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This problem that has a bounty on it, contains a reference to a solution using probability theory. I think the probability solution is quite elegant. It works though by showing an even stronger result than the one asked for. So, I suspect another proof must exist, but that proof might be more complex.


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Using the equation $$\sigma^2 = E[X^2]- \left( E[X] \right)^2,$$ we get $E[X^2]=25.$ For part (a), we expand to get $$E[(3+X)^2]=E[9]+6E[X]+E[X^2]=58.$$


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For every $a\in\mathbb R$, the complex exponential function $e^{iat}$ is positive-definite because $$ \sum_{k,l} e^{ia(t_k-t_l)}z_k\bar z_l = \left|\sum_{k} e^{iat_k} z_k\right|^2 \ge 0 $$ And $\cos^2 t$ can be written as a combination of complex exponentials with positive coefficients: $$ \cos^2 t = \frac12+\frac12 \cos 2t = ...


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I think your explanation is interesting. I think the person who is complaining about your explanation is confused about the effect of the reveal upon the probabilities. That the initial chance you chose correctly is $1/3$, and the open door -- post reveal -- has 0 likelihood of being the car, I think the confusion is where did the other $1/3$ go? You assert, ...


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After taking a sample of $n$ balls focus on unordered pairs of balls that are present in the sample. Give the balls in the sample randomly (i.e. without looking at their colors) numbers $1,\dots,n$ and for $i<j$ let $X_{i,j}=1$ if the balls numbered $i$ and $j$ are both white and $X_{i,j}=0$ otherwise. Then $\binom{X}{2}=\sum_{i<j}X_{i,j}$ so that ...


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As an example of how order makes no difference, let's look at the chance of drawing a $7$, say, going first or going tenth. Going first, your chance is $10\%$, clearly. But going last, your chance of getting the $7$ is the chance of everybody in front of you drawing something that is NOT a $7$. Which is ...


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Not answering directly OP's question - since the following is certainly not "without any complicated calculations". Just to verify computationally the correct result (that another answer reaches intuitively): ...


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As it has been mentioned already, the problem is ill-posed and the answer can be any number from $0$ to $1$ depending on the situation. I am writing it just to state it explicitly. The extreme points are Great Inquisitor (year 13..): Here are 4 cosmology questions. Your verdict will depend on what answers you choose (close to $1$) Teacher (any year you ...


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Whether one person gets a question wrong or right, the other has to get the same answer. If the answer was wrong, the second person needs to get the same answer to fit the condition that they get the same wrong answers. If the answer was right, the second person needs to get it right as well; otherwise, he would have answered a question wrong that the first ...


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I think it is a little confusing to talk of the probabilities changing. It is more usual to talk about conditional probabilities: what is the probability of one event given that some other event has occurred? You can avoid talking about conditional probabilities altogether by explaining the Monty Hall problem like this: if you picked the car initially, then ...


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The way to approach these types of problem is the following. We begin by noting that we can calculate the $z$-value as follows: $$z=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\frac{61-50}{7}=1.5714$$ And therefore we have that (using tables or an appropriate online tool): $$\mathbb{P}[Z<z]=\Phi(z)\approx 0.9420$$ Thus the $p$-value of this two-tailed test is ...


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$$ P\left({\mbox{Prize is behind chosen door}}\right){}\neq{}P\left({\mbox{Prize is behind chosen door}\,|\,\mbox{Prize is not behind door with goat}}\right)\,, $$ in general.


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You might want to nail down what probabilities you're talking about. In the first case, it's "the probability that the prize is behind that door, GIVEN what you know now." In the second, it's the same thing, given what you know AFTER the revelation. The numbers don't change -- they are different numbers all along. Consider a simpler experiment: I have two ...


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The simplest approach is for person $1$ to produce three parts which have equal probability measure under $\rho_1$. Person $2$ then takes those two parts with largest and second largest probability measure under $\rho_2$ and moves a bit from the former to the latter so they then have equal probability measure under $\rho_2$, each at least a third of the ...


