly holds for every . Assume that there is a color preserving endomorphism of which does not fix and consider the image of the previously chosen ray under , that is, let . Clearly has to lie either in a white sphere or in a sphere which has not yet been colored. We will look at those cases and show that all of them lead to a contradiction. So assume that for some .

• If , then must lie in since it must be a black neighbor of . For similar reasons and must hold. Now has to be a white neighbor of but only has black neighbors, a contradiction.

• If we get and by the same argument as above. Now would need to be a white neighbor of but only has black neighbors.

• If , where , then, for similar reasons as in the previous cases, and . Again has no white neighbors.

• If , then lies in one of , and . In the first case clearly has no white neighbors. In the other cases it may have a white neighbor in , but then has no white neighbors, because its neighbor in must have a different color.

• If , we can use an argument that is symmetric to the previous case.

Since there are no more cases left we can conclude that has to be fixed by every endomorphism which preserves this coloring.

However, we wish to prove that such endomorphism preserves all distances from , that is, that maps into itself for each .

We first show that any for must have its image in for some . Since is fixed, must be mapped to a vertex at even distance from . Furthermore, this vertex must be black and have a white neighbor, which again must have a white neighbor. It is easy to check that the only vertices for which all of this holds lie in for some .

Now assume that does not map into itself for every and consider the smallest such that . Then there must be some vertex such that . This immediately implies that and that , because otherwise a white vertex would be mapped to a black vertex or vice versa. In order to treat the remaining cases, consider a vertex whose predecessor in is , where is chosen to be minimal with respect to the properties , . The unique --path in must be mapped to a --walk with length at most . This implies that cannot lie in for . The --path does not contain two consecutive white vertices, hence the --walk cannot cross the two consecutive white layers and . So cannot lie in for . But this contradicts the fact that must lie in some for .

This completes the proof of the fact that all distances from are fixed by any endomorphism which preserves such a coloring.

For the second part of the proof, consider any enumeration of the vertices of such that, for all , we have for some . It is easy to see that such an enumeration is possible. Now color all vertices in whose predecessor is black and color all other vertices in this sphere white. Color the vertices of whose predecessor is white, and color all other vertices in this sphere black.

We claim that the so obtained coloring is not preserved by any endomorphism but the identity. We already know that every color preserving endomorphism maps every sphere into itself. Assume that there is a vertex which is not fixed by . Then it is easy to see that all vertices in whose predecessor is will be mapped to vertices whose predecessor is . Hence is not color preserving.

We conjecture that this result can be extended to uncountable trees. One does need a lower bound on the minimum degree though, see [10]. As we already noted, the fact that , together with the observations that and , supports the Endomorphism Motion Conjecture. Of course, a proof of the Endomorphism Motion Conjecture is still not in sight, not even for countable structures.

Finally, the computation of seems to be an interesting problem, even for finite cubes. Similarly, the computation of , where denotes the -th Cartesian power111For the definition of the Cartesian product and Cartesian powers see [8]. of , looks demanding.

\symbold

Acknowledgement We wish to thank one of the referees for numerous clarifying remarks and helpful comments.

