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1

The transition matrix is an $N \times N$ matrix given by $\mathbf{P} = \begin{pmatrix} 1/2 & 0 & 0 & \cdots & 0 &1/2 \\ 1/2 & 1/2 & 0 &\cdots &0 &0 \\ 0 & 1/2 & 1/2 & \cdots &0&0 \\ \vdots & & \ddots & \ddots && \vdots \\ \vdots &&&\ddots&\ddots &\vdots\\ 0 ...


1

If there are currently $0$, then you either go to $N$ with probability $1/2$ or stay at $0$ with probability $1/2$. If there are currently $0<n\leq N$, then you either go to $n-1$ with probability $1/2$ or stay at $n$ with probability $1/2$. (I honestly don't know how to give just a hint for this part, since I basically just read off the content of the ...


0

Okay, we need the true bounds of integration, which are a little different from what you mentioned. Here is a Wolfram Alpha plot of the bounds: If you integrate this with $x$ as the free variable, you end up with a two-stage integration, since the $y(x)$ function changes at $x=1$. As a result, I would recommend integrating with $y$ as the free variable. ...


0

The argument is OK, but where you base a conclusion on the fact that $C$ is a linear combination, I'd say specifically that it is their sum and they are independent. If they were not independent then there would be more work to do. Also, if this is homework, I'd write out the particular density function for the normal distribution with the particular mean ...


-3

If two RVs $X,Y$ are independent we have the following relation: $$E[XY]=E[X]E[Y]$$ So from what we've got here we can already see that $e^{it_1 X_1}$ and $e^{it_2X_2}$ are independent. These are the moment generating functions and for independent RVs we know that the moment generating functions have the following property $M(X+Y) = \langle e^{t(X+Y)} ...


0

consider with just 10 integers $$\large{x_1x_2x_3x_4x_5x_6x_7x_8x_9x_{10}}$$ where all $\large{x_i}$ is either $0$ or $1$. What is the chance of finding $\large{x_1x_2x_3x_4}$ all zeros ? It is just $\large{(\frac{1}{2})^4 = \frac{1}{16}}$. Now you just apply this to your question


0

Hint: If you never had $x$ zeros in a row, then (at least) one of the first $x$ numbers must be a one. If $n_z$ is the number of arrangements, then you can partition $n_z$ based on which digit (of the first $x$) is the first to be a one. This will give you a recurrence relation in terms of numbers $n_{z_0}$, where each $z_0<z$. Doing this will let you ...


1

No, continuity on $\bigcup_{n \in \mathbb{N}} \text{Img}(X_n)$ is not enough. (Counter)Example: Consider $([0,1],\mathcal{B}[0,1])$ endowed with the Lebesgue measure, $$g(x) := 1_{\{0\}}(x)$$ and $$X_n := \frac{1}{n}.$$ Then $X_n \to X := 0$ almost surely, $g$ is continuous on $(0,\infty) \supseteq \bigcup_n \text{Img}(X_n)$, but $g(X_n)=0$ does not ...


0

When the population distribution is assumed normal but the population standard deviation $\sigma$ is not known, it needs to be estimated from the sample. This estimate is $$s = \sqrt{\frac{1}{n-1} \sum_{i=1}^n (x_i - \bar x)^2}.$$ Then the standard error of the sample mean is $$SE = s/\sqrt{n}.$$ But once we use this estimate, the margin of error for the ...


1

I assume that $(abc)(def)(ghi)(jkl)$ is considered the same result as $(def)(ghi)(jkl)(abc)$ (i.e. rotations of tables is irrelevant) and $(bca)(def)(igh)(klj)$ (i.e. rotations within tables is irrelevant), but is considered different from $(bac)(def)(ghi)(jkl)$ (mirrors of tables considered different) and is considered different from $(def)(abc)(ghi)(jkl)$ ...


4

Suppose $i$ is an integer between $1$ and $24$, and let $A_i$ be the event that you roll an $i$ at least $4$ times in ten rolls. Notice that two of the $A_i$ could occur, but not three, since that would be twelve different rolls. By the inclusion/exclusion principle, you get that $$\mathbb{P}(\cup A_i) = \sum \mathbb{P}(A_i) - \sum_{i\ \neq j}\mathbb{P}(A_i ...


0

Outline: We use Inclusion/Exclusion. Find in one of the usual ways the probability of say $1$ coming up $4$ or more times. Multiply by $24$, and call the result $a$. However, this double-counts the ways in which $2$ faces come up $4$ or more times. (We cannot have more than $2$ faces coming up $4$ or more times.) So we will have to subtract the ...


