# Tag Info

3

The total number of ways to arrange $14$ people is $14!$. The number of ways to arrange $14$ people with a specific person last is $13!$. So in a single experiment, the probability of that person being chosen last is $\frac{13!}{14!}=\frac{1}{14}$. Hence in $3$ experiments, the probability of that person being chosen last every time is ...

2

Here is a guideline. Some details are left as smaller exercises, but I really encourage you to fill them up. I'll first recall how to prove that a sequence of random variables converges almost surely, before adapting the argument to the limsups. Convergence of sequences of random variables Let $(a_n)_{n \in \mathbb{N}}$ be a sequence of real numbers, and ...

2

Let $A_n$ be the natural numbers with the first $n$ natural numbers thrown out. I.e. $A_n=\mathbb N\setminus\{1,2\cdots n \}$. Then each $\mu(A_n)=\infty$ and $\cap A_n=\phi$ so that $\mu(\cap A_n)=0$. Can you adapt this argument to your example?

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Choose $x_k \in \Omega$ such that each $x_k$ is distinct. Let $A_n = \{x_k\}_{ k \ge n}$.

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There is a set $A \subset \Omega$ that can be identified as $\mathbb{Z}$ (in the sense that there is a bijection $f: \mathbb{Z} \to A$). Let $A_n = \mathbb{Z} \setminus \left\{ 1 ,\dots, n \right\}$. This sequence satisfies the claim.

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Try this, which I think is similar to your strategy. Let $C=\{x_1, x_2, \cdots \}$ be a countably infinite set of distinct points. Let $A_n:= \{x_i \in C : i\geq n\}$. So $A_1=C$, $A_2=\{x_2, x_3,\cdots \}$, etc. Then Clearly $$\bigcap_n A_n =\emptyset$$ and $A_n \downarrow$. However, for each $n$, $\mu(A_n)=\infty$, so $\mu(A_n)$ cannot tend to 0.

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To be true that $A_i \downarrow A \implies \mu(A_i) \downarrow \mu(A)$, you should have that $\mu(A_1) < \infty$. Hint: construct an example where $\mu(A_1) = \infty$.

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Start with a Poisson process $\Pi$ of rate $a+b$ on $(0,\infty)$. Paint each point of the Poisson process either red (with probability $a/(a+b)$) or green (with probability $b/(a+b)$), the colors being i.i.d. and independent of the locations of the points themselves. Let $X:=\min\{t\in\Pi: t$ is painted red$\}$ and $Y:=\min\{t\in\Pi: t$ is painted green$\}$. ...

1

You are correct. You are also correct that the probability of any discrete value from a continuous distribution is zero. In this case you can handle the problem by only looking at the probability of belonging to a small interval around $\bar{z}$, which is nonzero, and can be calculated as follows. $$P(x<X<x+\epsilon) \approx \epsilon f(x)$$ Let ...

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The first part of your logic is correct.   To find the individual probabilities, you can simply use combinatorics. \begin{align}p(9\textsf{ black}) ~=~& \dfrac{\dbinom{50}{9}}{\dbinom{100}{9}}\\[2ex]p(1\textsf{ white}, 8\textsf{ black})~=~&\dfrac{\dbinom{50}{1}\cdot\dbinom{50}{8}}{\dbinom{100}{9}}\end{align} ... and so on. PS: As a ...

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$(X,Z)$ are independent if forany $f,g$ $$\mathbb{E}(f(X)g(Z))=\mathbb{E}(f(X))\mathbb{E}(g(Z))$$ In your case : $$\mathbb{E}(f(X)g(Z))=\mathbb{E}\left(f(X)g\left(\frac{Y-pX}{\sqrt{1-p^2}}\right)\right)=\int_{\mathbb{R}^2}f(x)g\left(\frac{y-px}{\sqrt{1-p^2}}\right)p(x,y)~dx~dy$$ so this is a change of variable problem, you set : ...

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Your work appears to be correct. The probability would be: $$(1-\frac{13}{14})^{3}=(\frac{1}{14})^3$$

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Using Bayes' rule, for each value $x$ you have $$P(x | data) \propto P(data | x) prior(x)$$ Therefore, $P(x|data)$ is zero if $x \neq 0$ and so $P(0|data) = 1$, so your posterior is the same as the prior. No, the posterior is the same as the prior. Using this prior says that you have decided that the answer is zero and no evidence can convince you ...

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First, randi([1 2]) will give you a uniform distribution from 1 to 2. And you said that you will count the difference between number of heads and number of tails. In general, if the coin-flip experiment is fair, the probability of getting a head should be the same as the probability of getting a tail = 0.5. That means the difference should be zero. But in ...

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