# Tag Info

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Let $p \equiv P(S)$. The probability of all $n$ draws satisfying the condition is, in fact, $p^n$. The probability of exactly $k$ out of the $n$ draws satisfying the condition is given by $$\binom n k p^k(1-p)^{n-k}$$ In particular, the probability of exactly $1$ success is given by $$\binom n 1 p(1-p)^{n-1}$$ Finally the probability of at least one ...

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In over 20 years of applied work, this has never been a problem. As you put it, the amount of error is too small to be of concern. I can't imagine ever needing more than 6 decimals of accuracy in a summary statistic, and for most work 2 decimals are sufficient. The modeling error and noise in real-world problems far exceeds any issues in using rational ...

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I am a bit confused. Your example with $m = 2$ does not support your claim that the mean of the sequence exceeds $m/2 = 1$, since $4/7 < 1$. Furthermore, I should point out that under the assumption that the sequence is uniformly distributed, the mean would be $(m-1)/2$, not $m/2$, because no element of the sequence can attain $m$; all values are in ...

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If the major prize is given to one of $1000$ people uniformly randomly, then the probability of any one person receiving the prize is, as you correctly stated, $$\frac{1}{1000}$$ If the major prize is given to one person of the $998$ remaining people randomly after two other prizes are given randomly, then the probability of any one of the original $1000$ ...

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No. All this does is ensure that the person who won the first minor prize and the person who won the second minor prize cannot win the major prize.

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Let $X_i$ be standard $U(0,1)$ (translation/scaling doesn't affect the correlation coefficient). Use $\sigma_M=\sigma_m$ (symmetry) and $$E(m\mid M)=\frac M n\Rightarrow E(m)=\frac {E(M)}n\,\&\,\,E(Mm)=\frac {E(M^2)}n=\frac {E(M)^2+\sigma_M^2}{n}$$ to get$$\rho_{1,n}=\frac{E(Mm)-E(M)E(m)}{\sigma_M\sigma_m}=\frac{\frac ... 22 "No", in a particularly strong sense: The very reason limits came into mathematics was to extract numerical values from difference quotients$$ \frac{f(x) - f(x_{0})}{x - x_{0}} $$(which are, of course, algebraically indeterminate at x_{0}), in the limit as x \to x_{0}. The formal definition of L = \lim(f, x_{0}) ("For every \varepsilon > 0, ... 1 \lim_{x\to x_0} f(x)=L for a f:D\to\mathbb R is defined as For all sequences (a_n) with \lim_{n\to\infty} a_n = x_0 we find \lim_{n\to\infty} f(a_n) = L. To guarantee that L is uniquely defined we need at least one sequence (a_n) of the domain such as \lim_{n\to\infty} a_n = x_0. Therefore x_0 need to be an accumulation point of the ... 5 This is not at all necessary. For example:$$ \lim_{x\to 0} \frac {x^2}x =0, $$but \left.\dfrac{x^2}{x}\right\vert_{x=0} is undefined. If we do have that f is defined at a and \lim\limits_{x\to a}f(x)=f(a), then f is called continuous at x=a. 0 Think of the color options as 4 slots with "nothing" being a possible subtype. Selecting that option in, say, the blue slot would simply mean that there was no blue bonus attached to that card. By the specified terms you could choose "nothing" in all 4 slots or in any subset. Viewed this way, there are 21 possible subtypes in each of the four slots ... 0 You have 3 possibility : you are looking the First card's side 1, you are looking the first card's side 2 or you are looking the third card's side red. Sot the probability that the second side is red is \frac 2 3 1 Let B be the event that you selected the card with different sides, and R that you selected the red card. You are calculating$$P(\text{down red} | \text{up red})=P(B|\text{red up})P(\text{down red}|B)+P(R|\text{red up})P(\text{down red}|R)=\frac{P(B\cap\text{red up})}{P(\text{red up}}\cdot 1+\frac{P(R\cap\text{red up})}{P(\text{red up}}\cdot ...

