# Analyze a function defined in terms of an integral

Here is a question that really has puzzled me for quite a while. I happened to see this function defined in terms of an integral $$f(x):=\int_0^{\pi/2}\frac{2e^{x+e^x\cos y}}{1+\left(e^{e^x\cos y}\right)^2}dy.$$I want to analyze the behavior of the function when $$x \rightarrow \infty$$.

The strange thing is that when I used Mathematica to plot the function, the graph indicates that $$\lim_{x\rightarrow \infty} f(x)=0$$. However, it is easy to see that $$\liminf_{x\rightarrow \infty}f(x) \ge \frac{\pi}{4}$$, since $$\int_0^{\pi/2}\frac{2e^{x+e^x\cos y}}{1+\left(e^{e^x\cos y}\right)^2} \, dy \ge \int_0^{\pi/2}\frac{2e^{x+e^x\cos y}\sin y}{1+\left(e^{e^x\cos y}\right)^2}\, dy\\ =-\Big(\tan^{-1}\left(e^{e^x \cos y}\right)\Big)\Big|_{0}^{\pi/2}\\=\tan^{-1}\left(e^{e^x}\right)-\pi/4$$

Now I have two questions:

First, why the result from Mathematica is different from what I obtained?

Second, does $$\lim_{x\rightarrow \infty} f(x)$$ exist?

Maybe this question is not so suitable for mathoverflow, since it is just a calculus problem. However, I just feel so confused about the contradiction of numerical result and math. I want to understand the reason behind this situation. Any comments are really appreciated. Thank you very much.

Below is the code and picure I got from Mathematica....

• you probably made a coding error in Mathematica, when I plot it the limit is about 1.57 – Carlo Beenakker Jun 14 at 6:58
• You missed $2$ in antiderivative, so $\liminf$ is at least $\frac{\pi}{2}$ (which coincides with the number Carlo wrote). I bet that it won't be hard to prove that the the loss in multiplying by $\sin (y)$ is negligible as $x\to \infty$ and so the limit is actually $\frac{\pi}{2}$. – Aleksei Kulikov Jun 14 at 7:18
• @CarloBeenakker, thank you very much for the comment... As you can see from my updates, my code should be fine, but when I tried to plot the the graph for $x \in [0,20]$, something went wrong... – student Jun 14 at 13:31
• @AlekseiKulikov, exactly! Thank you very much! – student Jun 14 at 13:31

Mathematica seems to be plotting the function just fine...

If we look a bit at the integrand, it's clear that most of the mass is around $$y = \pi/2$$ as $$x$$ increases which should let us introduce a $$\sin y$$ term and use the antiderative you've already found.

We can try to cut the integral at $$\pi/2 - 1/x$$.

Let $$I = \int_0^{\pi/2 - 1/x} 2 e^x \frac{e^{e^x \cos y}}{1 + e^{2 e^x \cos y}} dy + \int_{\pi/2 - 1/x}^{\pi/2} 2 e^x \frac{e^{e^x \cos y}}{1 + e^{2 e^x \cos y}} dy = I_0 + I_1$$

Notice that $$f(u) = e^u / (1 + e^{2u})$$ is a decreasing function of $$u$$ and thus that $$I_0 < (\pi/2 - 1/x) 2 e^x \frac{e^{e^x \cos (\pi/2-1/x)}}{1 + e^{2 e^x \cos (\pi/2-1/x)}}$$

When $$x \rightarrow \infty$$ the $$\cos (\pi/2 - 1/x)$$ behaves as $$1/x$$ and the logistic function $$1-\sigma(u)$$ behaves as $$e^{-u}$$, so the right term behaves as $$\pi e^{x - e^x /x}$$ which converges to $$0$$.

For $$I_1$$, we note that if $$y \in [\pi/2-1/x,\pi/2]$$, $$1 - \frac{1}{2x^2} < \sin y \leq 1$$

$$I_1 \left(1-\frac{1}{2x^2}\right)< \int_{\pi/2-x}^{\pi/2} 2 e^x \frac{e^{e^x \cos y}\sin y}{1 + e^{2 e^x \cos y}} \leq I_1$$

The middle term converges to $$\pi /2$$ and thus so does $$I_1$$ and so does $$I$$.

• Thank you very much! By the way, I got the same picture for $x$ up to 10, but when I plotted the graph for $x \in [0,20]$, it gives me the limit going to $0$, as you can see from my updates.... – student Jun 14 at 13:26
• The support of the mass becomes too small and the numerical integration misses it. If you do a change of variable which blow up the region around pi/2 it'll be more stable. – Arthur B Jun 14 at 13:44


# Calculate variance from a stream of sample values

I'd like to calculate a standard deviation for a very large (but known) number of sample values, with the highest accuracy possible. The number of samples is larger than can be efficiently stored in memory.

The basic variance formula is:

$\sigma^2 = \frac{1}{N}\sum (x - \mu)^2$

... but this formulation depends on knowing the value of $\mu$ already.

$\mu$ can be calculated cumulatively -- that is, you can calculate the mean without storing every sample value. You just have to store their sum.

But to calculate the variance, is it necessary to store every sample value? Given a stream of samples, can I accumulate a calculation of the variance, without a need for memory of each sample? Put another way, is there a formulation of the variance which doesn't depend on foreknowledge of the exact value of $\mu$ before the whole sample set has been seen?

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This very same issue was discussed on dsp.SE, and the answers were very similar, with numerically unstable methods proposed in one answer, and more stable methods described by others. –  Dilip Sarwate Mar 4 '12 at 17:04

You can keep two running counters - one for $\sum_i x_i$ and another for $\sum_i x_i^2$. Since variance can be written as $$\sigma^2 = \frac{1}{N} \left[ \sum_i x_i^2 - \frac{(\sum_i x_i)^2}{N} \right]$$

you can compute the variance of the data that you have seen thus far with just these two counters. Note that the $N$ here is not the total length of all your samples but only the number of samples you have observed in the past.

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Wait -- shouldn't the N be inside the the expected values, such that the right side is divided by $N^2$? –  user6677 Feb 5 '11 at 22:05
e.g. $$\sigma^2 = \frac{(\sum_i x_i^2)}{N} - (\frac{\sum_i x_i}{N})^2$$ –  user6677 Feb 5 '11 at 22:27
@user6677: You are indeed right. Thanks for the correction. –  Dinesh Feb 5 '11 at 22:30
Note the sum of squares especially is at risk of overflow. –  dfrankow Dec 29 '14 at 18:06

I'm a little late to the party, but it appears that this method is pretty unstable, but that there is a method that allows for streaming computation of the variance without sacrificing numerical stability.

Cook describes a method from Knuth, the punchline of which is to initialize $m_1 = x_1$, and $v_1 = 0$, where $m_k$ is the mean of the first $k$ values. From there,

\begin{align*} m_k & = m_{k-1} + \frac{x_k - m_{k-1}}k \\ v_k & = v_{k-1} + (x_k - m_{k-1})(x_k - m_k) \end{align*}

The mean at this point is simply extracted as $m_k$, and the variance is $\sigma^2 = \frac{v_k}{k-1}$. It's easy to verify that it works for the mean, but I'm still working on grokking the variance.

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+1, excellent! I didn't feel a simple upvote would be sufficient to express my appreciation of this answer, so there's an extra 50 rep bounty coming your way in 24 hours. –  Ilmari Karonen Mar 4 '12 at 20:12
Isn't the final variance calculation should be Vk / k (not k-1)? –  errr Oct 10 '13 at 15:49

tldr: I think there is a missing $(k-1)$ term in @Dan's variance calculation. I think it should be $$\sigma^2_k = \frac{1}{k} \left( (k-1) \sigma^2_{k-1} + (x_k-m_{k-1})(x_k-m_k)\right)$$

Validation: I tried what Dan's formula but am getting something different from Matlab's var command. For example, take 1000 Gaussian numbers (mean 0, var 100) and compute their mean and var:

clear; clc
n=1000;
x=round(randn(1,n)*10);

% calcs
mm(1)=x(1);
vm(1)=0;
m(1)=x(1);
v(1)=0;
ma(1)=x(1);
va(1)=0;
for k=2:n
mm(k) = mean(x(1:k)); %matlab mean
vm(k) = var(x(1:k),1); %matlab var

m(k) = m(k-1) + 1/k * (x(k)-m(k-1)) ; %Knuth mean
v(k) = (1/k) * (v(k-1) + (x(k)-m(k-1)) * (x(k)-m(k))) ; %Knuth var

ma(k) = (1-1/k) * ma(k-1) + (1/k) * x(k); %Finch mean
%va(k) = (1-1/k) * (va(k-1) + (1/k) * (x(k)-ma(k-1))^2 ); %Finch 143
va(k) = (1/k) * ((k-1)*va(k-1) + (x(k)-m(k-1)) * (x(k)-m(k)) ); %Finch 143 rearranged to look like Knuth
end

% plots
figure(1); clf
subplot 211
plot(1:n, x, 'k--')
hold on
plot(1:n,mm, 'b-')
plot(1:n,m, 'g-')
plot(1:n,ma, 'r--')
hold off
ylabel('means')
legend('x','Matlab','Knuth','Finch')

subplot 212
plot(1:n,vm, 'b-')
hold on
plot(1:n,v, 'g-')
plot(1:n,va, 'r--')
hold off
ylabel('vars')
xlabel('steps')
legend('Matlab','Knuth','Finch')

As you can see I selected Matlab's (octave's) VAR(X,1) command which divides by $N$ and not by $(N-1)$, which goes back to @errr's comment - it has to do with using biased vs unbiased estimator (I usually go to biased one because it makes more sense to me)

So it seems like Knuth variance is ok for a couple of steps and then it's way below because of the missing $(k-1)$ term.

So what I tried then is based on Tony Finch's explanation of recursive variance (eq (143)) with $\alpha=\frac{1}{k}$. Admittedly it's a hack as my alpha changes with every step, whereas his is assumed to be constant (?), but in the end this formula agrees with matlab's var(X,1).