## References

• [1] M. O. Albertson and K. L. Collins, Symmetry breaking in graphs, Electron. J. Combin. 3 (1996) R18.
• [2] K. L. Collins and A. N. Trenk, The distinguishing chromatic number, Electron. J. Combin. 13 (2006) R16.
• [3] M. Conder and T. Tucker, The Motion Lemma and Distinguishability Two, Ars Math. Contemp. 4 (2011), 63–72.
• [4] J. Cuno, W. Imrich and F. Lehner, Distinguishing graphs with infinite motion and nonlinear growth, Ars Math. Contemp. 7 (2014), 201–213.
• [5] P. Erdős and A. Rényi, Asymmetric graphs, Acta Math. Acad. Sci. Hungar. 14 (1963) 295–315.
• [6] M. J. Fisher, G. Isaak, Distinguishing colorings of Cartesian products of complete graphs, arXiv:math.CO/0607465.
• [7] C. Godsil and G. Royle, Algebraic Graph Theory, Graduate Texts in Mathematics, Springer-Verlag, New York, 2001.
• [8] R. Hammack, W. Imrich and S. Klavžar, Handbook of product graphs, Second edition, Discrete Mathematics and its Applications (Boca Raton), CRC Press, Boca Raton, FL, 2011.
• [9] W. Imrich, J. Jerebic and S. Klavžar, The distinguishing number of Cartesian products of complete graphs, European J. Combin. 29 (2008) 922–929.
• [10] W. Imrich, S. Klavžar and V. Trofimov, Distinguishing infinite graphs, Electron. J. Combin. 14 (2007) R36
• [11] W. Imrich, S. M. Smith, T. Tucker and M. E. Watkins, Infinite motion and the distinguishing number of graphs, manuscript 2012.
• [12] S. Klavžar, T.-L. Wong and X. Zhu, Distinguishing labelings of group action on vector spaces and graphs, J. Algebra 303 (2006) 626–641.
• [13] V. Koubek and V. Rödl, On the minimum order of graphs with given semigroup, J. Combin. Theory Ser. B 36 (1984) 135–155.
• [14] A. Russell and R. Sundaram, A note on the asymptotics and computational complexity of graph distinguishability, Electron. J. Combin. 5 (1998), R23.
• [15] S. M. Smith, T. Tucker and M. E. Watkins, Distinguishability of Infinite Groups and Graphs, Electron. J. Combin. 19 (2012) P27.
• [16] T. Tucker, Distinguishing Maps, Electron. J. Combin. 18 (2011) R50.
• [17] M. E. Watkins and X. Zhou, Distinguishability of Locally Finite Trees, Electron. J. Combin. 14 (2007) R29.
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# Math Help - Surface integral vector field

1. ## Surface integral vector field

Evaluate the surface integral of the vector field F(x; y; z) = (siny, sinz, yz)over
the rectangle 0 y≤ 2, 0 ≤z ≤3 in the yz-plane with the normal pointing in the
negative x direction.

2. ## Re: Surface integral vector field

Hey apatite.

Can you show us what you have tried? (The definition of the integral?)

3. ## Re: Surface integral vector field

I get 3(cos2-cos0) I parametrize it and then use the surface integral formula. Is my answer correct? I appreciate your help!


text:
Non-linear transformation of univariate normal variables: $y_i=\frac{e^{x_i}}{\sum_{j=1}^{K}{e^{x_j}}}$

Let $\{x_i\in\mathbb{R}\}_{i=1}^{K}$ be a set of random univariate variables, each of which is distributed normally as $x_i\sim\mathcal{N}(\mu_{x,i},\sigma_{x,i}^2)$. Note that $x_i$'s are pairwise independent.

Also, let $\{y_i\in\mathbb{R}\}_{i=1}^{K}$ be the set of random variables, where each variable $y_i$ is given as the following non-linear transformation of $x_i$: $$y_i=\frac{e^{x_i}}{\sum_{j=1}^{K}{e^{x_j}}}.$$ How could we find $y_i$'s mean $\mu_{y,i}$ and variance $\sigma_{y,i}^2$?

• Nonnegative random variables distributed normally? This would be a surprise... – Did Oct 20 '16 at 11:51
• @Did correct, it's obviously not normal (I edit my question). But how could we compute $y_i$'s mean and variance? – nullgeppetto Oct 20 '16 at 11:57
• Without assuming joint normality, and for unspecified parameters $\mu_i$ and $\sigma^2_i$, I doubt there is a theoretical answer. For given values of the parameters, simulations might be an option. – Did Oct 20 '16 at 12:05
• OK, you did restrict the setting by assuming joint normality but you widened it by stopping to (implictely) assume independence. Anyway, even with joint normality and independence, "for unspecified parameters μi and σ2i, I doubt there is a theoretical answer. For given values of the parameters, simulations might be an option." – Did Oct 20 '16 at 12:42
• Specifications of this problem seem to be in flux. If this question is motivated by a practical application, you might want to look at Wikipedia on 'lognormal distributions' to see if anything there seems useful. – BruceET Oct 20 '16 at 16:03

You can use Unscented Transformation to find mean and covariance. Let $y = f(x)$ be a linear or nonlinear mapping from $x$ to $y$. Given that you know the mean and covariance of $x$, you can find mean and covariance of $y$ by generating sigma points corresponding to $x$.