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The first card has sides $R_1$ and $R_2$, two red sides. The second has $B_1$ and $B_2$ (two black sides), and the last has $R_3$ and $B_3$. You see either $R_1$, $R_2$ or $R_3$ and all are equally likely. But $R_1$ has $R_2$ on the other side, $R_2$ has $R_1$ on the other side, and $R_3$ has $B_3$ on the other side. So the other side is red with chance ...

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Say a $k$-long list is needed. (!) Start by generating a random list of $k$ elements, with duplicates, shorten it by deleting the duplicates; being left with a list of length $p$. Then, iteratively repeat step (!) generate lists of the size $k-p$ from the remaining candidate elements ($k-p$ replaces the $k$ from the previous step). Iterate until a step ...

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Strictly speaking no, if it's an uniform distribution over an interval of $\mathbb R$ it's almost certain that the outcome could not be encoded (with whatever coding method you've choosen). But if you as you say settle for "up to specified precision" then you've a situation where inclusion of irrationals doesn't make any difference. You just use a standard ...

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Under the hypothesis that $\pi$ is a normal number, you can shift the decimal expression of $\pi$: $$a_0=\pi, \qquad a_{n+1} = \operatorname{frac}(10a_n)$$ Instead of $\pi$, you can also use the Champernowne constant, which is known to be normal.

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pick a random rational number in (0,1) with the desired resolution. pick your favourite trigonometric function and your favourite range for the arguments. scale the rational to the range (such that (0,1) scales to your chosen range). (!) compute the (transcendental) result of applying the function to the scaled argument Accept the result with a ...

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Let $V=\{1,\dots,n\}.$ A vertex $v\in V$ is a sink if and only if none of the edges $uv$ exists where $1\le u\lt v.$ Let $q=1-p,$ the probability that an edge $uv$ does not exist. By symmetry, we have $$\text{expected value of number of sources}$$ $$=\text{expected value of number of sinks}$$ $$=\sum_{v=1}^n\text{probability}(v\text{ is a sink})$$ ...

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It is easy to write down a group with the desired properties; and, you should be able to see how to generalise this to provide infinitely many more examples. Take $G = C_2^2\times C_3$, and $P = C_2$ and $N = C_3$. The Sylow $2$-subgroup of $G$ is $C_2^2$, $P$ is a non-Sylow $2$-subgroup, and $N$ - with order coprime to $2$ - is certainly normal in $G$ ...

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In Excel 2003 Microsoft attempted to use the Wichmann-Hill generator. Their implementation was incorrect. They tried to fix it in Excel 2007 and again they did not get it right. The period of the generator, which is what you are asking about, has an unknown length. I hope the newer versions are better, but I wouldn't bet on it. You can read more detail ...

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Your friend is correct. The term random in common usage includes events that have different probabilities of occurring. Situations where all events have equal probabilities are just special cases. They are of interest, but random includes much more.

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Method 1 hint. $\Bbb E[W_t^n]=t^{n/2}\Bbb E[Z^n]$, where $Z$ is standard normal. You probably already know $\Bbb E[Z^n]$ for $n=1,2,3,4$. Method 2 hint. $(W_t^3-3tW_t)_{t\ge 0}$ is a martingale. (Why?) Likewise for the second equality.

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Initially there are $k$ seen balls which have been drawn from the bin, $d-k$ unseen balls in the bin, and $n-d$ unseen balls which never reached the bin. There is a probability $\frac{k}{n}$ of drawing from the $k$ seen balls and they are all equally likely so the probability of each one is $\frac{1}{k}\times \frac{k}{n}=\frac{1}{n}$. There is also a ...

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Yep, you could easily reconstruct a graph of either integral or derivative of the function. You could apply a numerical integration or differentian methods to obtain these graphs. Based on request to provide some examples. I will use a rather simple function $f(x) = x^2$ Suppose you would like to draw a graph of its dirivative on the following interval ...

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