# Tag Info

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Community wiki answer so the question can be marked as answered: As user76844 pointed out in a comment, the probability is $1$ since the $3$ people already seated cannot block all opportunities for $3$ people to sit next to each other.

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This can be solved using inclusion-exclusion. We have $n$ conditions for the $n$ couples sitting next to each other. If one of the conditions is fulfilled, we can regard the corresponding couple as a single person to be seated and include a factor $2$ for the two orders in which the couple can sit. So there are $2^k(2n-k-1)!$ different arrangements that ...

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For a sequence of real (or complex) numbers, bounded is the correct term. For a sequence of functions, the notions of bounded and uniformly bounded are distinct. Each individual function may be bounded without the sequence being uniformly bounded.

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You're calculating as if the three events whose probabilities you're multiplying were independent, but they're far from independent. The probability for a given edge to exist and be isolated is $p(1-p)^{2(n-2)}$, where the factor of $p$ is for the edge to exist and the $2(n-2)$ factors of $1-p$ are for the remaining $2(n-2)$ edges of the endpoints of the ...

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I think the correct sum is $$P(1)+P(3)+P(5)+\cdots=\sum_{k=0}^\infty 2\left(\frac{1}{3}\right)^{2k+1}=\frac{2}{3} \sum_{k=0}^\infty\left(\frac{1}{9}\right)^k=\frac{3}{4}~~,$$ because in that way you sum in the odd numbers only. It is not useful to think you are selecting a number "at random" from $S$ because you have the probabilities explicitly: they are ...

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By definition $$\mathbb{E}[Y^p] = \int_0^{M} y^p f_Y(y)dy\ ,$$ where $f_Y(y)$ is the pdf of $Y$, and thus equal to $f_Y(y)=\frac{d}{dy}\mathbb{P}[Y<y]=-\frac{d}{dy}\mathbb{P}[Y>y]$. Hence $$\mathbb{E}[Y^p] = -\int_0^{M} y^p \frac{d}{dy}\mathbb{P}[Y>y]dy\ ,$$ and using integration by parts $$\mathbb{E}[Y^p] = -\left[y^p ... 0 First of all you need for the probability density function f_X(x) \geq 0 and for it to integrate to 1.$$\int_{-\infty}^{\infty}f_X(x)\:\text{dx} = \int_{x \in D}{}f_X(x)\:\text{dx} = 1$$Then for the moments$$E[X^n] = \int_{-\infty}^{\infty}x^n f_X(x)\:\text{dx} = \int_{x \in D}{}x^nf_X(x)\:\text{dx} $$and since you used the moment-generating ... 0 As far as I am aware, the junction tree algorithm can be used as an approximation even in the case that the graphical model is not acyclic, since it creates a factor graph representation of the original model on which belief propagation can be performed (even though the results are only exact if the original model was itself acyclic). This is called ... 0 If X_1, X_2, ..., X_n are independent events then$$\Pr(\bigcap_{i=1}^n X_i) = \prod_{i=1}^{n} Pr(X_i)$$If the events are merely pairwise independent or threewise independent or not any kind of independent at all, use the chain rule of probability:$$\Pr(\bigcap_{i=1}^n X_i) = Pr(X_n | \bigcap_{i=1}^{n-1} X_i) Pr(X_{n-1} | \bigcap_{i=1}^{n-2} X_i) ... ...

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$Cor(X,Y)=\dfrac{E(XY)-E(X)E(Y)}{\sqrt{(E(X^2)-(E(X))^2)(E(Y^2)-(E(Y))^2)}}$ Now plug in the values.

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Proposition: Let $a,b,c,d$ be positive real numbers such that $ad-bc<0$. Then $$\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}$$ First, notice that the requirement is equivalent to $a/b<c/d$, so we can use either as the supposition in the proposition. $$ad-bc=ad+ac-(ab+bc)<0\Rightarrow\frac{a}{b}<\frac{a+c}{b+d}$$ by dividing both sides by ...