1

Let $W$ be the number of tagged fish caught. Then the number of untagged fish caught is $5-W$. Net earnings $Y$ are given by $$Y=10W+2(5-W)-25=8W-15.$$ It follows that $\text{Var}(Y)=8^2\text{Var}(W)$. I have not checked the correctness of your calculation of $\text{Var}(W)$. Note that the unit of variance is fishes$^2$.


0

I think you may have made a mistake in asking the question; in the distinction between independent and dependent. That is if I properly identified your intention if not disregard the above statement. My answer to you questions part A and B as they are written: Answer A: 2/3+1/4=11/12 The likelihood of one or more out of a series of independent happenings ...


1

$\newcommand{\E}{\operatorname{E}}$If $U$, $V$ are undependent and $\E U$ and $\E V$ both exist then $\E(UV)=(\E U)(\E V)$. Therefore $$ \E\Big((X_i-\E X_i)(X_j-\E X_j\Big) = \E(X_i-\E X_i) \E(X_j-\E X_j) $$ and $$ \E(X_i-\E X_i) = \E X_i - \E(\E X_i)) = \E X_i - \E X_i=0 $$ and similarly for $j$. \begin{align} & \sum_{k=2}^\infty k(k-1)q^{k-1}p = qp ...


1

The sample space with the (product) probabilities assigned to the elementary events: $$ \begin {matrix} \text{#}&\text{bus}&\text{train1}&\text{train2}&\text{prod. of probs.}&\text{resulting prob}\\ 1&\text{O}&\text{O}&\text{O}&\frac{1}{3}\frac{3}{4}\frac{3}{4}&\frac{9}{48}\\ ...


0

It will be $$ \dfrac{\binom{2}{2}\binom{8}{2}}{\binom{10}{4}}=\dfrac{4*7}{210}=\dfrac{2}{15} $$ I have not considered the order.


1

I think it is clearer to break up 1st and 2nd rolls into separate entities: $i)1st:$ sample space:${6\choose 1}=6$. Probability that is $4$:$\space {|(4)|\over 6}={1\over 6}.\space$$ 2nd:\space$sample space:$\space$all possible $1st$ rolls combined with all possible $2nd$ rolls:$\space {6\choose 1} {6\choose 1}=6*6=36.$ Probability that adds up to ...


1

For $i=1,\dots,N$ let $X_{i}$ take value $1$ if number $i$ is not drawn and value $0$ otherwise. Then $X:=X_{1}+\cdots+X_{N}$ equals the number of numbers that are not drawn. Then $\mathbb{E}X=\mathbb{E}\left(X_{1}+\cdots+X_{N}\right)=\mathbb{E}X_{1}+\cdots+\mathbb{E}X_{N}=\mathbb{P}\left(X_{1}=1\right)+\cdots+\mathbb{P}\left(X_{N}=1\right)$ An expression ...


0

The symmetry of this problem provides another way to look at it: you choose a coin at random and look at one of its faces at random. That face shows something (maybe heads, maybe tails). First question: What is the probability that the other face of the coin is the same as the face you can see? The answer to that question is clearly $\frac23$, since two ...


1

Hint: Satisfaction of condition $X_2=m$ means that a number in $\{1,2,3,6\}$ was thrown exactly $n-m$ times. What is the probability that $k$ of these times it was a number in $\{1,2,3\}$?


2

$P(X=5)=\dfrac{\binom{4}{4}}{\binom{100}{5}}$ $P(X=6)=\dfrac{\binom{5}{4}}{\binom{100}{5}}$ $P(X=7)=\dfrac{\binom{6}{4}}{\binom{100}{5}}$ $\dots$ And in general: $$\forall{n\in[5,100]}:P(X=n)=\dfrac{\binom{n-1}{4}}{\binom{100}{5}}$$ In words: Take ball #$n$, and choose another $4$ balls out of balls #$1,\dots,n-1$.


2

Count how many ways $n$ can be the largest number. If you replace the balls, there are $n^5$ ways they can be $\leq n$, minus $(n-1)^5$ ways they are all less than $n$. If you don't replace the balls, there are $n-1\choose4$ ways that the largest is $n$.


1

All of these statements can be false. What follows is more or less the standard counterexample of a local martingale that is not a martingale. I've stolen the details from an MO post of mine which constructs something slightly different. (To avoid search-and-replace I'm keeping my process called $Y$ instead of $\beta$.) Set $T=1$. Let $r(t)$ be any ...


0

Problems such as this can be solved by applying probability theory (see the other Answers). In this case a different approach is possible. We can use symmetry. Let us assume instead that the gambler has three coins in his pocket. The first one is fair, the second one is heads-heads and the third one is tails-tails. The gambler picks one coin at random from ...