text:
GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 18 Jun 2019, 06:01

### GMAT Club Daily Prep

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If John purchased goods for $1050, and sold $$\frac{1}{3}$$rd of these goods at a loss of 12%, then at what gain percent should remainder goods be sold to gain 18% on the whole transaction? A. 31% B. 33% C. 43% D. 67% E. 77% _________________ ____________________________ Regards, Chaitanya +1 Kudos if you like my explanation!!! Manager Joined: 14 Jun 2018 Posts: 223 If John purchased goods for$1050  [#permalink]

### Show Tags

27 Oct 2018, 00:42
2
Using alligation ,
$$\frac{x-18}{18-(-12)} = \frac{1}{2}$$
2x - 36 = 30
x = 33

Can be done with weighted average formula as well

or the classic

$$350 * \frac{88}{100} + 700 \frac{(100+k)}{100} = 1050 * \frac{118}{100}$$
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Re: If John purchased goods for $1050 [#permalink] ### Show Tags 27 Oct 2018, 03:14 1 eswarchethu135 wrote: If John purchased goods for$1050, and sold $$\frac{1}{3}$$rd of these goods at a loss of 12%, then at what gain percent should remainder goods be sold to gain 18% on the whole transaction?

A. 31%
B. 33%
C. 43%
D. 67%
E. 77%

Given: John sells 1/3rd of the overall stock at a loss of 12%

We are asked to find at which gain must he sell the remaining 2/3rd of the stock

Let's call that margin - x%. Since the ratio at which the stock is sold is 1:2,
we can deduce the following equatation -> $$1*0.88 + 2*x = 3*1.18$$

Simplifying that further, we get $$2x = 3.54 - 0.88 = 2.66$$ -> $$x = \frac{2.66}{2} = 1.33$$

Therefore, the remaining goods are sold at 33%(Option B) more to realize the 18% overall profit margin.
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text:
# A universal constant?

bomba923
Is the Ideal Gas Constant a universal constant?
*If not, then what is it relative to?

Last edited:

gravenewworld
I'm not sure what you mean by the Real Gas constant.
The ideal Gas constant R is universal
If you are refering to the vander waal equation that describes real gases and the constants a and b, then no, both a and b are not universal constants. You have to determine a and b experimentally and measure critical temperatures and critical pressures with respect to the compound in question.

Homework Helper
universal constant, e.g. the speed of light c, where it supposedly holds true at all inertial frames of reference...everywhere.

I don't believe that R is a universal constant.

gravenewworld
Why isn't R is a universal constant for an ideal gas? When does it ever change?

bomba923
Last edited:
Homework Helper
no I guess not, but before you go on to call it universal (as the speed of light is "universal) you might want to bring it up in one of the physics subforums

bomba923
GCT said:
no I guess not, but before you go on to call it universal (as the speed of light is "universal) you might want to bring it up in one of the physics subforums

!Why--exactly? How what does bringing up it a different forum have to do with the Ideal Gas Constant being universal/"calling it universal"??
------
But I agree, physics and engineering do often use the Ideal Gas Constant, and it easily could have been a question in one of the physics subforums . I posted this in chemistry because I worked with the Ideal Gas Constant most frequently in my chemistry courses.

Homework Helper
GCT said:
no I guess not, but before you go on to call it universal (as the speed of light is "universal) you might want to bring it up in one of the physics subforums

$$R=k_{B} N_{A}$$

Boltzmann's constant is universal, Avogadro's number is universal, ergo the ideal gases' constant is universal.

Daniel.


text:
anonymous one year ago Medal ** use polar coordinates to evaluate the integral

1. anonymous

$\int\limits_{0}^{1} \int\limits_{0}^{\sqrt{3}x} \frac{ 1 }{\sqrt{x^{2}+y^{2}}} dy dx$

2. amoodarya

|dw:1436659558096:dw|

3. amoodarya

|dw:1436659627501:dw|$0\le \theta \le \frac{\pi}{3}\\$

4. amoodarya

$x=rcos (\theta)\\x=1 \rightarrow rcos \theta =1 \rightarrow r=\frac{1}{\cos \theta}=\sec theta$

5. amoodarya

$\int\limits_{0}^{\frac{\pi}{3}}\int\limits_{0}^{\sec \theta}$

6. amoodarya

$\int\limits_{0}^{\frac{\pi}{3}}\int\limits_{0}^{\sec \theta} \frac{ 1 }{\sqrt{x^2+y^2} }dxdy=\\ =\int\limits_{0}^{\frac{\pi}{3}}\int\limits_{0}^{\sec \theta} \frac{ 1 }{ \sqrt{r^2} }*r dr d \theta$ it is simple to integrate


text:
### kostka's blog

By kostka, 11 months ago, ,

Just in a month from today, IOI in Iran will start.