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If you have a new question, please ask it by clicking the Ask Question button. Include a link to this question if it helps provide context. –  Mark Fantini Jul 14 '14 at 22:56
I guess it was unclear - I found a mistake in previous answer. added a small summary to say that up front. Thank you! –  alexey Jul 15 '14 at 0:01


# Probability Urn Problems

You don't know which balls where inserted, but as I go along picking balls and showing them to you, what can be said about the number of black and white balls in the urn? That's inverse probability, although it's an old term. An urn contains {eq}8 {/eq} white and {eq}6 {/eq} green balls. A ball is chosen at random from urn #1. , without any aids, but don’t worry about a time limit). Urn A contains 2 white and 4 red balls, whereas urn B contains 1 white and 1 red ball. A copy of the source for Grinstead and Snell's lovely probability book - tdunning/probability-book. Five marbles are randomly selected, with replacement, from the urn. For example, we assumed $$\p(B_1 \given \neg A) = 1/2$$ because the. Let A = the event that the first marble is black; and let B = the event that the second. A series of two-urn biased sampling problems Puza, Borek and Bonfrer, André 2018, A series of two-urn biased sampling problems, Communications in statistics - theory and methods, vol. 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Due to rapid growth in the field in recent years, this volume aims to promote interdisciplinary collaboration in the areas of quantum probability, information, communication and foundation, and mathematical physics. Show that the probability of rolling doubles with a non-fair (“ﬁxed”) die is greater than with a fair die. In SAS you can use the table distribution to specify the probabilities of selecting each integer in the range [1, c]. Two of the selected marbles are red, and three are green. A Collection of Dice Problems Matthew M. $\begingroup$ One way to do this is with the transfer matrix method. Are the events w = 3 and w= r dependent or independent? Problem 2 (10 points) Urn A contains 4 white balls and 8 black balls. However, let us now change the experiment and suppose that at 1 minute to 12 P. The probability that a customer will buy a product given that he or she has seen an advertisement for the product is 0. 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Urn 1 contains 2 white balls and 2 black balls. An urn contains n red and m black balls. Show that this is the same as the probability that the next ball is black for the Polya urn model of Exercise 4. A second urn contains 16 red balls and an unknown number of blue balls. Another urn contains 2 red chips. If your opponent draws first, what is the probability that you win?. The flippant juror. In the ﬁrst draw, one ball is picked at random and discarded without noticing its colour. For example, the probability of getting two "tails" in a row would be:. If a 4 appears, a ball is drawn from urn 1; otherwise, a ball is drawn from urn 2. Are the events w = 3 and w= r dependent or independent? Problem 2 (10 points) Urn A contains 4 white balls and 8 black balls. When TEST1 is done on a person, the outcome is as follows: If the person has the disease, the result is positive with probability 3/4. What's the probability of the event A = {the sum and the product of the numbers that come up are equal}? Problem 8 From an urn, which contains 10 white, 7 green and 6 red balls, 1 ball is taken out. The basic urn problem is to determine the probability of drawing one colored ball from an urn with known composition of differently colored balls. n(S) is the number of elements in the sample space S and n(E) is the number of elements in the event E. Introduction Drawing balls from an urn without replacement is a classical paradigm in probability (Feller 1968). If a red or white ball is chosen, a fair coin is flipped once. Two red, three white, and five black. The Type X urns each contain $$3$$ black marbles, $$2$$ white marbles. The Sock drawer. Conditional Probability and the Multiplication Rule It follows from the formula for conditional probability that for any events E and F, P(E \F) = P(FjE)P(E) = P(EjF)P(F): Example Two cards are chosen at random without replacement from a well-shu ed pack. Measure-valued Pólya urn processes Mailler, Cécile and Marckert, Jean-François, Electronic Journal of Probability, 2017 Optimal stopping rule for the no-information duration problem with random horizon Tamaki, Mitsushi, Advances in Applied Probability, 2013. What is the probability that the marble chosen is red? Example 8: Box #1 contains 9 violet and 7 white marbles. Question 290480: There are 3 urns containing 2 white and 3 black balls, 3 white and 2 black balls and 4 white and 1 black balls respectivelly. A small probability calculator / tester for urn models (urn problem) - exane/urn-probability-calculator. Note that to define a mapping from A to B, we have n options for f ( a 1), i. (4 balls are now in urn C. The probability the first is white is 6/15= 2/5. It is commonly used in randomized controlled trials in experimental research. The simplest experiment is to reach into the urn and pull out a single ball. Column B contains the six. If playback doesn't begin shortly, try restarting your device. What is the probability that the coin landed heads? So at first I figured the balls part of the question was irrelevant, and just. (b) 2marbles are selected without replacement. A tree diagram is a special type of graph used to determine the outcomes of an experiment. Find the probability of the event that at least 5 tosses are required. In Urn i there are i black balls and one. Introduction Drawing balls from an urn without replacement is a classical paradigm in probability (Feller 1968). (This will be the distribution after a long time if in every second a random urn is chosen, and a ball, if any, from that urn is moved into the clockwise neighboring urn. Second, an argument that generalizes from observed instances is similar to an urn problem, where we guess the contents of the urn by repeated sampling. If N is large enough (say, N=20), the probability of either of those events is much less than 1%. For example, if you have a bag containing three marbles -- one blue marble and two green marbles -- the. The locomotive problem. ) Jacob Bernoulli 1700. If there are N urns, two colours, and balls are coloured independently with probability 0. n = [3 × seed /9999] + 1. The Urn Problem with Indistinguishable Balls. Find the probability that at least one ball of each color is chosen. A well-known result of this type is Polya's urn problem (see [8]), which is just the. An urn contains 75 balls, some red, some blue. The Type X urns each contain $$3$$ black marbles, $$2$$ white marbles. To determine whether to accept the shipment of bolts,the manager of the facility randomly selects 12 bolts. A basic problem first solved by Jakob Bernoulli is to find the probability of obtaining exactly i red balls in the experiment of drawing n times at random with replacement from an urn containing b black and r red balls. For example, here is a three-part problem adapted from mathforum. So the probability of moving from state 3 to state 4 is 2=5 2=5. Naturally, the problems on this site are perfect for a blog such as this, so this is the first of many interesting such problems that I will post 🙂 This particular exercise is a probability problem that will appeal to anyone that likes games involving dice. Problems and Complete explanatory solutions to problems on probability involving drawing, picking, selecting, choosing two or more balls from a box, bag, urn, container. Please read our cookie policy for more information about how we use cookies. One ball is chosen randomly from the urn. Durrett, The Essentials of Probability, Duxbury Press, 1994 S. Each turn, we pick out a ball, note it's colour then replace it and add d more balls of the same colour into the urn. What's the probability of the event A = {the sum and the product of the numbers that come up are equal}? Problem 8 From an urn, which contains 10 white, 7 green and 6 red balls, 1 ball is taken out. lf X = 2 choose with replacement 6 balls from Urn B. For each new ball, with probability p, create a new bin and place the ball in that bin; with probability 1 − p, place the ball in an existing bin, such that the probability the ball is placed in a bin is proportional to m γ,wheremis the number of balls in that bin. Examples 8-1 through 8-10 deal with some of the more impor-tant sorts of questions one may ask about drawings without replacement from such an urn. Container in many probability-theory problems is a crossword puzzle clue that we have spotted 1 time. of selecting 1 st urn × prob. org 21 Even Knight’s a priori probabilities—those based on some symmetry of a problem—are sus-pect. The probability that it is either a black ball or a green ball is ; A box contains 150 bolts of which 50 are defective. When TEST1 is done on a person, the outcome is as follows: If the person has the disease, the result is positive with probability 3/4. You can also express this relationship as 1 ÷ 6, 1/6, 0. Therefore, one student must have solved at least 5 problems. But anyways using the binomial theorem. Ron chooses three re©hree blue marbles 3a3 + [2 marks] [2 marks] 3) Which of the following numbers cannot be the probability of some. You and your friend take turns randomly picking a ball from the urn. Problem 1. Defining Risk November/December 2004 www. Durrett, The Essentials of Probability, Duxbury Press, 1994 S. Let X be the number of red balls you pull before any black one, and Y the. Example An urn contains 6 red marbles and 4 black marbles. There is equal probability of each urn being chosen. Example : Probability to pick a set of n=10 marbles with k=3 red ones (so 7 are not red) in a bag containing an initial total of N=100 marbles with m=20 red ones. Math: Conditional Probability. To see this, fix an urn. 1702-1761) was a Presbyterian minister. If the probability of an event is 0, it is impossible for that event to occur. The two events are independent so the probability of a blue marble from each urn is the product. At each step we randomly choose a ball from Urn 1, throw it away, and move a red ball from Urn 2 into Urn 1. Question 1157976: Urn A contains 5 red marbles and 3 white marbles Urn B contains 2 red marbles and 6 white marbles a. Exercises in Probability Theory Nikolai Chernov All exercises (except Chapters 16 and 17) are taken from two books: R. Four balls are drawn at random. One ball is drawn at random. This Web site is a course in statistics appreciation; i. Time is the main factor in competitive exams. What is the probability that they are both of the same color? b. Originally, the urn contains 6 white and 9 black balls, total of 15. After the nth step, Urn 2 is empty. Events that are unlikely will have a probability near 0, and events that are likely to happen have probabilities near 1. However, these are not all given equal probability when you take 20 objects from an urn with 10 of each color. This activity shows the classic marble example of elementary probability. Problem 1. The objective probability that a random draw from an urn yields a black ball changes over time if the urn has a hole through which its mixture of black and red balls spills. So, in a Blackjack game, to calculate the chances of getting a 21 by drawing an Ace and then a face card, we compute the probability of the first being an Ace and multiply by the probability of drawing a face card or a 10 given that the first was an Ace: $1/13 \t imes 16/51 \a pprox 0. Problem 1 An urn contains n+m balls, of which n are red and m are black. choose a ball from an urn and record its color, then do it again; flip a coin and record Head or Tail, then choose a ball from an urn and record its color The branches emanating from any point on a tree diagram must have probabilities that sum to 1. An urn contains n+m balls, of which n are red and m are black. Urn contains red balls and black balls. Category:Probability theory. I am trying to solve this problem: r balls are randomly assigned into n urns. (Some of them might appear on our problem sheets. What is the probability that the urn contains three balls of each color? 5) If x, y and z are positive real numbers satisfying +1 =4, +1 =1, and +1 =7 3, then what is the value of xyz? Relation Problems 6) Create a set A of people you know well. Divide the number of events by the number of possible outcomes. To win, all six numbers must match those chosen from the urn. What's the probability that the ball, which was taken out is: a) white b) green c) red Problem 9 An urn contains 8 white and 4 black balls. And indeed, the induced frequencies do converge to the Dirichlet distribution with k equal parameters. One ball is picked at random from urn 1 and, without. Container in many probability-theory problems is a crossword puzzle clue that we have spotted 1 time. What is the probability that they are both of the same color? b. The topic of statistics is presented as the application of probability to data analysis, not as a cookbook of statistical recipes. Suppose that there are 71 urns given, and that balls are placed at random in these urns one after the other. what is the probability that the equipment will fail before the end of one year? 2. The multinomial theorem is a statement about expanding a polynomial when it is raised to an arbitrary power. Please justify your answers and don't simply give the answer. 6 balls are randomly drawn from the urn in succession, with replacement. Sample Problem 1: A six-sided die is rolled six times. Xi;Xj/with i 6Dj has the same distribution. One urn contains 2 blue chips. Conditional Probability 4. An urn contains 1 red ball and 10 blue balls. You choose an urn and your opponent chooses an urn from the remaining ones. Supposethat we win$2 for each black ball selectedand we lose $1 for each white ball selected. What is the probability that a white ball is drawn?. From the first urn to the second 4 balls are moved. lf X = 2 choose with replacement 6 balls from Urn B. If one ball is drawn at random, what is the probability of getting a red or a white ball? 3. 5 and placed in urns uniformly with probability 1/N then the expected number of trials to get a collision is order √ πN. This consists. Suppose that each of Barbara’s shots hits a wooden duck target with probability p1, while each shot of Dianne’s hits it with probability. We move a ball from urn 1 to urn 2. At least I think so. Suppose that there are 71 urns given, and that balls are placed at random in these urns one after the other. b) Find the probability that urn 1 was used given that a red ball was drawn. 24) Don't Lose Your Marbles! 1. Dodgson) is used to illustrate the nature, standing and understanding of probability within the wider English mathematical community of his time. Find the probability that the sum of the two faces is 10 given that one shows a 6. If the composition is unknown, then it is called uncertain. I recommend studying rst, using previous homeworks, exams and exam practice problems. Keywords: Urn problems, drawing without replacement, enumerative combinatorics, Scrabble, R. At least 1 + 2 + 3 = 6 problems were solved by the students mentioned in the problem statement. What is the probability that the coin landed heads? So at first I figured the balls part of the question was irrelevant, and just. Probabilities of drones with replacement question if your day is rolled five times what is the probability of - Answered by a verified Math Tutor or Teacher We use cookies to give you the best possible experience on our website. 