0

An alternate method to compute $E[X]$ (where $X$ is the non-negative random variable "distance of second-furthest point from the origin") is to note that $$E[X]=\int_0^\infty P(X>r)\,\mathrm dr =\int_0^1 P(X>r)\,\mathrm dr =\int_0^1 \bigl(1-P(X\le r)\bigr)\,\mathrm dr $$ If $p_r$ is the probability that a single random point is at distance $\le r$ ...


0

For the update equations, they treat alpha, beta, and bj as independent of aij. If you see it this way I am sure you will see how they get the partials. I think their derivation could have been better myself. For example, they could have arrived at (27) from (26) by moving lambda to the other side without multiplying by aij and summing over j. Maybe I'm ...


2

The problem is equivalent to Coupon collector's problem. The expected value is $$\mathbb{E}(X) = N H_N \approx N \, ln \, N$$ where $H_N$ is $N$-th harmonic number. Here $\mathbb{E}(X) \approx 2364.64$ The idea in solving this problem is calculating the expected number of people such that the number of different birthdays we have written increases ...


2

Using the formula for conditional probability: $$p(\text{fair}\mid\text{head})$$ $$=\frac{p(\text{fair and head})}{p(\text{head})}$$ $$=\frac{1/4}{3/4}$$ $$=1/3$$


2

Let $A$ be the event : "The coin is fair" Let $B$ be the event : "Heads appears." We have $P(B)=\frac{1}{2}\times \frac{1}{2}+\frac{1}{2}\times 1=\frac{3}{4}$ and $P(A\cap B)=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$ So, we have $P_B(A)=\frac{1}{3}$


1

Outline: Inetead of $\mathbf{P} \left( - S < \frac{Y - \mu}{z/ \sqrt{n}} \leqslant S \right),$ use $\mathbf{P} \left( - z < \frac{\bar X - \mu}{S/ \sqrt{n}} = Y \leqslant z \right).$ Find the limit of $Y$ using Slutsky's Theorem (see Wikipedia, if not in your text). Use the CLT on $\bar X$ and the fact that $S$ tends to constant $\sigma$. For an ...


0

If $Z$ is independent of $X$ and independent of $Y$, then $(X,Z )\sim (Y,Z)$. Indeed, if $s,t\in\mathbf R$, then \begin{align} \mathbb P(X\leqslant s,Z\leqslant t)&=\mathbb P(X\leqslant s)\mathbb P(Z\leqslant t)\quad\mbox{ since $X$ and $Y$ are independent}\\ &= \mathbb P(Y\leqslant s)\mathbb P(Z\leqslant t)\quad \mbox{ becauseĀ }X\sim Y \\ ...


0

The probability of being within distance $z$ from the center is $z^2$ and the probability of being more than $z$ from the center is $1-z^2$. For $n-1$ points to be within distance $z$ from the center, it can happen in 2 ways: Some point $P_0$ is more than $z$ from the center (probability $1-z^2$) and the remaining $n-1$ points within $z$ from the center ...


0

Define $$f(a_1,\dots,a_n):=\mathsf E \left[\left(X_0-m_0\right)+a_1\left(X_1-m_1\right)+\ldots+a_n\left(X_n-m_n\right)\right]^2;$$ this can be rewritten in terms of $\lambda_{i,j}$, namely, $$f(a_1,\dots,a_n)=\lambda_{0,0}^2+2\sum_{j=1}^n\lambda_{0,j}a_j+\sum_{i,j=1}^n\lambda_{i,j}a_ia_j. $$ We have to find the critical points of $f$. Notice that ...


0

Hint: Use that $$(1_{[0,\tau]}(t))^2 = 1_{[0,\tau]}(t)$$ and $$1_{[0,\tau]}(t) = 1_{[0,t]}(\tau).$$ So, $$\mathbb{E}(X_t) = \mathbb{E}(1_{[0,t]}(\tau)) = \dots$$


2

Regardless of whether or not Plura is at the gym, Carla has a 20% chance of showing up at the gym. Therefore, the probability that Plura will meet Carla at the gym is 20%. Similarly, the probability that Plura shows up at the gym is 15%, so the probability that Carla meets Plura at the gym is 15%. The probability that Plura AND Carla show up is ...


0

Thanks to the comment given by David Z, I got the answer. There can only be two possibilities, first is three cameleons and two cameleons, and second is five cameleons. The first case is already calculated in the question. Here I will only focus on second case. Assume that the five chameleons are named $1,2,3,4,5$. Suppose we pick one chameleon randomly, ...