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$\dfrac{\alpha+y}{\alpha+n+\beta}=\dfrac{\alpha+\beta}{\alpha+n+\beta}\cdot\dfrac{\alpha}{\alpha+\beta}+\dfrac{n}{\alpha+n+\beta}\cdot\dfrac{y}{n}$ Take $\gamma=\dfrac{n}{\alpha+n+\beta}\in(0,1)$ then $\dfrac{\alpha+y}{\alpha+n+\beta}=(1-\gamma)\dfrac{\alpha}{\alpha+\beta}+\gamma\dfrac{y}{n}$. It is a convex combination of $\dfrac{\alpha}{\alpha+\beta}$ ...

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$\lim_{x \to \infty} 2e^{-2x} = 0$ And we have $$P(X>1) = \int_1^{\infty} 2e^{-2x} = \frac{1}{e^2}$$ You must have missed something when you did the integration Finally $$\int_0^{\infty} 2e^{-2x} = 1$$

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If I am not mistaken, $\left\{Y=X \right\}$ and $\left\{Y \neq X\right\}$ partition the space. So $\left\{X=m \right\} \subset \left\{ Y=m \right\} \cup \left\{Y\neq X\right\}$ and the symmetric statement (change the roles of $X$ and $Y$) give your inequality.

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I saw the question and answer from one of the website about the b) from https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.408149.html Note: The probability of x successes in n trials is: In this case p = .05 & q = .95 a) P(exactly 3 of 20 being defective computers) = 1140(.05)^3(.95)17 = ...

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Let X_n be the random variable corresponding to the number of +1 steps in n turns. $X_n$ will follow Binomial Distribution with parameters $n$ and $p$. If $k$ is the number of +1 steps. Then, $n-k$ is the number of +2 steps. Then, $k+2(n-k) =t$. On solving, we get $k = 2n - t$. Thus, $P(T_n = t) = P(X_n = 2n-t) = {n \choose 2n-t}p^{2n-t}(1-p)^{t-n}$. Now, ...

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Note that $\frac{S_i+2}{3}$ is distributed as $B(1,p)$, giving that $\frac{T_n+2n}{3}$ is distributed as $B(n,p)$. Thus: $$\mathbb{E}\left(\frac{T_n+2n}{3}\right)=np, \mathbb{V}\left(\frac{T_n+2n}{3}\right)=np(1-p)$$ from which you can derive: $$\mathbb{E}(T_n)=n(3p-2), \mathbb{V}(T_n)=9np(1-p)$$ Furthermore, knowledge of the binomial distribution gives ...

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$P(T_n=t)=P(\sum_{i=1}^n S_i=t)$ so this is essentially the $n$-fold convolution of $S_1,S_2,...,S_n$. Not sure if a neat answer exists. $E(T_n)=E(\sum_{i=1}^nS_i)=\sum_{i=1}^nE(S_i)=\sum_{i=1}^n[1\times p-2\times(1-p)]=n(3p-2)$ $Var(T_n)=Var(\sum_{i=1}^nS_i)=\sum_{i=1}^nVar(S_i)=nVar(S_1)$ due to iid $S_i$. Find $Var(S_1)=E(S_1^2)-(E(S_1))^2$. Simplify. ...

