New answers tagged

0

Let $X$ be the number of people with disease. It's clear that $X$ has binomial distribution with $n=10000$ and $p=\frac{1}{1200}$. We can calculate the exact probability using the formula $$P(X=8) = {10000\choose 8}\biggl(\frac{1}{1200}\biggr)^8\biggl(1-\frac{1}{1200}\biggr)^{(10000-8)}$$ The exact calculation of the above quantity is hard. We can ...


0

(this answer assumes "flush" is meant to also include the possibility of straight flush or royal flushes) From earlier example, we know $Pr(A)=\frac{4\binom{13}{5}}{\binom{52}{5}}$ and we know $Pr(B)=\frac{4^5\binom{13}{5}}{\binom{52}{5}}$ Let us attempt to calculate $Pr(A\cap B)$. Our sample space will be all ways in which Abe and Bernard can be dealt ...


1

This random variable has the binomial distribution. Hence, the probability that an island with $10000$ inhabitants has precisely $8$ people born with that particular disease is given by $$ {10000\choose 8}\biggl(\frac1{1200}\biggr)^8\biggl(1-\frac1{1200}\biggr)^{(10000-8)}\approx0.1387. $$


2

\begin{align}P(WW)&=\sum_{i=1}^{11}P(WW\mid bin=i)P(bin=i)=\sum_{i=1}^{9}\frac{\dbinom{10-(i-1)}{2}}{\dbinom{10}{2}}\frac{1}{11}\\[0.2cm]&=\frac{1}{11\cdot45}\sum_{i=1}^9\dbinom{11-i}{2}=\frac{1}{11\cdot 45}\sum_{i=0}^{8}\dbinom{2+i}{2}\\[0.4cm]P(BB)&=\sum_{i=1}^{11}P(BB\mid ...


0

Let $X$ represent the state of becoming ill. Thus, $X \sim \text{Binomial}(3,0.5)$. Then, $$ P(X = 1) = {3 \choose 1}0.5^1\cdot 0.5^2 = 0.375 \to 37.5\%$$


2

No, the statement is not true. Let for example $Ω=\{1,2,3,4\}$ with $p(ω)=1/4$ for each $ω$ and let $X=\{1,2\},\, Y=\{2,3\}, \, Z=\{3,4\}$. Then $$P(X\cap Y\cap Z)=P(\emptyset)=0$$ but you can check that $X,Y,Z$ are pairwise independent with $P(X)=P(Y)=P(Z)=\frac12>0$.


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The equation as stated without additional assumptions on the independence of $A$ and $B$ is false. Consider the following counterexample. Consider the experiment where we flip two fair coins in sequence. We have the equiprobable sample space $\{HH, HT, TH, TT\}$ Let $A$ be the event that the first coin is a head. $Pr(A)=0.5$ Let $B$ be the event that ...


0

The first mistake is when you calculated the probability that exactly one die lands on $6$, as you included the case when both dice are $6$ twice (first when the loaded die is $6$, and secondly when the fair die is $6$) - so you need to subtract the case when they were both $6$ two times:- $$\begin{align}P(\text{exactly one die lands on 6}) &= ...


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In short, you erroneously divided by $\Pr\{\text{sum is 9}\}$ when calculating $\Pr\{\text{sum is 9 and exactly one die lands on 6}\}$. $$ \begin{align*} \Pr\{\text{sum is 9 and exactly one die lands on 6}\} & = \Pr\{(6,3),(3,6)\} \\ & = \Pr\{(6,3)\} + \Pr\{(3,6)\} \\ & = \left(\frac{1}{2}\right)\left(\frac{1}{6}\right) + ...


1

Let $E$ be the event the workout ends early. Then $$\Pr(E\mid W^c)=\frac{\Pr(E\cap W^c)}{\Pr(W^c)}.$$ You have calculated $\Pr(W^c)$. We now need to find $\Pr(E\cap W^c)$. The event $E\cap W^c$ can happen in two ways: (i) we drink coffee and the workout ends early, or (ii) we drink the commercial drink and the workout ends early. The probability of (i) is ...


2

You are describing the multi-armed bandit problem. You have $N$ slot machines (decks), each with some unknown expected payoff (win %). You want to maximize your payoff, which demands a careful balance between exploration (gathering data about each slot machine/deck) and exploitation (using the slot machine/deck that appears to be best so far). There are a ...