Below you can find the complete list of participants (based on registration and data provided by Codeforces community): http://weaselcrow.com/pro/cf/ioi2017/

Some random facts:

• 326 students from 84 countries will participate (this will be a new record, the previous record was at IOI 2015 in Kazakhstan — 322 contestants from 83 countries)
• El Salvador, Iceland, Morocco, and Palestine will send their contestants for the very first time, welcome to IOI!
• 114 (almost 35%) participants were already in Russia, including golden medalists: reyna, yutaka1999, pacu, natsukagami, also Cyprus and Bangladesh will have exactly same team, as in previous year.

•
• +83
•

 » 11 months ago, # |   +3 What do you think about adding topcoder/atcoder accounts?
 » 11 months ago, # |   +18 Thanks, I will definitely use your list to double-check the past contestants matching.I now have access to the official registration data (and not just the scoreboard) and will start slowly processing it. Expect the main batch to be released on IOI Stats over the weekend.
•  » » 10 months ago, # ^ | ← Rev. 2 →   +32 Over the weekend *cough cough* 2 weeks ago...
•  » » » 10 months ago, # ^ |   0 Yeah, sorry... Work got in the way and then the servers are behaving really weirdly in the last couple of days. Should be done as soon as we can get the servers to a stable state.
•  » » » » 10 months ago, # ^ |   0 Awesome, thanks for the update!
•  » » » » » 10 months ago, # ^ |   +3 I hope we fixed the issue causing the downtime. I have already started entering some data, and I'll aim to finish it completely in the next couple of days. Again, sorry for the delay.
 » 11 months ago, # |   0 Wow, Jordan is so desperate to get a medal they are breaking the rules by allowing someone who finished high school and is enrolled in college to participate. Shame on you Jordan. Hope you get disqualified.Anyone who wants to say hiasat hasn't been attending college: don't make a fool of yourself, I, including many others, can easily prove that. Framing your participant as a high school student is never a moral thing to do. Does that conform to your Islamic values, Jordan?
•  » » 11 months ago, # ^ |   0 I don't mean anything bad to Islam. I'm Muslim myself. I mean it's a shame for such reputed country to do that
•  » » 11 months ago, # ^ |   -12 You can buy Africans folks
•  » » 11 months ago, # ^ |   0 The rule is as follows:2.5. A Contestant is a student who was enrolled at a school for secondary education, in the Country they are representing, during the period September to December in the year before IOI’n, and is not older than twenty years on the 1st of July of the year of IOI’n. Students who are studying abroad may represent the Country of their nationality.So if he started attending college in this calendar year — everything is fine.
•  » » » 11 months ago, # ^ |   +7 He has been attending university since September 2016.
•  » » » » 11 months ago, # ^ | ← Rev. 2 →   +6 well, isn't that during the period September to December in the year before IOI’nEDIT: I didn't know Secondary Education != university
•  » » » 11 months ago, # ^ |   +5 Without commenting at all at the discussion in hand, I want to add a small comment because it came up last year and is not obvious from that definition in my opinion. A student has to be enrolled at a school for secondary education during the ENTIRETY (not AT SOME POINT) of the period from September to December in the year before IOI. This can and has created confusion in the past where contestants would travel to IOI to find out they are actually illegible to compete (and they have not been disqualified but rather omitted from contestant list completely).
•  » » 11 months ago, # ^ |   +12 You should send proof to IOI organisers, nothing will happen if you just write on CF (especially not using a fake account — that's like asking not to be taken seriously).
•  » » » 11 months ago, # ^ |   +5 I'm collecting formal proof to send to the organizers. This comment is just for pointing that out to the community.
•  » » » » 11 months ago, # ^ | ← Rev. 5 →   +30 *grabs popcorn
•  » » » » » 11 months ago, # ^ |   +14 plot twist: the one who went 5th in Jordanian OI is the owner of the account.
•  » » » » » » 11 months ago, # ^ | ← Rev. 3 →   +6 That wouldn't be unexpected enough. If CheaterKiller_vIOI was an evil twin brother of hiasat, who just wouldn't want his brother to participate, that would be real plot twist!
