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1

You choose $n$ numbers. Success occurs when you choose at least one is a five (event $F$) and at least one even number (event $E$). Failure occurs when neither or only one event happens. You wish to find: $\mathsf P(E\cap F)$ You should the law of complements, the principle of inclusion and exclusion, and that: $$\begin{align} \mathsf P(E^c) &= \frac ...


2

Not sure this is the most efficient, but: The probability it is not divisible by $5$ is $(1 - 1/9)^{n}$. The probability it is not divisible by $2$ is $(1 - 4/9)^{n}$. The probability it is not divisible by $2$ and $5$ is $(1 - 5/9)^{n}$. The quantity you seek is $1 - ((1 - 1/9)^{n} +(1 - 4/9)^{n} -(1 - 5/9)^{n})= 1 - (8/9)^n -(5/9)^n + (4/9)^n$.


3

Let $X$ and $Y$ be independent random variable with distribution uniform on $[0,1]$, and let $W=\max(X,Y)$. For $0\le w\le 1$, we have $$F_W(w)=\Pr(W\le w)=\Pr\left((X\le w)\cap(Y\le w)\right)=w^2$$ For the density function $f_W(w)$ of $W$, differentiate $F_W(w)$. We get $2w$ on the interval $(0,1)$, precisely as the simulation suggests.


3

Corrigendum $$\mathbb{var}[\overline{x}]=\mathbb{var}\left[ \frac{1}{N}\sum_{i=1}^Nx_i \right]\stackrel{\text{iid}}{=} \frac{1}{N^2}\sum_{i=1}^N\mathbb{var}\left[x_i \right]=N \cdot \frac{m_{c_2}}{N^2}=\frac{m_{c_2}}{N}$$ $Cov({x_i},{x_j}) = E\left[ {{x_i}{x_j}} \right] - E\left[ {{x_i}} \right]E\left[ {{x_j}} \right] = 0$ Since $\{x_1,\ldots, x_N\}$ iid : ...


1

$$ 𝔼[\frac{1}{N^2}\sum_{i = 1}^n\sum_{j \neq i}x_ix_j] = \frac{1}{N^2}\sum_{i = 1}^n\sum_{j \neq i}𝔼[x_i]𝔼[x_j] $$ And this is not equal to zero.


0

UUID v 1 will produce collision if the tiestamps are the same, so this depends on generation rate and timer resolution. The node-id should be the MAC of a built-in NIC, otherwise it should come from a special address block requested fro IEEE or based on a good random number (http://tools.ietf.org/html/rfc4122#section-4.5). The way I read that spec however is ...


1

Consider a distribution with $$P(X=m\delta)=1-\delta$$ $$P\left(X=m\tfrac{1-\delta+\delta^2}{\delta}\right)=\delta$$ then $E[X]=m$ but $P(X\gt a )$ can be made arbitrarily small by choosing $\delta \lt \frac{a}{m}$ and if necessarily smaller. So to get something useful, you would need some constraint on $X$ such as a given maximum or a given variance.


0

let's for simplicity generate points within an angle ALPHA of the north pole P = (1 - cos(ALPHA))/2 U = random(0,1) a random value in the range (0,1) PHI = acos(1-2*P*U) THETA = random (-PI,PI) PHI is the angle from the north pole


0

Looks an awful like a version of a jump-diffusion process, especially since its brownian between oscillations.


0

Hint: In the random walk, $X_i$ is step #$i$ (to right if $+1$, left if $-1$). Since $\alpha > 1/2$, this is a biased random walk. What does the condition on $\sum_{k=1}^n Z_k$ say about $\sum_{k=1}^n X_k$ (which is where the random walk ends up after $n$ steps)?


0

The proposition you provide is correct if each of the $X_i$ can only take on non-negative values. (I'm sure there are other more complicated conditions that would also make it correct, but for general distributions, the inequality does not hold.) A counterexample with $n=2$: Let $X_1$ and $X_2$ each be uniformly disributed on $-3,-2$, and choose $k = ...


0

It is true if your random variables are nonnegative: $$ X_1+X_2+\cdots+X_n\leq k\implies X_1\leq k\;\&\;X_2\leq k\;\&\;\ldots\;\&\;X_n\leq k. $$ So $$ \Pr(X_1+X_2+\cdots+X_n\leq k)\leq\Pr(X_1\leq k\;\&\;X_2\leq k\;\&\;\ldots\;\&\;X_n\leq k)=\prod_{i=1}^n\Pr(X_i\leq k). $$ The equality above uses independence.


1

The notation $F_X$ is absurd since $X$ is a random process, not a random variable. For every $t$ and $x$ positive, $[X(t)\leqslant x]=[A^t\leqslant x]=[A\leqslant x^{1/t}]$ hence $F_{X(t)}(x)=F_A(x^{1/t})=\min\left\{1,\frac18x^{3/t}\right\}$. Equivalently, $F_{X(t)}(x)=0$ for every $x\leqslant0$, $F_{X(t)}(x)=\frac18x^{3/t}$ for every $0\leqslant ...


0

each bus has $(1-\frac{1}{100})^{100}\sim\frac{1}{e}$ to be left empty. By linearity of expectaions the answer is $100(1-\frac{1}{100})^{100}\sim\frac{100}{e}$.



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