1

The number of permutations without fixed points in $S_5$ is given by: $$ 5!\left(\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}\right)=\color{red}{44}\tag{1} $$ by the inclusion-exclusion principle, hence the wanted probability is: $$ \frac{44}{4^5} = \color{red}{\frac{11}{256}}\approx 4,3\%.\tag{2}$$


1

Define events: $$F_1 = \text{"first product is faulty"} \\ F_2 = \text{"second product is faulty"} \\ A = \text{"products chosen from $A$"} \\ B = \text{"products chosen from $B$"}.$$ Then, we require \begin{eqnarray*} P(F_2\mid F_1) &=& \dfrac{P(F_2\cap F_1)}{P(F_1)} \\ && \\ &=& \dfrac{P(F_2\cap F_1 \mid A)P(A) + P(F_2\cap F_1 ...


2

Let $X$ be the event of a faulty element (and $X_n$ the $n^{th}$ element being faulty). And let $A$ and $B$ be the event of picking both elements from the manufacture A and B, respectively. We look for $P(X_2\mid X_1)$. Since nothing's said, let's assume that all events are independent, since otherwise we miss information. Note that since $A$ and $B$ are ...


1

One way to approach it is as follows. Firstly, $Y$ is either $X_2$ or $X_3$ and with equal probability because $f_{X_1,X_2,X_3}$ is symmetric wrt those two variables. So we begin by conditioning on the event that $X_2=Y$: \begin{eqnarray*} E(X_2\mid X_1=x, Y=y) &=& E(X_2\mid X_1=x, Y=y, X_2=y)P(X_2=y\mid X_1=x, Y=y) + \\ && E(X_2\mid X_1=x, ...


2

HINT: Since the company opens at most one store in a block, if a town has $n$ blocks, we can represent any possible way of opening stores by a string of $n$ zeroes and ones: a $0$ represents a block in which the company does not open a store, and a $1$ represents a block in which the company does open a store. The other restriction means that we want only ...


0

Consider leftmost part: there must be either store, then gap, then something or gap then something. So $f(n)=f(n-2)+f(n-1)$.


1

Here are some computations that sort of ended in a rather messy result. However, here it is written down anyway. Will have to see if I (or someone else) can bring it to a conclusion. Let the distribution $\cal{D}(n,k)$ denote the number of distinct values when $k$ values are drawn from $\{1,\ldots,n\}$: i.e., $A_k\sim \cal{D}(n,k)$, while ...


0

Call the people $1,2,3,\dots,N$. Define random variable $X_i$ by $X_i=1$ if Person $i$ is good at both mathematics and computer science, and by $X_i=0$ otherwise. Then the number $Y$ of people good at both is $X_1+\cdots+X_N$, and by the linearity of expectation $E(Y)=E(X_1)+\cdots+E(X_N)$. Note that $E(X_i)=\Pr(X_i=1)$. By independence, this is ...


2

Note that if $X=4$ is the event that the final score is four, $N=1$ is the event that the die is rolled once, and $N=2$ is the event that the die is rolled twice, then $$P(X=4) = P(X=4\mid N=1)P(N=1) + P(X=4\mid N=2)P(N=2).$$ Solving for $P(X=4\mid N=2)$, $$P(X=4\mid N=2) = \frac{P(X=4) - P(X=4\mid N=1)P(N=1)}{P(N=2)}.$$ You calculated that $P(X=4) = ...


3

In terms of conditional probabilities, $$p(4|\text{twice})=\frac {p(4\cap \text{twice})}{p(\text{twice})}$$ $$=\frac{3\times(\frac {1}{36})}{\frac 12}=\frac 16$$


4

Your answers and reasoning are correct for parts 1 and 2. Given that the die is thrown twice implies that the first die throw is either a 1 a 2 or a 3. In each of these cases there is only one result for the second die that will cause your final total score to be 4 (if the first die was a 1, second die must be a 3; if first die was a 2, second die must be ...


2

In fact $\lim\limits_{\varepsilon\,\downarrow\,0} (a+\varepsilon) =a$, so there is no separate number $a^+$ that is equal to $\lim\limits_{\varepsilon \, \downarrow\,0} (a+\varepsilon) =a$. The notation $F(a^+)$ does not mean the value of $F$ at a number called $a^+$; rather it is just a shorthand for ...


0

CDF continuous. If the random variable $X$ is 'continuous' then the CDF $F_X(x)$ is a continuous function. Examples: (1) The CDF of $X \sim Unif(0,1)$ is $F_X(x) = x,$ for $0 \le x \le 1$ (and $0$ for smaller values of $x,$ and $1$ for larger values of $x$). For a continuous random variable $X$, we have $P(X = a) =0.$ Positive probability is assigned to ...



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