•  » » » » » » » 11 months ago, # ^ |   +51 I think the greatest plot twist is CheaterKiller_vIOI is hiasat himself, and he is forced to go to IOI by the Jordan goverment.
•  » » » » 10 months ago, # ^ |   0 Link to Facebook post announcement about IOI team Jordan From google translate: The procedures for the travel of the Jordanian team at the 29th International Olympiad for Informatics have been completed during its twenty-ninth session which will be held in the city of Tehran during the period between the 28th of July and the fourth of August with the participation of more than 80 countries. The co-chairs are as follows: First: Sarah Abu Alian from the Jubaiha secondary school for girls. Second: Mohamed Kilani from the Omani schools Amman. Third: Ahmed Nasr from King Abdullah Schools of Excellence Blue. We wish everyone good results, God willing.  tl;dr: hiasat is not going. Have a nice day!
•  » » » » » 10 months ago, # ^ |   0 P.S. I'm glad this was resolved, but I'm worried these comments can only cause further harms to the people involved. If MikeMirzayanov could delete the parent comment, I would be highly relieved.
•  » » » » » » 10 months ago, # ^ |   +16 These comments should not have been made public in the first place. There were millions of ways to resolve this situation without involving anonymous public shaming with dubious intentions.To the best of my knowledge, this situation arose due to genuine ambiguity of the regulations, IC was asked for a clarification on the regulations and the decision has been made.
•  » » » » » » » 10 months ago, # ^ | ← Rev. 4 →   0 I hate to say that publicly, but since you mentioned it: it really wasn't due to genuine ambiguity. The committee knew that that was illegal, and they tried to get away with it. In fact, the participant has started lying to everyone about dropping out of university since October, although all of his friends at university knew that he's still enrolled and attends regularly. They decided not to let him go only after their actions have been brought into light and public complaints have aroused. Nevertheless, it was indeed a childish way of resolving the issue. I shouldn't have publicly commented; it is greatly irritating to see someone go to IOI illegally whilst you tried really hard for years and couldn't (legally) go. I apologize for my behavior and I hope the administration agrees on deleting the comments.
•  » » » » » » » » 10 months ago, # ^ |   0 At no point in the current IOI regulations being a university student outright disallows you to compete in the IOI (at least in my interpretation).
•  » » » » » » » » » 10 months ago, # ^ |   0 In Jordan, going to university == graduating from high school and not going to it
 » 11 months ago, # |   +27 The number of IOI participants is increasing!
•  » » 11 months ago, # ^ |   +92 Mystery of Ioi Participants Number
 » 11 months ago, # |   0 He has made a mistake, Cuba will not participate with the same team. This year, will participate David_Garcia, the previous year I participate Ernestico.
•  » » 11 months ago, # ^ |   +3 Touche. I don't know why I wrote Cuba, sorry!
 » 11 months ago, # | ← Rev. 3 →   0 No issue
•  » » 11 months ago, # ^ |   0 Nvm seems it changed just now :)
•  » » 11 months ago, # ^ | ← Rev. 3 →   +1 I don't know why they weren't visible for you, maybe some cache issues? I really have no clue.
•  » » » 11 months ago, # ^ |   0 Maybe I sorted in the wrong order and they were not together. Thank you for this list.
•  » » 11 months ago, # ^ | ← Rev. 2 →   0 After quickly checking, there are only 4 contestants missing from this list, which is the Israel's delegation, because of their participation at the off-site competition as discussed during IOI 2016 GA meetings.Also, keep in mind that the contestant lists can actually change (and I will update them on IOI Stats as the updates become available). The reason why I haven't released any data earlier is because it was still changing largely 2-3 weeks ago.
•  » » » 11 months ago, # ^ |   0 Isreal was added with a note about offsite participation.
 » 10 months ago, # |   0 Will be there a online mirror?
•  » » 10 months ago, # ^ |   +1 There were IOI2015 and IOI2016 online mirror in Yandex, probably they will continue on tradition :)Also, you can find IOI problems starting from 2003 and virtual participate in some years: https://contest.yandex.com/ioi/
 » 8 months ago, # | ← Rev. 2 →   +6 Kazakhstan still on first place by number of student