3 MB Law of Large Numbers - Urn Problems - Low Resolution. Introduction Drawing balls from an urn without replacement is a classical paradigm in probability (Feller 1968). At each step we randomly choose a ball from Urn 1, throw it away, and move a red ball from Urn 2 into Urn 1. Scott Boras: Astros players don’t need to apologize. IBPS Clerk, RBI Assistant I Probability Basic Concept & Tricks I Video देखने के बाद ये Topic Easy है - Duration: 37:30. Electronic Journal of Probability, 16, 1723-1749, 2011. If you summarize the results, you will find that the outcome "2 red, 2 white" occurs (almost exactly) 6 times as often as the outcome "4 red" or "4 white". Example : Probability to pick at least once each card from a deck of N=50. b) 2 heads and a tail. The person who selects the third WIN ball wins the game. (2 points each) In each of the scenarios below, indicate whether the distribution of X is binomial, poisson, negative binomial (r >1), geometric, or neither. In what follows, S is the sample space of the experiment in question and E is the event of interest. Probability Definitions: Example #1. You have select each urn with probability$0. • Put your NAME on each sheet. Urn A has 4 white and 16 red balls. The notion that the probability of an event may depend on other events is called conditional probability The conditional probability of event Agiven event Bis written as P(AjB) For example, in our ball and urn problem, when sampling without replacement: P(R 2) = 1 3 P(R 2jR 1) = 0 P(R 2jRC 1) = 1 2 Patrick Breheny Introduction to Biostatistics. CAT Probability Questions is a sample set of problems that is asked from this topic in the CAT Probability Section. For example, if you have a bag containing three marbles -- one blue marble and two green marbles -- the. ) A pair is not drawn 3) Four balls are selected at random without replacement from an urn containing three white balls and five blue balls. A Collection of Dice Problems Matthew M. Electronic Journal of Probability, 16, 1723-1749, 2011. One ball is picked at random from urn 1 and, without. Hints help you try the next step on your own. (This will be the distribution after a long time if in every second a random urn is chosen, and a ball, if any, from that urn is moved into the clockwise neighboring urn. For more practice, I suggest you work through the review questions at the end of each chapter as well. The basic urn problem is to determine the probability of drawing one colored ball from an urn with known composition of differently colored balls. Every minute, a marble is chosen at random from the urn, and then returned to the urn, together with another marble of the same colour. Show that this is the same as the probability that the next ball is black for the Polya urn model of Exercise 4. • Time limit 110 minutes. Note that to define a mapping from A to B, we have n options for f ( a 1), i. Probability Problems for Group 1 (Due by EOC Mar. Pull balls from the urn one by one without replacement. Question 3 Solution Urn 1: seven red and three green balls. Thomas Bayes was an English minister and mathematician, and he became famous after his death when a colleague published his solution to the “inverse probability” problem. Let N be the number of throws of a usual six-sided die that is needed for the sum of the scores on these throws to be at least 3. What strategy maximizes your chance of victory? Problems from Rosen 7. Probability Urn simulator This calculator simulates urn or box with colored balls often used for probability problems and can calculate probabilities of different events. Access-restricted-item true Addeddate 2011-06-22 16:08:09 Bookplateleaf 0002 Boxid IA1398805 Camera Canon EOS 5D Mark II City Upper Saddle River, NJ Donor. Let the probability that the urn ends up with more red balls be denoted. Savage offered the example of an urn that contains two balls: Both may be white; both may be black; or one may be white and. ) I'm happy to discuss about them on my oﬃce hours (unless we are too many on those). Bayes Theorem Practice Problems With Solutions Genetics. Question 3 Solution Urn 1: seven red and three green balls. This limit distribution is the negative binomial distribution with parameters and (the corresponding mathematical expectation is , while the variance is ). The formula is. What is the probability that the ball you drew is green? (b)You then look at the ball and see that it is green. The geometry problems are treated in a separate post. YOU are responsible for studying all the sections to be covered on the midterm. If your opponent draws first, what is the. That is, we seek a random integer n satisfying 1 ≤ n ≤ 3. , that the coin ﬂip was Tails) Let T be the event that the coin ﬂip was Tails. • Bose-Einstein distribution. Let X be the number of red balls removed before the first black ball is chosen. Thus, the probability that they both give the same answer is 39. The probability of an event is a measure of the likelihood that the event will occur. 4 Practice Problems Problem 34. After a severe winter, potholes develop in a state highway at the rate of 5. A risk, on the other hand, is defined to be a higher probability event, where there is enough information to make. I manage to calculate the individual probabilities on a per problem basis, but I need to find a way to phrase a general solution. Alice selects one ball at random (each of the 7 balls can be selected with equal probability) and takes it out of the urn. Ghahramani, Fundamentals of Probability, Prentice Hall, 2000 1 Combinatorics These problems are due on August 24 Exercise 1. A primary concern with PoS systems is the “rich getting richer” phenomenon, whereby wealthier nodes are more likely to get elected, and hence reap the block reward, making them even wealthier. We're interested in the probability of a certain observed outcome given a known process. The trials are independent; that is, the outcome on one trial does not affect the outcome on other trials. The field of Probability has a great deal of Art component in it - not only is the subject matter rather different from that of other fields, but at present the techniques are not well organized into systematic methods. This paper designs some uncertain urn problems in order to compare probability theory and uncertainty theory. This is exactly the binomial experiment. I'm trying to answer the following question using a simple Monte Carlo sampling procedure in R: An urn contains 10 balls. In these cases, we will need to use the counting techniques from the chapter 5 to help solve the probability problems. Urn A contains 2 white and 4 red balls, whereas urn B contains 1 white and 1 red ball. k indistinguishable balls are randomly distributed into n urns. A room contains four urns. Remarkable selection of puzzlers, graded in difficulty, that illustrate both elementary and advanced aspects of probability. An urn contains pink and green balls. There are 40 marbles in an urn: (PROBABILITY) There are 40 marbles in an urn: 11 are green and 29 are yellow. Let the probability that the urn ends up with more red balls be denoted. What is the probability that both marbles are the same color if: a. b) Find the probability that urn 1 was used given that a red ball was drawn. Edited by Bernard R. 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. Two marbles are drawn without replacement from the urn. In the case of rolling a 3 on a die, the number of events is 1 (there’s only a single 3 on each die), and the number of outcomes is 6. Let Event A = {The Ball From Urn No. The occupancy problem in probability theory is about the problem of randomly assigning a set of balls into a group of cells. Clas-sical mathematicians Laplace and Bernoulis, amongst others, have made notable contributions to this. An urn contains 10 balls: 4 red and 6 blue. What strategy maximizes your chance of victory? Problems from Rosen 7. Probability Definitions: Example #1. What is the probability a red marble is drawn?. This formula relies on the helper table visible in the range B4:D10. A Probability trees in probability theory is a tree diagram used to represent a probability space. Acknowledgements This work was made possible by a grant from NSF-DUE and the support of Cornell University and the statistics department at Stanford. If there are N urns, two colours, and balls are coloured independently with probability 0. What is the probability that the coin landed heads? So at first I figured the balls part of the question was irrelevant, and just. The Type X urns each contain $$3$$ black marbles, $$2$$ white marbles. JANOS FLESCH, DRIES VERMEULEN AND ANNA ZSELEVA. Calculate the number of blue balls in the second urn. 1 Is Black}; Event B= {The Ball From Urn. Such problems have been considered by Nishimura and Sibuya [13, 14] and Selivanov [17]. Urn and Beads Problem The next family of problems was modeled after Edwards (1968). What are your chances of getting exactly 4, 5, or 6 matches? Many lotteries and gambling games are based on this concept of picking from mixed good and bad balls. Suppose you have n ball- lled urns, numbered 1 through n. We'll look at a number of examples of modeling the data generating process and will conclude with modeling an eCommerce advertising simulation. In the two parameter case, the matrix of transition probabilities has N+1 distinct eigenvalues λ j =1−2j/N, where j=0, 1,…, N. Can You Solve This Intro Probability Problem? An urn contains 10 balls: 4 red and 6 blue. Heads, a ball is drawn from Urn 1, and if it is Tails, a ball is drawn from Urn 2. Home-Learning Problems 3. Intro to Probability - Homework Assignment 2 Please solve the problems and type the solutions up using LateX and the template provided. So Urn B is more. $$\frac{\textrm{Probability of picking the 2nd urn and picking a blue ball}}{\textrm{Probability of picking a blue ball}}$$ Since there are equal numbers of balls in each urn, and each urn is equally likely to be picked, you can do this by counting. 2 Real Life Is More Complicated. Practice online or make a printable study sheet. If we draw 5 balls from the urn at once and without peeking,. What's the probability that the ball, which was taken out is: a) white b) green c) red Problem 9 An urn contains 8 white and 4 black balls. three marbles are drawn from the urn one after the other. An urn contains 10 balls: 4 red and 6 blue. Question 1157976: Urn A contains 5 red marbles and 3 white marbles Urn B contains 2 red marbles and 6 white marbles a. What is the probability that neither is red, given that neither is white? 2) A basketball player makes free throws with a 0. 1 2 ⋅ 1 2 = 1 4. Thus in this experiment each time we sample, the probability of choosing a red ball is $\frac{30}{100}$, and we repeat this in $20$ independent trials. Barbara and Dianne go target shooting. Probability of second ball being red = 3/9 (because there are 3 red balls left in the urn, out of a total of 9 balls left. The probability of any event can range from 0 to 1. Sample Problem 1: A six-sided die is rolled six times. A card is randomly selected from an ordinary pack of 52 playing cards. 625 subscribers. Access-restricted-item true Addeddate 2011-06-22 16:08:09 Bookplateleaf 0002 Boxid IA1398805 Camera Canon EOS 5D Mark II City Upper Saddle River, NJ Donor. An urn contains 10 red and 8 white balls. what is the probability that the equipment will fail before the end of one year? 2. • Put your NAME on each sheet. Then another ball is drawn and its color is recorded. (This will be the distribution after a long time if in every second a random urn is chosen, and a ball, if any, from that urn is moved into the clockwise neighboring urn. I've tried two approaches and neither work. An important problem in the Ehrenfest chain is the long-term, or equilibrium, distribution of , the number of balls in urn A. Math 29 — Probability Practice Second Midterm Exam 1 Instructions: 1. The ticket shows your six good balls, and there are 50 bad balls. I manage to calculate the individual probabilities on a per problem basis, but I need to find a way to phrase a general solution. The Philosophy of Statistics [S]tatistical inference is firmly based on probability alone. An urn contains three red balls numbered 1, 2, and 3, three black balls numbered 8, 9, and 10 and four white balls numbered 4, 5, 6, 7. The probability of this happening is =~ 0. ) are represented as colored balls in an urn or other container. Savage offered the example of an urn that contains two balls: Both may be white; both may be black; or one may be white and. 3, 794-814, 2012. Find P(N = 1), P(N = 2) and P(N = 3). (b) Find EN and Var(N). Probability basics 50 xp Queen and spade 50 xp. We recommend you review today's Probability Tutorial before attempting this challenge. Roll a fair 6-sided die until your first roll of a “3. The user may control the total number of balls in the urn (N), the number of red balls (R) and the number of balls sampled from the urn (n). A ball is selected at random from the ﬁrst urn and placed in the second. You can also express this relationship as 1 ÷ 6, 1/6, 0. Imagine two urns. Let us suppose that the urns are labelled with the numbers 1,2,, n and let tj be equal to k if the j-th ball is placed into the k-th urn. A Ball Is Randomly Drawn From Urn No. What is the probability that a white ball is drawn?. In probability and statistics, an urn problem is an idealized mental exercise in which some objects of real interest (such as atoms, people, cars, etc. An urn contains 5 red, 3 green, and 4 white balls. What is the probability that neither is red, given that neither is white? 2) A basketball player makes free throws with a 0. Two marbles are selected at random without replacement. 625 subscribers. After the four iterations the urn contains six balls. same answer. |A) is a probability function multiplicative formula: P(B and A) = P(B|A)P(A) Oct 28 Monty Hall problem §3. Bayes probabilities can also be obtained by simply constructing the tree. Then the number of tables is bounded by this multiple, so for large n, the probability of joining one of the k (fixed) tables is roughly , so this should behave roughly like the standard Polya Urn. What is the probability that the urn contains three balls of each color? Solution A tree diagram is ideal to deal with the problem, up to a sample space of 4 balls (so the first two pulls): So, we have a probability of (1/2)*(2/3) + (1/2)*(2/3) = 2/3 of ending up with 3 balls of one color. Then treat this like a mock exam (i. Home-Learning Problems 3. Show that the probability of rolling doubles with a non-fair (“ﬁxed”) die is greater than with a fair die. Gelbaum DOVER PUBLICATIONS, INC. We write P (heads) = ½. Find P(N = 1), P(N = 2) and P(N = 3). What strategy maximizes your chance of victory? Problems from Rosen 7. The basic urn problem is to determine the probability of drawing one colored ball from an urn with known composition of differently colored balls. What is the probability the ball drawn from urn C is black? I've figured out that P(K\A) = 2/3 and P(K\B) = 5/6. , TTHor HHHT). The probability a black ball is drawn is 9/13. In the example shown, the formula in F5 is: =MATCH(RAND(), D$5:D$10) How this formula works. If you pick 5 balls from the urn at random, what is the probability that x of them will be. If a 4 appears, a ball is drawn from urn 1; otherwise, a ball is drawn from urn 2. There will be total 10 MCQ in this test. What is the probability that the coin landed heads? So at first I figured the balls part of the question was irrelevant, and just. In the two parameter case, the matrix of transition probabilities has N+1 distinct eigenvalues λ j =1−2j/N, where j=0, 1,…, N. of getting white marble). From a given ball's perspective, the probability of being matched is $\frac{M}{N}(1-(1-\frac{1}{M})^N)$. The first problem-book of a similar kind as ours is perhaps Mosteller's well-known Fifty Challenging Problems in Probability (1965). Once again, balls are chosen at random with equal probability. At each step, an urn is selected according to their weights. What's the probability that the ball, which was taken out is: a) white b) green c) red Problem 9 An urn contains 8 white and 4 black balls. 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. The urn model and the Pólya process, in which the Pólya distribution and the limit form of it arise, are models with an after effect (extracting a ball of a particular colour from the urn increases the probability of extracting a ball of. He titled it The Two Children Problem, and phrased. User Account. of selecting 2 nd urn × prob. Walk through homework problems step-by-step from beginning to end. There are three ways to measure the average: the mean, median, and mode. Hume’s Problem. Create a set B of foods. Six balls are picked from the 56 balls in an urn. Urn #2 contains 5 black and 2 red balls. Probability of second ball being red = 3/9 (because there are 3 red balls left in the urn, out of a total of 9 balls left. Publication date 1987 Topics Probabilities, Probabilités, Probabilités Publisher Internet Archive Books. Then the probbilities that the urn from which I have drawn the tickets is A or B, are by the principle given above, as K: K'. If one ball is drawn at random, what is the probability of getting a red or a white ball? 3. The Type X urns each contain $$3$$ black marbles, $$2$$ white marbles. The first case,. By convention, statisticians have agreed on the following rules. Probability Trees: Many probability problems can be simplified by using a device called a probability tree. The probability for the dice to yield 1 or 2 is 2/6. 