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For any $i=1,\ldots,n$, we have, using the Total Law of Expectation, conditioning on the value of $U_i$, \begin{eqnarray*} && \\ E(U_i\vert \max\{U_1,..,U_n\}=t) &=& E(U_i\vert U_i=t\cap \max\{U_1,..,U_n\}=t)P(U_i=t\vert \max\{U_1,..,U_n\}=t) \\ && + E(U_i\vert U_i\neq t\cap \max\{U_1,..,U_n\}=t)P(U_i\neq t\vert \max\{U_1,..,U_n\}=t) ...


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$\left\{ X=m\right\} \subseteq\left\{ Y=m\right\} \cup\left\{ Y\neq X\right\} $ leading to: $$P\left\{ X=m\right\} \leq P\left\{ Y=m\right\} +P\left\{ Y\neq X\right\} $$ Likewise: $$P\left\{ Y=m\right\} \leq P\left\{ X=m\right\} +P\left\{ Y\neq X\right\} $$


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There's no contradiction. Everything you write is true. Let's define $$ \alpha = D_{n,0}, $$ $$ \beta = \sum_{m=1}^nD_{n,m}, $$ $$ \gamma = 0 D_{n,0} = 0, $$ $$ \delta = \sum_{m=1}^nmD_{n,m}. $$ Then you showed that $$ \alpha+\beta = \sum_{m = 0}^nD_{n,m} = \sum_{m=0}^nmD_{n,m} = \gamma+\delta. $$ This looks surprising at first glance - we expect that ...


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Observe that $$P(X=m)=\sum_y P(X=m,Y=y)=P(X=m,Y=m)+P(X=m,Y\ne X)\\ \le P(Y=m)+P(Y\ne X)$$Similarly, $P(Y=m)\le P(X=m)+P(Y\ne X)$


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Question 1: This one is actually pretty easy. It helps to start by considering a single light. The number of times the light is toggled is binomially distributed, with parameters $n = 32$ and $p = 1/64$, and the probability that it ends up being lit at the end of the process is the sum of the probabilities that it is toggled an odd number of times, i.e.: ...


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I suppose that you can make the expresion shorter using $(1+a)(1-a)=1-a^2$ so $$A=\frac{1}{(1+x)(1−x)(1+x^2)(1−x^2)(1+x^3)(1−x^3)\cdots}=\frac{1}{(1−x^2)(1−x^4)(1-x^6)\cdots}$$ and now use the fact that $$\frac{1}{1-y}=\sum_{i=0}^{\infty}y^i$$ and replace successively $y$ by $x^2$, $x^4\cdots$,$x^{2n}$ before computing the overall product. Doing so, for a ...


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Remember that $E(I_A) = P(A)$ for any event $A$ and $E(I_AI_B)=P(A \text{ and } B)$. $F_X(x)=P(X\le x)$ and $F(x,y)=P(X \le x\text{ and } Y\le y)$.


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$$\frac{1}{1-x^k}=1+x^k+x^{2k}+x^{3k}\cdots$$ $$\frac{1}{1+x^k}=1-x^k+x^{2k}-x^{3k}\cdots$$


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If you set $$f(x)=\prod_{n=0}^\infty (1-x^n)^{-1}=\sum_{n=0}^\infty p(n)x^n$$ we see that yours is just $$f(x^2)=\sum_{n=0}^\infty p(n)x^{2n}.$$


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You are right in saying that the teacher did not strictly define the experiment and thus it is hard to tell whether or not there is a bias. For example you could also ask was the selection process random. However I think the idea the teacher was trying to get across was that if I have $n$ objects and randomly partition the set into $A$ and $B$ with the same ...


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This appears in the oeis, wherein it is given that the generating function is $$\prod_{k>0}1-x^{2k-1}=\prod_{k>0}\frac{1}{1+x^k}$$ There are also many references there, I highly recommend that link.


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Assuming the two events are independent (being two different lotteries, I assume), the probability of both events happening is the product of probabilities, that is $$ \frac{1}{175,223,510}\cdot \frac{1}{258,890,850} = \frac{1}{45,363,763,443,883,500} $$ The denominator is about 45 quadrillion.