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[Assuming the dice is otherwise fair - that is, if you just rolled $x$, the next roll is discrete uniform on $\{1,2,3,4,5,6\}-x$] Say the rolls are ordered $X_1,X_2,\cdots$ and $x_i\in{1,2,3,4,5,6}$ for all $i$. Then the joint pdf is: $$\mathbb{P}(X_1=x_1, X_2=x_2,\cdots,X_n=x_n)=\frac{1}{6}\left(\frac{1}{5}\right)^{n-1}\prod_{i=1}^{n-1}\mathbb{I}\{x_i\neq ... 2$$\mathbb{E}\left(y-\text{sign}\left(s-0.5\right)\right)=\mathbb{E}y-\mathbb{E}\text{sign}\left(s-0.5\right)=\mathbb{E}y-\frac{1}{2}\left(1+\left(-1\right)\right)=\mathbb{E}y=\mathbb{E}\mathbb{E}\left(y\mid s\right)=\mathbb{E}(2s-1)=0$$0 Yes, the process regarding the purchases will be a Poisson Process too with the new rate being \lambda = 7 * 0.65 = 4.55. Let N(t) be the purchases made till time t. Using the formula for number of arrivals till time t, Pr[N(t) = n] = \frac{e^{-\lambda t}(\lambda t)^n}{n!} . In your case, n is 3 and t is 1. In case, you are wondering about ... 0 We can use the concept of expected value, and do it in one step. We expect an increase of \frac49 black balls in bag Y and, of course, an increase of 1 is certain in the total number of balls in bag Y Thus Pr = \frac{6 + 4/9}{13+1} = \frac{58}{126} 1 You're right that E[Y]=E[X], but I don't quite follow how you calculated E[Y]. I'd phrase the probabilistic argument as follows: Your weighted selection of a person can be expressed as uniformly picking a friendship and then uniformly picking one of its two friends. If we now uniformly select a friend of this person, we're uniformly selecting a ... 1 You have to consider two cases 1) A black ball is drawn from bag X (B_x) 2) A white ball is drawn fram bag X (W_x) And let the event that a black ball is drawn from bag Y denoted as B_y First case The probability that a black ball is drawn from bag X is \frac{4}{9}. The black ball is put into bag Y. In bag Y there are now 7 white balls and 7 ... 1 Split it into disjoint events, and add up their probabilities: The probability of drawing a black ball from bag X and then a black ball from bag Y:$$\frac{4}{5+4}\times\frac{6+1}{7+6+1}=\frac{28}{126}$$The probability of drawing a white ball from bag X and then a black ball from bag Y:$$\frac{5}{5+4}\times\frac{6}{7+6+1}=\frac{30}{126}$$Hence ... 3 Given that X,Y and Z are independent random variables, so: E(X+Y+Z)^3=E(X^3+Y^3+Z^3+3(X+Y)(Y+Z)(X+Z)) = E(X^3+Y^3+Z^3+3X^2Y + 3X^2Z + 3XY^2 + 3Y^2Z + 3XZ^2 + 3YZ^2 + 6XYZ) = E(X^3)+E(Y^3)+E(Z^3)+E(3X^2Y) + E(3X^2Z) + E(3XY^2) + E(3Y^2Z) + E(3XZ^2) + E(3YZ^2) + E(6XYZ) = E(X^3)+E(Y^3)+E(Z^3)+3E(X^2)E(Y) + 3E(X^2)E(Z) + 3E(X)E(Y^2) + 3E(Y^2)E(Z) ... 1 Use linearity of expectation and the fact that all powers X^a, Y^b and Z^c are also mutually independent so$$\mathbb{E}(X^aY^bZ^c)=\mathbb{E}(X^a)\mathbb{E}(Y^b)\mathbb{E}(Z^c).$$4 Note that Y=|X| is a random variable that only takes on non-negative values. So the lemma holds for the random variable Y. Because \lim_{y\to\infty}y\Pr(Y\gt y)=0, there is a B\ge 1 such that if y\ge B then \Pr(Y\gt y)\le \frac{1}{y}. Let p=1-\epsilon. By the lemma,$$E(Y^{1-\epsilon})=\int_0^B (1-\epsilon)y^{-\epsilon}\Pr(Y\gt y)\,dy+ ...

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For ordinary simulation, a box-muller/polar marsaglia/ziggurat algorithm with cholesky decomposition allows you to simulate a bivariate normal very efficiently. However as you said you want to apply the central limit theorem, which seems that you want to simulate a lot of random variables and show the convergence. Assume you already have a random number ...

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Split it into disjoint events, and add up their probabilities: The probability of $\color\red0$ heads in each part is $\frac{\binom{3}{\color\red0}}{2^3}\cdot\frac{\binom{3}{\color\red0}}{2^3}$ The probability of $\color\red1$ heads in each part is $\frac{\binom{3}{\color\red1}}{2^3}\cdot\frac{\binom{3}{\color\red1}}{2^3}$ The probability of $\color\red2$ ...