8--Conditional Probability With Urns and Marbles Glenn Olson. The probability of a sample point is a measure of the likelihood that the sample point will occur. To Probability Practice Problems for Exam # 2 1. One ball is drawn at random. 8--Conditional Probability With Urns and Marbles Glenn Olson. For each new ball, with probability p,createanewbin and place the ball in that bin; with probability 1−p,placetheballin an existing bin, such that the probability the ball is placed in a bin is proportional to mγ,wheremis the number of balls in that bin. An Example Suppose that a 6-sided die is rolled, what is the probability that the result is a 1? From our earlier results, if A is the event that a 1 is rolled, then P(A) = 1 6. Urn A contains 2 white and 4 red balls, whereas urn B contains 1 white and 1 red ball. The probability the first is white is 6/15= 2/5. It is concluded that uncertainty theory is better than probability theory to. 6 balls are randomly drawn from the urn in succession, with replacement. What is the probability of getting exactly k red balls in a sample of size 20 if the sampling is done with replacement (repetition allowed)? Assume 0 ≤ k ≤ 20. One ball is drawn at random and its color noted. What is the probability of winning the. Fully worked-out solutions of these problems are also given, but of course you should ﬁrst try to solve the problems on your own! c 2013 by Henk Tijms, Vrije University, Amsterdam. Binomial probability distribution; Hypothesis testing, types of error, and small samples; Introduction to statistics and the relative frequency histogram; Measures of variability and relative standing ; Other graphical methods and numerical methods; Poisson probability distribution and the urn model; Probability; Random variables. From a given ball's perspective, the probability of being matched is $\frac{M}{N}(1-(1-\frac{1}{M})^N)$. For example, if the probability of picking a red marble from a jar that contains 4 red marbles and 6 blue marbles is 4/10 or 2/5, then the probability of not picking a red marble is equal to 1 - 4/10 = 6/10 or 3/5, which is also the. So in a more conventional forward probability problem. What is the probability that it gets a) all 4 aces b) at least 1 ace; Hearts 5 cards are chosen from a standard deck of 52 playing cards (13 hearts) with replacement. Applications of probability arise everywhere: Should you guess. A series of two-urn biased sampling problems Puza, Borek and Bonfrer, André 2018, A series of two-urn biased sampling problems, Communications in statistics - theory and methods, vol. CAT Probability Questions cover all the different ways in which question can be asked. ) Jacob Bernoulli 1700. of selecting 1 st urn × prob. 1 Preliminaries This document is designed to get a person up and running doing elementary probability in R using the prob package. What strategy maximizes your chance of victory? Problems from Rosen 7. Another urn contains 2 red chips. Suppose that a white ball is selected. Column B contains the six. But what is the probability that when you throw a 6 sided unbiased die twice, the second output is 1 given that the first output is 5? Answer is 1/6, because P[first output is 5]=1 since it has already occurred. But anyways using the binomial theorem. Above the plane, over the region of interest, is a surface which represents the probability density function associated with a bivariate distribution. What is the probability that the first ball was also red?. This list of problems should serve as a good place to start studying, and it should not be considered a comprehensive list of problems from the sections we’ve covered. Now there are 13 balls, 4 white and 9 black. So the probability that urn iis empty is 1 11 i 1 i+1 1 1 i+2 1 1 n. A ball is drawn from Urn A and then transferred to Urn B. Every minute, a marble is chosen at random from the urn, and then returned to the urn, together with another marble of the same colour. A ball is also chosen at random from urn #2. It does not matter who picked the first three WIN balls. One ball is drawn from an urn chosen at random. (d) The variance of X. The formula is. Annals of Probability, 41, no. 4 Conditional Probability and Independence 1. 2012 John Wiley & Sons, Inc. Two balls are drawn at random from each one of the urn A and urn B, and are placed into urn C. Let A = { a 1, a 2, a 3,, a m }, B = { b 1, b 2, b 3,, b n }. It is concluded that uncertainty theory is better than probability theory to deal with uncertain urn problems. Four balls are drawn at random. 2: 19, 23, 29, 33. If the composition is unknown, then it is called uncertain. urn problem: when to stop sampling? Then you find the probability that you would have drawn fewer than j white marbles if your urn had a certain proportion, say k, of black marbles. The hypergeometric distribution is implemented in the Wolfram Language as HypergeometricDistribution[N, n, m+n]. Find the expected number of stages needed until there are no more black balls in. It is useful in practice, and many elementary statistics textbooks use it to introduce the binomial and hypergeometric distributions. We have included a number of Discussion Topics designed to promote critical. For more practice, I suggest you work through the review questions at the end of each chapter as well. The topic of statistics is presented as the application of probability to data analysis, not as a cookbook of statistical recipes. Find the probability of the event that the ﬁrst bin contains balls of both colors. In the ﬁrst draw, one ball is picked at random and discarded without noticing its colour. The flippant juror. Introduction Drawing balls from an urn without replacement is a classical paradigm in probability (Feller 1968). Please justify your answers and don't simply give the answer. I recommend studying rst, using previous homeworks, exams and exam practice problems. Here's how. There are in nitely many Urns which we call Urn 0, Urn 1, Urn 2 etc. One ball is picked at random from urn 1 and, without. 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. CAT Probability Questions is a sample set of problems that is asked from this topic in the CAT Probability Section. An urn contains {eq}8 {/eq} white and {eq}6 {/eq} green balls. Probability of second ball being red = 3/9 (because there are 3 red balls left in the urn, out of a total of 9 balls left. Question 1157976: Urn A contains 5 red marbles and 3 white marbles Urn B contains 2 red marbles and 6 white marbles a. falling apart on inspection. The probability that a consumer will see an ad for this particular product is 0. A ball is drawn at random from an urn containing 15 green, 25 black, 16 white balls. What is the probability that the ball you drew is green? (b)You then look at the ball and see that it is green. If a set of 3 balls is randomly selected, what is the probability that each of the balls will be (a) of the same color; (b) of different colors?. (You'd think the urns would be empty by now. A risk, on the other hand, is defined to be a higher probability event, where there is enough information to make. This contains all the class notes for the Introduction to Probability class currently taught at Stanford using these, and other web-resources. Probability Trees: Many probability problems can be simplified by using a device called a probability tree. Combinatorics: The Fine Art of Counting. Bonus Problem: Problem 7. ) are represented as colored balls in an urn or other container. If the outcome is heads, then a ball from urn A is selected, whereas if the outcome is tails, then a ball from urn B is selected. Find P ( B C | A) from the Venn diagram: 3. Problem One. Suppose that this experiment is done and you learn that a white ball was selected. In this post, I’m going to discuss some of the non-geometry problems. Clue: Container in many probability-theory problems. If you know how to manage time then you will surely do great in your exam. JANOS FLESCH, DRIES VERMEULEN AND ANNA ZSELEVA. The locomotive problem. We shall now apply this principle to the solution of several problems. It is used to solve problems of the form: how many ways can one distribute indistinguishable objects into distinguishable bins? We can imagine this as finding the number of ways to drop balls into urns, or equivalently to arrange balls and. The problem of induction is to find a way to avoid this conclusion, despite Hume's argument. The simplest experiment is to reach into the urn and pull out a single ball. Suppose that a machine shop orders 500 bolts from a supplier. Probability is the likelihood or chance of an event occurring. Urn i has exactly i 1 green balls and n i red balls. If the composition is unknown, then it is called. ♦ BALL AND URN (AoPS calls this "Stars and Bars") The classic "Ball and Urn" problem statement is to find the number of ways to distribute N identical balls into 4 distinguishable urns, for example. But our solution to the urn problem relied on random sampling. The Crossword Solver found 21 answers to the Container in many probability theory problems crossword clue. Wolfram Education Portal ». In the last lesson, the notation for conditional probability was used in the statement of Multiplication Rule 2. The Associated Press. Users that wish to investigate especially large or intricate problems are encouraged to modify and streamline the code to suit their individual needs. The theorem is also known as Bayes' law or Bayes' rule. Publication date 1987 Topics Probabilities, Probabilités, Probabilités Publisher Internet Archive Books. k indistinguishable balls are randomly distributed into n urns. Show that this is the same as the probability that the next ball is black for the Polya urn model of Exercise 4. This consists. 3, 794-814, 2012. • Time limit 110 minutes. This paper designs some uncertain urn problems in order to compare probability theory and uncertainty theory. EXAMPLE 1 A Hypergeometric Probability Experiment Problem: Suppose that a researcher goes to a small college with 200 faculty, 12 of which have blood type O-negative. (a) Draw marbles from a bag containing 5 red marbles, 6 blue marbles and 4 green marbles without replacement until you get a blue marble. Once you have decided on your answers click the answers checkboxes to see if you are right. 4 Conditional Probability and Independence 1. Hence the probability of getting a red ball when choosing in urn A is 5/8. You decide to actually count the balls in the urn, that you friend has in problem 6, and discover that there are 6 green, 6 yellow, 7 blue, 4 black, and 2 red balls. Some of these matrices will be used in calculation in subsequent posts. In Problem 3, suppose that the white balls are numbered, and let Y i equal 1 if the i th white ball is selected and 0. The hypergeometric distribution is implemented in the Wolfram Language as HypergeometricDistribution[N, n, m+n]. Bayes Theorem Practice Problems With Solutions Genetics. a) 4 ⁄ 7 b) 3 ⁄ 7 c) 20 ⁄ 41 d) 21 ⁄ 41 View Answer. For example, if you have a bag containing three marbles -- one blue marble and two green marbles -- the. In the first urn, there are $20\%$ red balls, so the probability to draw a red ball is $0. Winning an unfair game. The first limitation is that the ravens we observe in real life aren't randomly sampled from nature's "urn". Thomas Bayes was an English minister and mathematician, and he became famous after his death when a colleague published his solution to the “inverse probability” problem. Ask Question Asked 5 years, 5 months ago. (The two marbles might both be black, or might both be white, or might be of different colors. When we have multiple event probabilities, as in the problem. Now there are 14 balls, 5 white and 9 black. 57 % of the children in Florida own a bicycle. An urn contains 9 red, 7 white and 4 black balls. ) Let N be the number of games played. Probability measures the likelihood of an event occurring. That is, after each draw, the selected ball is returned. balls numbered 1 through 10 are placed in the urn and ball number 1 is withdrawn; at 1 2 minute to 12 P. Each turn, we pick out a ball, note it's colour then replace it and add d more balls of the same colour into the urn. If the probability of an event is 0, it is impossible for that event to occur. Probability (MATH 4733 - 01) Fall 2011 Exam 2 - Practice Problems: Selected answers and hints Due: never Here are some sample problems for you. Among the balls are R red and N-R white balls. These urn models are also excellent practice problems on thinking about Markov chains and deriving the transition probability matrices. Two balls are chosen randomly from an urn containing8 white, 4 black, and 2 orange balls. What is the probability that they come from urns I, II or III? 36. Recommended Problems: Problems 1. Combinatorics: The Fine Art of Counting. The probability of any event can range from 0 to 1. The probability of a sample point is a measure of the likelihood that the sample point will occur. What is the probability that the second card. Again, one ball is drawn at random from the urn, then replaced along with an additional ball of its color. Supposethat we win$2 for each black ball selectedand we lose \$1 for each white ball selected. Depending on whether we sample with or without replacement, the chance (or probability) of getting m red balls (successes) in a sample of n balls changes. Find the probability that both the first and last balls drawn are black. Two of the selected marbles are red, and three are green. There are related clues (shown below). Can You Solve This Intro Probability Problem? An urn contains 10 balls: 4 red and 6 blue. ) are represented as colored balls in an urn or other container like box. In these cases, we will need to use the counting techniques from the chapter 5 to help solve the probability problems. Chapman-Kolmogorov Equations We have already defined the one-step transition probabilities [pic]. Construct an 80% confidence interval for the proportion of red marbles…. Then treat this like a mock exam (i. Euler became professor of physics in 1731, and professor of mathematics in 1733, when. The probability X failing during one year is 0. An urn contains n+m balls, of which n are red and m are black. Lindley The Statistician: Journal of the Royal Statistical Society. Recommended Problems: Problems 1. An urn has 4 green balls, 5 yellow balls, and 6 red balls. Wolfram Education Portal ». In other words, in order to get a new value of seed, multiply the old value by 7621, add 1, and, finally, take the result modulo 9999. One ball is drawn at random. Let B 1 W 2 denote the outcome dial the first bull drawn is B 1 and the second ball drawn is W 2. Hence a probability of 0. Urn problems and how to approach them Probability and real-life situations (lottery, poker, weather forecasts, etc. Practice Problem: A certain lottery has a hat with the numbers 1 through 10 each written on a single scrap of paper. What is the probability the ball drawn from urn C is black? I've figured out that P(K\A) = 2/3 and P(K\B) = 5/6. You can also express this relationship as 1 ÷ 6, 1/6, 0. So, if the probability that they actually had a ﬂat tire is p, then the probability that they both give the same answer is 1 4 (1−p)+p = 1 4 + 3 4 p. In SAS you can use the table distribution to specify the probabilities of selecting each integer in the range [1, c]. Do Problem 2 ONLY please. Urn 2 contains three red balls and four white balls. Two red, three white, and five black.
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# If $A$ and $B$ are sets of real numbers, then $(A \cup B)^{\circ} \supseteq A^ {\circ}\cup B^{\circ}$