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As stated, the statement seems false. Tacking $X$ to be a Bernoulli with parameter $p=1$ ($X=1$ a.s.; $\mathbb{E}[X]=\mathbb{E}[X^2] = 1$), that would mean that for all $a\geq 1$ $$ 0 = \Pr[X > a] \geq (a-1)^2 \geq 0\ . $$


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The expected number of flips can be infinite. I first show how to make it larger than any finite number. Choose a large integer $n$. Divide $[0,1]$ into $2^n$ diadic intervals, which we indexed by $W_n:=\{ 0, 1 \}^n$. Choose real numbers $\alpha < \beta$. Let $X$ be the union of those dyadic intervals indexed by $w \in W_n$ for which at most $n/2+ ...


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Generally, this is not true: the assumption $o(w^2h^2)$ as $wh\to 0$ does not yield the conclusion. For example, let $k(w)=c/(1+w^4)$ with $c>0$ chosen so that the integral of $k$ is $1$. The function $w^4h^4$ satisfies the assumption $o(w^2h^2)$ as $wh\to 0$, but $$\int k(w) w^4h^4\,dw = \infty \quad \text{ for all } h>0$$ So, you need some ...


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HINT: Label the vertices of $K_{n+1}$ with the integers $0,\ldots,n$, and add a loop at each vertex to get a graph $G_n$. The edges of $G_n$ correspond in an obvious way to the dominoes of a double-$n$ set. A domino train corresponds to a path in $G_n$ with no repeated edge. Note that each vertex of $K_{n+1}$ has degree $n$, so each vertex of $G_n$ has ...


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For the symmetric case where , ℙ(St+1−St=1)=0.5,andℙ(St+1−St=−1)=0.5, the solution is $\frac{T-\frac{[T]}{2}}{2^{T}}$.


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This can be described by a markov chain with the following transition diagram: The associated transition matrix is: (with order 3H, 2T, S, 1H, 2H, 1T) $$A = \begin{bmatrix} 1 & 0 & 0 & 0 & .5 & 0\\ 0 & 1 & 0 & 0 & 0 & .5\\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & .5 & 0 & 0 & .5 \\ 0 ...


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I'd post this response as a comment, but I don't have enough reputation to (why do new accounts require 50 reputation to comment? Shouldn't it be the other way around? But I digress...). So please forgive any lack of mathematical rigor, as these are just intuitive thoughts. In terms of convergence in probabilistic sense, it's clear that there is no ...


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Continued fractions: \begin{align} \frac{11235}{100000}&=\frac{2247}{20000}\\ &=\cfrac{1}{8+\cfrac{2024}{2247}}\\ &=\cfrac{1}{8+\cfrac{1}{1+\cfrac{223}{2024}}}\\ &=\cfrac{1}{8+\cfrac{1}{1+\cfrac{1}{9+\cfrac{1}{13+\cfrac{2}{17}}}}}\\ &=\cfrac{1}{8+\cfrac{1}{1+\cfrac{1}{9+\cfrac{1}{13+\cfrac{1}{8+\cfrac{1}{2}}}}}}\\ &=[0;8,1,9,13,2] ...


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Like you hint, we can observe that the value $$0.11235955056\ldots = \sum_{k = 1}^{\infty} 10^{-k} F_k,$$ where $F_k$ is the $k$th Fibonacci number (with the convention that $F_0 = 0$, $F_1 = 1$), is simply the value of the series $$F(x) := \sum_{k = 1}^{\infty} F_k x^k = x + x^2 + 2x^3 + 3x^4 + 5x^5 + 8x^6 + 13x^7 + \cdots$$ at $x = \frac{1}{10}$. Now, ...


2

An exhaustive search (by computer) of all fractions with numerator and denominator $< 100$ shows there is only one whose decimal representation starts 0.11235: $$ \frac{10}{89} $$ The decimal expansion continues $\ldots 955056179775\ldots$ (I haven't bothered finding aprogram that can calculate enough that we can see the period), that's completely ...


0

These are fractions $f$ with $11235/100000\le f<11236/100000.$ A direct, if slow, method for counting these would be $$ \sum_{d=1}^{100}\left\lfloor\frac{100000d}{11235}\right\rfloor-\left\lfloor\frac{100000d}{11236}\right\rfloor $$ or sum(d=1,100,100000*d\11235-100000*d\11236)


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If $X$ is any continuous random variable (e.g. normally distributed), then $P(X=c) = 0$ for all $c$ in $\mathbb{R}$, no matter what distribution $X$ has or what point $c$ is. Points are too small to have non-zero probability of being hit by $X$. If by the quantity $b$ you mean the variance -- normally, we write $X \sim N(\mu,\sigma^2)$ -- then $b$ is zero ...