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Since probability is being asked for, we needn't distinguish between people. Ways with first row full = ways to fill residue in rows $2$ and $3 = \binom{17}{13} = A$ (say) Ways with first and second row full = ways to fill residue in row $3 = \binom{10}6 = B$ (say) Ways with first row full and second row not full = $A - B = C$ (say) Unrestricted ways = ...

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Hint: Let $X$ be the number of heads in the first three rolls, and $Y$ be the number of rolls in the last three. Then $X$ and $Y$ are independent and follow what kind of distribution? Further, we are asked to calculated $$P(X = Y) = \sum_{k = 0}^3P(X = k, Y=k).$$

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You will have to apply inclusion-exclusion in this manner: All strings - strings with at least one from $1,2,3$ absent + at least two of $1,2,3$ absent - all from $1,2,3$ absent Allowing zeroes at start, it would work out as: $10^m - \binom31 9^m + \binom32 8^m - 7^m$ Added Oh, you wanted the probability, so , of course, you'd have to divide by $10^m$ ...

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What matters here is which seats are occupied, not who sits in which seat. Therefore, we can take our sample space to be the $$\binom{5 + 7 + 10}{18} = \binom{22}{18}$$ ways the patrons can occupy $18$ of the $22$ available chairs. What is the probability the first row is full? If the first row is full, $5$ of the $18$ people sit in the first row, ...