I have a proof for this question, but I want to check if I’m right and if I’m wrong, what I am missing.

Definitions you need to know to answers this question: $\epsilon$-neighborhood, interior points and interiors. Notation: $J_{\epsilon}(a)$ means a neighborhood formed around a (i.e. $(a-\epsilon, a+\epsilon)$. An interior point in some set $A$ is a point where an $\epsilon$-neighborhood can be formed within the set. The set of all interior points in $A$ is denoted as $A^0$ and is called the interior.

My proof for the question: I’m proving based on most subset proofs where you prove that if an element is in one set, then it must be in the other. Say $x \in A^0 \cup B^0$. Then there is a $\epsilon > 0$, where $J_{\epsilon}(x) \subseteq A$ or $J_{\epsilon}(x) \subseteq B$. This implies that $J_{\epsilon}(x) \subset A \cup B$ which then implies $x \in (A \cup B)^0$. We can conclude from here that $(A \cup B)^0 \supseteq A^0 \cup B^0$.

#### Solutions Collecting From Web of "If $A$ and $B$ are sets of real numbers, then $(A \cup B)^{\circ} \supseteq A^ {\circ}\cup B^{\circ}$"

Your proof is good- there is no problem with it. I have reworded a bit to make things a little more clear, specifically, where you say:

Then there is $\epsilon>0$, where $J_\epsilon (x)\subseteq A$ or $J_\epsilon (x)\subseteq B$.

Instead it would be more appropriately stated as: If $x\in A^0\cup B^0$ then $x\in A^0$ or $x\in B^0$. Then you can assume $x$ is in $A$, prove that $x\in (A\cup B)^0$, by symmetry the same argument will work if $x\in B^0$.

I have included an edited proof- but as I originally stated your proof is good,
this just more directly reflects the definitions involved.

Let $x\in A^0\cup B^0$. Then $x\in A^0$ or $x\in B^0$. Assume $x\in A^0$. Then there is, by definition, $\epsilon>0$ such that $J_{\epsilon}(x)\subseteq A$. Since $A\subseteq A\cup B$, we also have $J_{\epsilon}(x)\subseteq A\cup B$. Thus by definition we have $x\in (A\cup B)^0$. If instead $x\in B^0$, then by symmetry the same argument works. Thus, $A^0\cup B^0 \subseteq (A\cup B)^0$.

Here is a slightly shorter proof:

We have $A^\circ \subset A \cup B$ and $B^\circ \subset A \cup B$, so we must have
$A^\circ \cup B^\circ \subset A \cup B$.

Since $A^\circ \cup B^\circ$ is open and the interior is the largest open set contained in a set, we must have
$A^\circ \cup B^\circ \subset (A \cup B)^\circ$.