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That depends if $X$ has a probability density function or not. If it has, no difference, elsewhere it can be different (for example if $X=0$ with probability one).


1

The probability that their first successes are simultaneously is $\frac{p}{2-p}. \quad \checkmark$ The probability that their first successes are not simultaneously is the converse probability: $$1-\frac{p}{2-p}=\frac{2-p}{2-p}-\frac{p}{2-p}=\frac{2-2p}{2-p}$$ Then the probability that Nick’s first success precedes Penny’s is the half of it.


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Here is an example. The question is how to show that $$\binom{n}{k}^{-1}=(n+1)\int_0^1 x^k (1-x)^{n-k} \, dx. $$ To make this self-contained, I'll paste this answer below: Let's do it somewhat like the way the Rev. Thomas Bayes did it in the 18th century (but I'll phrase it in modern probabilistic terminology). Suppose $n+1$ independent random variables ...


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Thomas Bayes showed that $${n \choose k}\int_0^1 x^k (1-x)^{n-k}\mathrm dx = \frac{1}{n+1}$$ by pure thought, without using calculus (for all integers $k,n$ with $0 \leq k \leq n$). His argument, known as the Bayes' billiards argument, uses two equivalent probabilistic stories about picking random points on a number line from $0$ to $1$. That's just one ...


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I'm answering my own question to keep the number of unanswered questions low. Let $g(X) = X^2$, then \begin{align} E(X^3) &= \lambda E(g(X+1)) \\ &= \lambda E((X+1)^2) \\ &= \lambda (E(X^2) + 2E(X) + 1) \\ &= \lambda (\lambda+\lambda^2 + 2\lambda + 1)\\ &= \lambda^3 + 3\lambda^2 + \lambda \end{align}


0

Take the probability that each ball lands in a box that doesn't have a ball in. For the first ball, this is $1$ (as none of the boxes have a ball in). For the second ball, the probability that it lands in a box without another ball in is $\frac{99}{100}$. For the third, $\frac{98}{100}$ etc etc. So to find the over probability that there is exactly one ball ...


1

The number of ways to choose an ordering of the $100$ boxes (to put the balls in one by one) is $100!$. The number of possible ways they can land in the boxes is $100^{100}$. Therefore the probability is $$\dfrac{100!}{100^{100}}$$


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The probability that the $1$st ball will land in an empty box is $\frac{100}{100}$ The probability that the $2$nd ball will land in an empty box is $\frac{99}{100}$ The probability that the $3$rd ball will land in an empty box is $\frac{98}{100}$ The probability that the $k$th ball will land in an empty box is $\frac{101-k}{100}$ Hence the answer is ...


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The appropriate model has the five dice distinguishable. A more familiar example is tossing two coins. A model that treats two heads, two tails, and one of each as equally likely gives answers that do not match reality. There are thus $6^5$ equally likely outcomes. We now need to count the favourables. We divide into cases. (i) All dice show the same ...


0

The 252 options are not equally likely. One option has all dice rolling 1 - the probability is $(1/6)^5$. Another option has one 1, one 2, one 3, one 4 and one 5. But there are 120 ways this can happen, from 12345, 12354, 12435 and so on, depending on the value of die 1, which has 5 choices, and so on. So the probability of getting all numbers 1 to 5 is ...


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Each boy should get G number of candies and each girl should get B number of candies. This would ensure that sum of the candies received by boys and girls is equal.


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Conditional probabilities: for the probability measure $\mathbb{P}$ (appropriately defined now, we don't like zero divisions), $$ \mathbb{P}\left(B \,|\, A\right){}={}\frac{\mathbb{P}\left(B \,\cap \, A\right)}{\mathbb{P}\left(A\right)} $$ Choose the probability measure $\mathbb{P}\left( \dot\,\,|\,\, C\right)$ and apply it to this definition above, ...



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