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The clue is that you are off by 3! which is the number of ways to arrange 3 objects... To flesh this out: Consider when $m=3$, you get $3 \choose 3$$(1/10^3) =1/1000 but you know that there are 6 ways - namely every possible combination of 123 (123, 132, 213, 231, 321, 312). What your formula is giving is the probability that 1,2 and 3 appear in one ... 1 This implies that sales equal demand if demand is less than the quantity of loaves available, i.e. Y=X when X\le k-1 (noting that k is integer valued). But for demand greater than k, demand cannot be met, hence sales then equal the total supply of loaves, k, i.e. Y = k for X\ge k. I don't know how to use this to find a pmf for Y. ... 0 In total there are 6^5 ways to display five digits, selected without bias but allowing repetition (ie: independently). Next we count the (equally-probable) favored outcomes. First we choose the two digits for the pair, and the singleton, \binom 6 2\binom 4 1, then we choose which from the five dice to host them, \binom 5 2\binom 3 2\binom 1 1. ... 3 When you have n objects to choose from, and you want to choose r, if the order doesn't matter, and repetition is not allowed, the answer is \displaystyle{n\choose r}. If the order does matter, then you'd get n(n-1)\cdots(n-r+1)=P(n,r) by the multiplication rule. However, the order doesn't matter, so that means that every selection of r objects ... 0 Well, both are wrong. Take 100 brand new disks, each with an expectancy of 100,000 hours to the first failure. How long does it take until the first one fails? Certainly before 100,000 hours, but you'd need to know at least the variance, and better the distribution. You might have disks that all without exception fail after 99,000 to 101,000 hours, then ... 0 The 1-\alpha confidence interval for \hat p is$$\large{\left[ \hat p-z_{(1-\frac{\alpha}{2})} \cdot \sqrt{\frac{\hat p(1-\hat p)}{ n}} , \hat p+z_{(1-\frac{\alpha}{2})} \cdot \sqrt{\frac{\hat p(1-\hat p)}{ n}} \right]}$$z_{(1-\frac{\alpha}{2})} is the z-value of the standard normal distribution. 1-\alpha is the confidence level. In your case ... 1 As Ian said, T_n for n>1 is actually Gamma Distributed with k=1. In fact, the Exponential Distribution is a special case with k=1. You did mention you were using T_n for the interarrival times, but that isn't the standard notation - T_n - T_{n-1} is the interarrival time, where T_n is simply the total arrival time for x_n. I think you're ... 1 S = \{1,2,3,\ldots,20\} A_n = \{x \in S: x \: \text{is a multiple of} \:n\} \:\:\:n=1,2,3,\ldots A_1 = \{1,2,3,\ldots,20\} A_2 = \{2,4,6,\ldots,20\} because the products of n=2 with the elements in S are 2,4,6,\ldots,20 up to 20 because we have the restriction that x \in S. So 1 \leq x \leq 20 A_3 = \{3,6,9,\ldots,18\} and ... 1 A_2=\left\{2,4,6,8,\ldots,20 \right\}, the even numbers.$$P(A_2)=\frac{10}{20}=\frac{1}{2}$$0$$P(R)=1/5$$For each forecast i=1,2,3$$P(H_i = R|R)= 3/4$$and$$P(H_i=R|\bar{R})=1/4$$We need to find$$P(R|H_1=R, H_2= R, H_3 = \bar{R}) = \frac{P(R) P(H_1=R, H_2=R, H_3=\bar{R}|R)}{P(H_1=R, H_2=R, H_3=\bar{R})} = \frac{P(R) P(H_1=R, H_2=R, H_3=\bar{R}|R)}{P(R) P(H_1=R, H_2=R, H_3=\bar{R}|R)+P(\bar{R}) P(H_1=R, H_2=R, H_3=\bar{R}|\bar{R})} ... 0 As you have pointed out, $$\mathbb{E}[X|Y = y] = \int_{-\infty}^{\infty} x f_{X|Y}(x|y)dx.$$ The main difference between$\mathbb{E}[X|Y=y]$and$Z = \mathbb{E}[X|Y]$is that$\mathbb{E}[X|Y]$is a random variable while$\mathbb{E}[X|Y=y]$is not. Therefore, you can not really 'compute'$\mathbb{E}[X|Y]since it is a r.v. Of course, you can compute the ... 0 \begin{align} \ell(\theta) = \log L(\theta) = \log \left( \theta^n \bigg( \prod_{i=1}^n x_i \bigg)^{\theta - 1} \right) & = \log(\theta^n) + \log \left( \left( \prod_{i=1}^n x_i \right)^{\theta-1} \right) \\[10pt] & = n\log\theta + (\theta - 1) \log\prod_{i=1}^n x_i. \end{align} You could then go on to write\displaystyle\log\prod_{i=1}^n x_i$as ... 0 Yes, this is also reasonable. The generalisation of these two solutions is $$\textsf{Pr}[n]=\begin{cases}\frac23(1-q)q^{(n-2)/2}&n\text{ even}\;,\\\frac13(1-q)q^{(n-1)/2}&n\text{ odd}\;,\end{cases}$$ with$q=\frac12$for your solution and$q=\frac14$for the other solution. 0 Assuming they are real valued:$P(X=Y)=\int_{-\infty}^\infty P(Y=x|X=x) \ P(dx)=\int_{-\infty}^\infty P(Y=x) \ P(dx)=0$, because$P(Y=x)=0$for any$x\in\mathbb{R}$since$Y$has a continuous distribution (pdf). If you meant for the cdf to be continuous, then the answer changes. In general, though$P(X=Y)>0$is possible, but you have to relax some of ... 2 Let$T$be the count of tails on the coin toss (ie$0$for heads,$1$for tails). Thus can partition the space over these two disjoint events, using the Law of Total Expectation. Because it is a fair coin: $$\mathsf P(T=0)~=~\mathsf P(T=1)\\~~~=~\tfrac 12$$ Let$X$be the sum of the dice results, and$X_1,X_2$be the result of the individual ... 4 If$p_n$is the probability that the$n$th roll is a six, the generating function for this sequence is $$\phi(s)=\sum_{n\geq1} p_n s^n={120s\over(6-s)(5-s)(4-s)(3-s)(2-s)(1-s)}.$$ A partial fraction expansion gives us the explicit formula $$p_n=\sum_{j=0}^5 {5\choose j}{(-1)^j\over (j+1)^n}.$$ The first few values are$\$p_1={1\over 6},\quad p_2={49\over ...

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