Question

# If $$\tan\dfrac {\theta}{2}=\mathrm{cosec}\theta-\sin\theta$$, then

A
sin2θ2=2sin218o
B
cos2θ+2cosθ+1=0
C
sin2θ2=4sin218o
D
cos2θ+2cosθ1=0

Solution

## The correct options are A $$\sin^2\dfrac {\theta}{2}=2 \sin^2 18^o$$ D $$\cos 2\theta+2 \cos\theta-1=0$$Simplifying, we get  $$\dfrac{\sin \dfrac {\theta}{2}}{\cos \dfrac {\theta}{2}}=\dfrac{1-\sin ^{2}\theta}{\sin \theta}$$ $$\dfrac{\sin \dfrac {\theta}{2}}{\cos \dfrac {\theta}{2}}=\dfrac{\cos ^{2}(\theta)}{2\sin \dfrac {\theta}{2}.\cos \dfrac {\theta}{2}}$$ $$2\sin ^{2}\dfrac {\theta}{2}=\cos ^{2}\theta$$ $$2\sin ^{2}\dfrac {\theta}{2}=(1-2\sin ^{2}\dfrac {\theta}{2})^{2}$$ Let, $$2\sin ^{2}\dfrac {\theta}{2}=t$$$$\implies t=(1-t)^{2}$$ $$\implies t=t^{2}-2t+1$$ $$\implies t^{2}-3t+1=0$$ $$\implies t=\dfrac{3\pm\sqrt{9-4}}{2}$$ $$\implies t=\dfrac{3\pm\sqrt{5}}{2}$$ Now $$|\sin \theta|\leq 1$$ $$\implies t=\dfrac{3-\sqrt{5}}{2}$$ Or  $$\sin ^{2}\dfrac {\theta}{2}=\dfrac{3-\sqrt{5}}{4}$$ $$\implies 2\left(\dfrac{3-\sqrt{5}}{8}\right)=2\sin ^{2}(18^{0})$$ Hence, option A is correct. $$\cos ^{2}\dfrac {\theta}{2}=1-\sin ^{2}\dfrac {\theta}{2}=\dfrac {1+\sqrt{5}}{4}$$  Option D  $$\cos 2\theta +2\cos \theta -1=0$$ $$-2{ \sin }^{ 2 }\theta +2\cos \theta =0$$ $$=-2+2{ \cos }^{ 2 }\theta +2\cos \theta =0$$ $$=-1+{ \cos }^{ 2 }\theta +\cos \theta =0$$ $${ \cos }^{ 2 }\theta =1-\cos \theta =2{ \sin }^{ 2 }\cfrac { \theta }{ 2 }$$ which is true (as proven in the first part)Hence, option A and D are correct.Maths

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# Find missing number from sum of first few natural numbers

A child was asked to add the first few natural numbers $1+2+3+...$ as long as his patience permitted. As he stopped, he gave the sum as $575$. When the teacher declared the result wrong, the child discovered that he had missed a number in the sequence during addition. What was the number he missed?

-
Does the teacher know which number the child was adding up to? –  Michael Albanese Dec 25 '12 at 14:47
lol. I think its a way to frame the question. I too doubt on that same as you. =) –  harish.raj Dec 25 '12 at 14:49
If the teacher doesn't know the number, then my answer is accurate. If the teacher does know the number, the inequality $\frac{n(n+1)}{2} - 575 < n$ should be replaced by $\frac{n(n+1)}{2} - 575 \leq n$ as the reasoning in the bracket following the original inequality no longer applies. However, this does not change the final answer. –  Michael Albanese Dec 25 '12 at 15:12

We have $$1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}.$$ The numbers obtained from such sums are called Triangular numbers.

As the child's answer was $575$, the correct answer must be greater than $575$ (as he missed a number). The first triangular number greater than $575$ is $595$ corresponding to $n = 34$. If this was the sum he was trying to evaluate, the child must have missed $20 = 595 - 575$. However, if the correct answer was the next triangular number, $630$, corresponding to $n = 35$, the only way the child could have obtained $575$ is if he missed $55 = 630 - 575$, but this is not in the list of numbers $1, \dots, 35$ so he could not have been trying to add up the first $35$ natural numbers.

More generally, we need $\frac{n(n+1)}{2} > 575$ (the actual sum needs to be greater than the sum he obtained) and $\frac{n(n+1)}{2} - 575 < n$ (this difference tells us which value he missed, and it can't equal $n$ because then he would have written $1 + \dots + (n - 1)$ which is a triangular number, so the teacher could not have declared it wrong). Rewriting, we must find $n$ such that $$0 < \frac{n(n+1)}{2} - 575 < n\quad (\ast)$$ which only has solution $n = 34$. Therefore, we must be in the situation of the previous paragraph, so he missed the number $20$.

Added later: I just thought I'd add some details here about finding natural numbers $n$ which satisfy $(\ast)$.

First of all, consider the left hand inequality which rearranges to $f(n) := n^2 + n - 1150 > 0$. Solving $f(n) = 0$, we obtain $n = \frac{-1\pm\sqrt{4601}}{2}$; as $n > 0$, we ignore $n = \frac{-1-\sqrt{4601}}{2}$. Note that $\left\lfloor\frac{-1+\sqrt{4601}}{2}\right\rfloor = 33$, and we have $f(33) < 0$ and $f(34) > 0$. So $f(n) < 0$ for $n \in \{1, \dots, 33\}$ and $f(n) > 0$ for $n \geq 34$.

The right hand inequality rearranges to $g(n) := n^2 - n - 1150 < 0$. Solving $g(n) = 0$, we obtain $n = \frac{1\pm\sqrt{4601}}{2}$; as $n > 0$, we ignore $n = \frac{1-\sqrt{4601}}{2}$. Note that $\left\lfloor\frac{1+\sqrt{4601}}{2}\right\rfloor = 34$, and we have $g(34) < 0$ and $g(35) > 0$. So $g(n) < 0$ for $n \in \{1, \dots, 34\}$ and $g(n) > 0$ for $n \geq 35$.

Hence, $n = 34$ is the only natural number which satisfies $(\ast)$.

-
great explanation. Thank you a lot. –  harish.raj Dec 25 '12 at 14:31
+1 Michael: nice explanation! –  amWhy Dec 25 '12 at 14:35

Suppose he tried to sum the first $n$ positive integers. Omitting a term means this sum is less than the sum of the first $n$ positive integers but not less than the sum of the first $(n-1)$ positive integers. Hence, $\frac{n(n-1)}{2}\leq 575<\frac{n(n+1)}{2}$ so that $n=34$. This means the missing term is $595-575=20$.

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awesome. can you elaborate a little about how you got the value of n? (it will be great if you do it in your answer itself) –  harish.raj Dec 25 '12 at 14:15


# Math Help - complex numbers

1. ## complex numbers

Hi all,

I am having trouble getting the answer to:
[(1-sqrt(3)i)/ (sqrt(3) + i)]^7

where i = complex number

ArTiCk

2. Originally Posted by ArTiCK
Hi all,

I am having trouble getting the answer to:
[(1-sqrt(3)i)/ (sqrt(3) + i)]^7

where i = complex number

ArTiCk
My advice is to first convert the numbers into polar form.

3. Originally Posted by mr fantastic
My advice is to first convert the numbers into polar form.
I have already converted the numbers into polar form...
i have it = e^(-7i*pi/3)/ [e^(7i*pi/6)]

The trouble i have is that i tried converting e^(-7i*pi/3) into Cartesian form as well as the denominator... but i am having trouble with the nominator. What i did was:

e^(-7i*pi/3) = cos(-7*pi/3) + i*sin(-7*pi/3) = cos (-2*pi - pi/3) + i*sin(-2*pi -pi/3) = cos(-pi/3) + i*sin(-pi/3)

I think it is there where i have gone wrong, help would be appreciated

Thanks, ArTiCk

4. Hello, ArTiCK!

Simplify: . $\left(\frac{1-i\sqrt{3}}{\sqrt{3}+i}\right)^7$
Ratinalize the "inside" first . . .

. . $\frac{1-i\sqrt{3}}{\sqrt{3} + i}\cdot\frac{\sqrt{3}-i}{\sqrt{3}-i} \;=\;\frac{\sqrt{3} - i - 3i +i^2\sqrt{3}}{3-i^2} \;=\;\frac{\sqrt{3} -4i-\sqrt{3}}{3+1} \;=\;\frac{-4i}{4} \;=\;-i$

So we have: . $(-i)^7 \;=\; i$

5. Originally Posted by ArTiCK
I have already converted the numbers into polar form...
i have it = e^(-7i*pi/3)/ [e^(7i*pi/6)]

The trouble i have is that i tried converting e^(-7i*pi/3) into Cartesian form as well as the denominator... but i am having trouble with the nominator. What i did was:

e^(-7i*pi/3) = cos(-7*pi/3) + i*sin(-7*pi/3) = cos (-2*pi - pi/3) + i*sin(-2*pi -pi/3) = cos(-pi/3) + i*sin(-pi/3)

I think it is there where i have gone wrong, help would be appreciated

Thanks, ArTiCk
$\left(\frac{1-i\sqrt{3}}{\sqrt{3}+i}\right)^7 =
$
$\left[ \frac{2 \, \text{cis} \left( -\frac{\pi}{3}\right) }{2 \, \text{cis} \left( \frac{\pi}{6}\right)} \right]^7 = \left[ \text{cis} \left( -\frac{\pi}{3} - \frac{\pi}{6} \right)\right]^7 = \left[ \text{cis} \left( -\frac{\pi}{2} \right)\right]^7 = (-i)^7$.

6. Hello, Mr. F !

Originally Posted by mr fantastic

$\left(\frac{1-i\sqrt{3}}{\sqrt{3}+i}\right)^7 =
$
$\left[ \frac{2 \, \text{cis} \left( -\frac{\pi}{3}\right) }{2 \, \text{cis} \left( \frac{\pi}{6}\right)} \right]^7 = \left[ \text{cis} \left( -\frac{\pi}{3} - \frac{\pi}{6} \right)\right]^7 = \left[ \text{cis} \left( -\frac{\pi}{2} \right)\right]^7 = (-i)^7$.

Lovely!


# Determine Jordan block size.

The following question is from as System Theory test.

Let the system matrix $A$ be given as $A = \begin{bmatrix} 0&0&0&1\\0&-1&1&3\\0&1&-1&-1\\0&-1&1&2 \end{bmatrix}$

Which has the eigenvalues $\lambda_i=0$ for $i = 1,2,3,4$. Which of the following matrices is the Jordan form of $A$.

$A) \quad J=\begin{bmatrix} 0&1&0&0\\0&0&1&0\\0&0&0&1\\0&0&0&0 \end{bmatrix}$

$B) \quad J=\begin{bmatrix} 0&1&0&0\\0&0&1&0\\0&0&0&0\\0&0&0&0 \end{bmatrix}$

$C) \quad J=\begin{bmatrix} 0&1&0&0\\0&0&0&0\\0&0&0&1\\0&0&0&0 \end{bmatrix}$

$D) \quad J=\begin{bmatrix} 0&1&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0 \end{bmatrix}$

My approach:

The eigenvalues are given so the nullspace of $\left[A-\lambda I\right]$ can be found to be $\begin{bmatrix} 0\\1\\1\\0 \end{bmatrix}$ and $\begin{bmatrix} 1\\0\\0\\0 \end{bmatrix}$.

The dimension of this nullspace is $2$ which indicates that there are 2 Jordan blocks. So answer $B$ or $C$.

Also I've found that : $\operatorname{rank}(\ker(A-\lambda I)^2)=3$ and $\operatorname{rank}(\ker(A-\lambda I))=2$. Subtracting these outcomes gives $3-2=1$. So there's one Jordan block of size 2 or larger. And thus only answer $B$ is correct.

Right or wrong? thanks in advance.

• It's perfect for me. – Bernard Aug 9 '18 at 13:10

Your argument is correct, but your notation is innapropriate. Where you wrote $\operatorname{rank}(\ker\cdots)$, you should have written $\dim(\ker\cdots)$.
I like to find the actual Jordan form, especially including the matrices that give $R^{-1}A R = J.$ It is not bad when the eigenvalues are integers, and makes concepts concrete. This begins with choosing a column vector i called $w$ such that $A^2 w \neq 0.$ That becomes the far right column.
$$\frac{1}{8} \; \left( \begin{array}{cccc} 8&-4&-4&0\\ 0&1&3&0\\ 0&2&-2&0\\ 0&-4&4&8\\ \end{array} \right) \left( \begin{array}{cccc} 0&0&0&1\\ 0&-1&1&3\\ 0&1&-1&-1\\ 0&-1&1&2\\ \end{array} \right) \left( \begin{array}{cccc} 1&2&1&0\\ 0&2&3&0\\ 0&2&-1&0\\ 0&0&2&1\\ \end{array} \right) = \left( \begin{array}{cccc} 0&0&0&0\\ 0&0&1&0\\ 0&0&0&1\\ 0&0&0&0\\ \end{array} \right)$$


# Trying to prove that a group is not Cyclic

Given the following Euler groups : \begin{align*} U_{12} &= \{1,5,7,11\}\\ U_{16} &= \{1,3,5,7,9,11,13,15\} \end{align*}

I want to prove that they are not cyclic. I used the following theorem :

A group of order $n$ is cyclic if and only if it has an element of order $n$.

Let's take for example $U_{12}$: (I will use the notation of $o(x)$ to denote the order of the element $x\in G$)
\begin{align*} o(5)&\colon 5^2=25 \to 25\bmod 12 = 1\to o(5)=2\\ o(7)&\colon 7^2= 49 \to 49\bmod 12 = 1 \to o(7)=2\\ o(11) &\colon 11^2 = 121 \to 121\bmod 12=1 \to o(11)=2 \end{align*}

Then by using the above theorem , this group is indeed not a cyclic group.

Question : do I really have to check each element in the group for its order ?

Regards

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Fixed. thanks ! – ron Mar 9 '12 at 11:58
BTW, the structure of the multiplicative group of integers modulo n $U_n$ is will known. In particular, it is cyclic iff $n$ is 2, 4, any power of an odd prime or twice any power of an odd prime. – lhf Mar 9 '12 at 12:08
Please learn some basic $\LaTeX$ instead of waiting for others to pretty up your posts for you. Thank you. – Arturo Magidin Mar 9 '12 at 16:45
Comment: Stop using $\to$ to mean "and since" or some other connectives. The arrow has specific meanings in mathematics; students who use $\to$ and $=$ as general purpose connectives to mean something like "and then I did some thinking and this is what I came up with" drive professors crazy. – Arturo Magidin Mar 9 '12 at 16:49
@Arturo, I don't think driving professors crazy is the real problem - for most of us, crazy is within walking distance, anyway - I think the real problem is the student who writes that way is pretty much guaranteed to get things wrong. – Gerry Myerson Mar 10 '12 at 6:58

Well really the "theorem" is the definition of cyclic group...it is a group generated by one element.

Now if I give you a group and you check that all but one element isn't a generator, how do you know that the one you didn't check isn't a generator?

So yes, you must check ALL elements.

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... or you can use interesting theorems. For example, there's an easy upper bound on the number of elements of order 2.... – Hurkyl Mar 9 '12 at 14:28

Some shortcuts are available. For example, when testing $3$ in $U_{16}$, you find its powers are $3,9,11,1$, so you don't have to test $9$ or $11$. Do you see why?

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I didn't get your meaning .What do you mean by "its powers" ? 3^3 = 27 --> 27 modulo 16= 11 .... so I generated using 3 , the element 11 . What does it mean ? – ron Mar 9 '12 at 12:11
You're trying to decide whether the order of $3$ is $8$. So you look at $3$, then at $3\times3=9$, then at $3\times3\times3=3\times9=27=11$, then at $3\times3\times3\times3=3\times11=33=1$. Those are the powers of $3$, the numbers $3^r$, $r=1,2,3,\dots$. Using $3$, you generate $3$, $9$, $11$, and $1$. – Gerry Myerson Mar 9 '12 at 12:34
Okay , so now we know that the order of |3|=4 . Why does it help me ? now I won't need to check the generated elements using 3 ? – ron Mar 9 '12 at 12:42
"Why does it help me ?" This is an exercise for the reader -- use knowledge about the order of $3$ in order to deduce something about the order of $3^2$.... – Hurkyl Mar 9 '12 at 14:26
You should see if you can prove that if $g$ doesn't generate $G$ and $h$ is in the subgroup generated by $g$ then $h$ doesn't generate $G$ either. – Gerry Myerson Mar 10 '12 at 6:59

As I mentioned yesterday in reply to a question of yours: a cyclic group has at most one element of order $2$.

In your first two lines, you've shown that $5$ and $7$ both have order $2$ in $U_{12}$. Therefore, the group cannot be cyclic.

Similarly, since $9^2\equiv 1\pmod{16}$ and $15^2\equiv 1\pmod{16}$, $U_{16}$ has at least two distinct elements of order $2$, and therefore cannot be cyclic.

In general, a cyclic group of order $n$ has exactly $\phi(d)$ elements of order $d$ for each divisor $d$ of $n$.

-

What of if your set is infinite? I think the best idea is to find the general form of the elements in the set (if possible) and pick arbitarary element, check your assertion and then generalised. But in your case, since the elements are finite, you have to check it for all the elements in the set.

-


# Average Case Analysis for finding max and min value on an array

Given the following algorithm, to find the maximum and minimum values of an array - don't mind the language:

MaxMin(A[1..n])
max = A[1];
min = A[1];
for (i = 2; i<=n; i++)
if (A[i] > max)
max = A[i];
else if (A[i] < min)
min = A[i];
print(max, min);

I need to do a probabilistic analysis for the average case of comparisons that will be made on its execution.

So far, my solution is:

Given an indicator random variable: $$X_i = \begin{cases} \text{1, if max > A[i]}\\ \text{0, if max < A[i]}\\ \end{cases}$$

and assuming an uniform distribution for $A[1..n]$, the expected probability is: $$E[x] = \sum\limits_{i=1}^{n} Pr(X_i)$$

where $$Pr(X_i)$$ is the probability of the i-th element be the $max$ element in $A[1..n]$. It's possible to determine that: $$Pr(x_1) = 1, Pr(x_2) = 1/2, Pr(x_3) = 1/3 ,..., Pr(x_n) = 1/n$$ and thus for an array of size $n$ the expected value can be calculated as:

$$E[x] = 1 + 1/2 + 1/3 + ... + 1/n = \sum\limits_{i=1}^{n}{\frac{1}{i}} \approx \log{n}$$

And the same goes for $min$, which will give the same result $\log{n}$. (1)

My questions are: is it correct? Does the complexity for the average case of the given algorithm is $\theta(\log{n})$? Can I use the argument pointed by (1), just modifying $X_i$ so: $$X_i = \begin{cases} \text{1, if min < A[i]}\\ \text{0, if min > A[i]}\\ \end{cases}$$

I already read this (unfortunately the link to the video isn't available), but it only explains for $max$ and my analysis must be for both $max$ and $min$.

• What do you mean "average case of comparisons"? Do you mean the average quantity of comparisons? That can't possibly be less than $n-1$, so $\log n$ is wrong. Perhaps you should edit the question to clarify? – Wildcard Mar 27 '18 at 1:46
• Of course your analysis assumes the array is in random order. Many times arrays are not in random order. Like sorted arrays. – gnasher729 Mar 29 at 19:59

First, note that the first comparison will always compare $$n-1$$ times, independent of the distribution of the input. So, what you really want to know is how many times the second part of the if compares. For this, you can use a indicator random variable like this:

$$X_i = \begin{cases} \text{1, if A[i] \le \max}\\ \text{0, if A[i] > \max}\\ \end{cases}$$

So, just like you did, assuming an uniform distribution for $$A[1..n]$$, the expected probability is:

$$E[x] = \sum\limits_{i=1}^{n} Pr(X_i)$$

But we do not want $$Pr(i=\max)$$, we want $$\overline{Pr(i=\max)}$$, like we stated before. So, using your probability of the $$i-th$$ element to be the $$max$$ element:

$$Pr(i = \max) = 1/i$$

We have this complement:

$$\overline{Pr(i=\max)} = (1 - 1/i)$$

From which we can put in the expected probability:

$$E[x] = \sum\limits_{i=2}^{n} (1 - 1/i) = \sum\limits_{i=2}^{n}1 - \sum\limits_{i=2}^{n}1/i = n - 1 - (\ln|n| - 1 + \Theta(1))$$

So, to get the total quantity of comparisons, we can just sum the number of comparisons of the two parts. Note that the $$\Theta(1)$$ can absorb a constant values.

$$(n - 1) + n - 1 - \ln|n| + 1 + \Theta(1) = 2n - 1 - \ln|n| + \Theta(1)$$

And we get the expected total number of comparions:

$$2n - \ln|n| + \Theta(1)$$

(1) You are looking for either min or max. Then,

(1-1) is it correct? Yes, it is correct.

(1-2) Does the complexity for the average case of the given algorithm is $\Theta(logn)$? Indeed it is. My experience in practice is that it does not work that way (or at least the is a large constant involved).

(1-3) Can I use the argument pointed by (1), just modifying $X_i$? Yes, you can do that too.

(2) You are looking for both min and max. Then, the scenario is different. You should have something like: $$Z_i = \begin{cases} \text{1, if max > A[i] and min < A[i]}\\ \text{0, if max < A[i] and min > A[i]}\\ \end{cases}$$

Then you can calculate $E[Z]$. It is also possible to calculate $E[XY]$, where $X$ is for max, and $Y$ is for min. Try different techniques and see which one works easier.

• In my case, I'm interested on your second explanation (for $Z_i$). Really liked your explanation, but @LionsWrath explanation tackled my question right on the spot! Tough, thank you for your explanation :) – woz Mar 27 '18 at 11:09


# Prove: $\text{ord}(a^m) = \frac{\text{lcm}(m,n)}{m}$ [duplicate]

This is Pinter $$10.G.5$$

Let:

$$a \in G$$

$$\text{ord}(a) = n$$

Prove: $$\text{ord}(a^m) = \frac{\text{lcm}(m,n)}{m}$$

Use $$10.G.3$$ and $$10.G.4$$ to prove this.

Here is $$10.G.3$$:

Let $$l$$ be the least common multiple of $$m$$ and $$n$$. Let $$l/m = k$$. Explain why $$(a^m)^k = e$$.

Here is $$10.G.4$$:

Prove: If $$(a^m)^t = e$$, then $$n$$ is a factor of $$mt$$. (Thus, $$mt$$ is a common multiple of $$m$$ and $$n$$.)

Conclude that: $$l = mk \leq mt$$

OK, let's begin.

By $$10.G.3$$:

$$(a^m)^{\text{lcm}(m,n)/m} = e \tag{5}$$

If $$\text{lcm(m,n)}/m$$ is the lowest number such that (5) is true, then:

$$\text{ord}(a^m) = \text{lcm}(m,n)/m \tag{9}$$

Let's assume that there is a number

$$q < \text{lcm}(m,n)/m \tag{7}$$

such that:

$$(a^m)^q = e \tag{6}$$

By $$10.G.4$$ with (6):

$$n \ \big|\ mq$$

$$l \lt mq$$

$$\text{lcm}(m,n) \leq mq$$

Isolate q:

$$\text{lcm}(m,n)/m \leq q \tag{8}$$

So (9) must be true.

That's how I approached it. If anyone notices any issues, I'd be happy to know about them.

Even if it is considered correct, do you feel there is a better way?

• You should use \le rather than <=. – Angina Seng Aug 13 '19 at 2:52
• @LordSharktheUnknown Thanks! Updated! – dharmatech Aug 13 '19 at 2:54
• @JyrkiLahtonen Can you explain how this is a duplicate of the question you linked to? There is no requirement in this question that G be cyclic. Whereas the question you linked to is for cyclic groups. – dharmatech Aug 13 '19 at 8:49
• @José Carlos Santos Do you really think that it's a duplicate? – Michael Rozenberg Aug 13 '19 at 9:06
• @YuiToCheng Once again not a proper dupe target becauyse - among other things - the OP is working with LCMs not GCDs, and is working from specific lemmas. Please be more careful in choosing proper dupe targets. – Bill Dubuque Aug 14 '19 at 13:32

Yes, $$\,(a^{m})^{k} = 1\!\! \overset{(1)\!\!}\iff n\mid mk \iff m,n\mid mk\!\! \overset{(2)\!\!}\iff {\rm lcm}(m,n)\mid mk\iff {\large{\frac{{\rm lcm}(m,n)}m\,\mid}}\, k$$
$$\!(1)$$ and the (omitted) final conclusion is by Corollary' here, and $$(2)$$ is the LCM Universal Property
• @BillDubuque Do you really think this question has not been handled on our site? The OP's complaint about my choice of dupe was mostly because "their version is not only about cyclic groups". Yet they only work within the cyclic group generated by $a$. – Jyrki Lahtonen Aug 14 '19 at 6:18


# Open and closed implication formula

So I am a newbie in mathematical logic and one of the problems I have faced throughout this one week of study is the one concerning open or closed statement. Let me know if I have used some terms such as "statement", "sentence", "formula", etc in this question the wrong way.

I posted probably a related question before here: Connection between universal Quantifier and implication, and have stumbled upon new terms such as "open", "closed", "free", and "bound". I still need some confirmation about this but I state a different question.

So, am I right to say that $\forall x[x\in \mathbb{R}\implies x^2\geq0]$ is a closed formula since I know it has a no free variable?

Does every closed formula have a truth value?

I know that $x\in \mathbb{R}\implies x^2\geq0$ is a formula and is true for every assignment of $x$, or IS it? Why can't I just conclude that $x\in \mathbb{R}\implies x^2\geq0$ is just true and why should I consider its hidden universal quantifier (which, of course, it does not sound the same using existential quantifier)? Is it just for the sake of the existing rule of making sentence in FOL?

Probably the same question: Why can't all quantifiers be bounded quantifiers, and be written that way, considering the existence of domain of discourse? Why don't we explicitly write that domain in the sentence? Meaning, instead of saying "In the domain of natural numbers, $\forall x[x\geq0]$", why not simply $"\forall x\in \mathbb{N}[x\geq0]"$?

Hope my confusion is understood as a newbie. Thanks for the help! :D

• Correct; the formula $∀x[x∈R ⟹ x^2≥0]$ is a closed formula (or sentence) because it has no free occurrences of variables. Jun 9 '18 at 9:16
• Yes; a closed formula has a definite truth value in an interpretation. $\forall x (x=0)$ is FALSE in $\mathbb N$, while an open one, like e.g. $(x=0)$ may change truth value (for a specific interpretation) according to the value assigned to the free variable $x$. Jun 9 '18 at 9:18
• You can see Classical Logic and Model Theory and Logical Truth and Logical Constants for an introduction. Jun 9 '18 at 9:43
• The last point is that usually the "pure" presentation of first-order logic has no predicate constant, like $\in$. Thus the way to formalize $∀x \in \mathbb N [x≥0]$ is $∀x[N(x) \to x≥0]$ where $N(x)$ is a unary predicate. Now again, the truth-value of the sentence depends on the way to interpret $N(x)$. Jun 9 '18 at 10:04
• Thus, from the "pedagogical" point of view, we have reduced restricted quantification to quantification of a conditional: thus, in any case, we have to first learn how to manage quantifiers. Jun 9 '18 at 10:11

The "founding" idea is that logic must be formal.

This idea was firstly investigated by Aristotle : the modern way to investigate the concept "formal" is to define it through syntax :

a formula [a grammatically correct expression] is an expression built-up according to specific rules.

Then we have semantics :

the way to give meaning (and truth value) to an expression through an interpretation.

The interaction of syntax and semantics is the key-point of modern formal logic.

You can see : John MacFarlane, WHAT DOES IT MEAN TO SAY THAT LOGIC IS FORMAL (2000):

What does it mean, then, to say that logic is distinctively formal?

Three things: logic is said to be formal (or “topic-neutral”)

(1) in the sense that it provides constitutive norms for thought as such,

(2) in the sense that it is indifferent to the particular identities of objects, and

(3) in the sense that it abstracts entirely from the semantic content of thought.

Thus, "topic neutral" and "to abstract entirely from the semantic content" mean to separate semantics from syntax : the domain of interpretation is obviously semantical.

• I get it now. The idea is to formalize that sentence formed with bounded quantifier. So, my last point, how do you think about introducing
– bms
Jun 9 '18 at 13:17
• get it now. The idea is to formalize that sentence formed with bounded quantifier. So, my last point, what is a reasonably good way to introduce my student the way of proving an implication? Do you think it is fine to assume (with careful context reading) that there are always hidden universal quantifiers before the implication statement that bound the variables in the hypothesis, and tell them that? Hence, they will always formally start with "Take an arbitrary x. If p(x) is true, then.... Thus q(x)". I ask this since my usual understanding is
– bms
Jun 9 '18 at 13:24
• that "for all" sentence is equivalent to "If, then" sentence because the way we prove them is essentially the same.
– bms
Jun 9 '18 at 13:26


# Why are the probability of rolling the same number twice and the probability of rolling pairs different?

Two scenarios:
1. Using one die, roll a 6 twice.
$\frac16\times\frac16=\frac1{36}$

1. Rolling two dice roll the same number (a pair).
$\frac6{36}=\frac16$

Why are these two probabilities different? Because the events are independent, isn't rolling a pair the same as rolling a die twice?

In a sense, rolling two dice at once is the same as rolling 1 die twice at the same time? How does this "timing" issue affect the probability?

In the second case, your pair can be any one of $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)$ and you'd satisfy "obtaining a pair".

That gives you six possible pairs, each of which one has probability of occurring $\dfrac 1{36}$ gives us $$6\times \frac 1{36} = \frac 16$$

Now, if you want to know what the probability of rolling two dice simultaneously and obtaining two sixes (one prespecified pair of the six possible pairs), that would be $\dfrac 1{6\cdot 6} = \dfrac 1{36}$.

With this distinction made, yes, the probability of obtaining two sixes when rolling one die twice, and the probability of rolling two sixes simultaneously are equal.

• I understand how the sample space is defined. But isn't it the same thing as rolling two dice back to back? it just happens that in the two dice sample, the rolls are happening together. Whereas in the one die example, the rolls happen with a small delay in between. – William Falcon Mar 8 '15 at 18:28
• You specified the probability of rolling two sixes. On the other hand, the probability of rolling consecutively (one roll, then another), and obtaining the same number twice is $1/6$. The first roll can be any number (probability of rolling a number $1-6$ is equal to $1$, and then the probability of rolling that particular number on the second roll is $\frac 16$ for overall probability of rolling the same number twice being $1/6$. – amWhy Mar 8 '15 at 18:32

Note that you are asking two different questions. In the first problem, you are asking the probability of rolling a particular number twice, while in the second problem, you are asking the probability of rolling one of the numbers twice.

The probability of rolling a $6$ twice is $$\frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}$$ but so is the probability of rolling a $1$ twice, a $2$ twice, a $3$ twice, a $4$ twice, or a $5$ twice. Hence, the probability of rolling the same number twice is $$6 \cdot \frac{1}{36} = \frac{1}{6}$$

I think the simplest answer for the second, pairing, scenario is:

Your first rolling does not have any possibility expected, which means you will have any numbers. However, your second roll should be pairing the first number and the possibility should be 1/6.

so to roll a pair, the odds is 1 x 1/6 = 1/6


# Closed Bounded but not compact Subset of a Normed Vector Space

Consider $\ell^\infty$ the vector space of real bounded sequences endowed with the sup norm, that is $||x|| = \sup_n |x_n|$ where $x = (x_n)_{n \in \Bbb N}$.

Prove that $B'(0,1) = \{x \in l^\infty : ||x|| \le 1\}$ is not compact.

Now, we are given a hint that we can use the equivalence of sequential compactness and compactness without proof.

However, I don't understand how sequences of sequences work? Do I need to find a set of sequences and order them such that they do not converge to the same sequence?

Does the sequence $(y_n)$ where $y_n$ is the sequence such that $y_n = 0$ at all but the nth term where $y_n =1$ satisfy the requirement that it does not have a convergent subsequence?

I think I have probably just confused myself with this sequence of sequences lark. Sorry and thanks.

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You have to find a sequence in $B'(0,1)$ which does not contain a convergent subsequence. Because the existenc of such a sequence implies (right from the definition) that $B'(0,1)$ isn't sequentially compact.

Hint Consider the sequence $(y^n)_n$ defined by $$y^n_k := \begin{cases} 1 & n=k \\0 & n \not= k \end{cases} \qquad (k \in \mathbb{N})$$ Then $y^n \in B'(0,1)$ for all $n \in \mathbb{N}$ and $\|y^n-y^m\|_{\infty}=1$ for $m \not= n$. This means that $(y^n)_n$ doesn't contain a Cauchy-subsequence, hence in particular no convergent subsequence.

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Let $E = \{ e_n \}_n$ where $e_n$ is the vector with $1$ in the $n$th position and zero everywhere else. Clearly $E \subset \overline{B}(0,1)$, hence it is bounded. Since $\|e_n - e_m\| = 1$ whenever $n \neq m$, it is clear that each point in $E$ is isolated. Hence $E$ is closed.

Now let $U_n = B(e_n, \frac{1}{2}$). Then $\{ U_n\}_n$ is an open cover of $E$ that had no finite subcover. Hence $E$ is not compact.

If you would rather use a sequential argument, choose the sequence $x_n =e_n$. As above, we have $\|x_n -x_m\| = 1$ whenever $n \neq m$, hence no subsequence can be Cauchy. Hence no subsequence can converge, hence $E$ is not sequentially compact.

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These are very intriguing examples but what condition of the Heine-Borel Theorem is absent? B-T says closed and bounded is equivalent to compact in at least all n dimensional real or complex sets. What specifically is missing in these examples for them to not contradict Heine-Borel?

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In answer to my own question, my first three attempts to state the Heine-Borel Theorem in n dimensional real space. Certainly a complete metric space would work! Can anyone soften the topological requirements of the Theorem?

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