0

Suppose $T_1,T_2$ are the times until the first and second machines fail, respectively. With the exponential distribution, thinking about $\Pr(T_1>t)$ is simpler than thinking about $\Pr(T_1\le t)$. One has \begin{align} & \Pr(T_1>t) = e^{-t/10} \quad (\text{for }t>0) \text{, and} \\ & \Pr(T_2>t) = e^{-t/15}. \end{align} Let $T_3$ be ...


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The question is phrased as whether $$ \sum_{x\in\mathcal X, y\in\mathcal Y}p(x,y)\log\,p(x) = \sum_{x \in \mathcal X} p(x)\log\,p(x) \quad \text{?} $$ I disapprove of using the same symbol, $p$, for two different functions. If one instead writes $p_X(x)$ with capital $X$ and lower-case $x$ in the appropriate places, one can then understand such things as ...


1

As the OP points out, each ordered triple of rolls has the same probability, namely $\frac 1{6^3}$. Assume the last roll is a $4$ (so event $A_4$ has happened). Let's proceed to enumerate the rolls that comprise $A_4\cap B$, working by cases according to the first roll. Clearly, if we specify the first and third rolls, the only thing left to specify is the ...


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Let $X\sim \exp(10)$, $Y\sim \exp(15)$ So for the first question, we are trying to find $P(X<Y)$. Notice we can take an integral over the values of $Y$ to find the result: $$ P(X<Y)=\int P(X<y)\cdot P(Y=y\pm \epsilon)\, dy=\int F_X(y)f_Y(y)\,dy $$ Where $F_X$ is the cdf of $X$, and $f_Y$ the density function of $Y$. For the second, we are trying ...


0

I would use Markov inequality $P(Y\ge n)<\frac{\mathbb E(X)} n$ because either $Y=1-X$ and $n$ are greater than zero. This way we have $$\lim_{n\to\infty} P(-\sum_n{\log(1-X_n)}\ge n)<\frac{\mathbb E(-\sum_n{\log(1-X_n)})} n =\frac{\sum_n{\mathbb E(\log(1-X_n)})} n$$ Using Jensen's inequality for the concave function $\log(x)$ we have ...


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There are $2^{m+n}$ outcomes in total. Now treat the string of $m$ consecutive heads as a single head with $n$ more tosses. So our string of $m$ heads can be in $n+1$ different positions relative to the other $n$ coin tosses. The outcomes of those $n$ tosses do not matter since your question asks for at least a string of $m$ heads. So the probability should ...


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Method 1: Abbreviate $Y:=E[X|\mathcal F]$. Let $g(x)$ denote the right-hand derivative of $\varphi$ at $x$. Because $\varphi$ is strictly convex, we have $\varphi(x)>g(m)(x-m)+\varphi(m)$ for all $x\not=m$. Thus, $$ \varphi(X)\ge g(Y)(X-Y)+\varphi(Y) $$ with strict inequality off $\{X=Y\}$ (almost surely). Taking conditional expectations in the ...


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Simply because $$ \sum_{y\in\mathcal Y}p(x,y)= p(x). $$ Really, you had done all difficult work already.


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In general, if you have $k_1$ items of type 1, $k_2$ of type 2, ..., $k_n$ items of type $n$, the number of orderings is given by the multinomial coefficient: $\begin{align} \binom{k_1 + k_2 + \dotsb + k_n}{k_1, k_2, \dotsc, k_n} = \frac{(k_1 + k_2 + \dotsb + k_n)!}{k_1! \, k_2! \, \dots \, k_n!} \end{align}$ To prove this, take each of the items as ...


0

Your answer is correct. You have ten positions to fill. You can fill five of them with white tokens in $\binom{10}{5}$ ways. You can fill the remaining five of them with black tokens in $\binom{5}{5} = 1$ way. Hence, the number of ways you can arrange the tokens is $$\binom{10}{5}\binom{5}{5} = \binom{10}{5}$$ If you instead had five red tokens, three ...


0

I would count the number of three digits numbers (which is 900) and then count how many of them are a multiple of 7 (which is the greatest integer less than or equal to 900/7).


0

$$IE^i_A\subseteq B, \,\,\,\forall 1\le i\le k_A \implies \bigcup_{i=1}^{k_A}IE^i_A\subseteq B \implies A\subseteq B \implies P(A)\le P(B) $$


2

Yes, except that: I would replace $\frac{100}{7}$ with $\frac{99}{7}$ (to see why this is important consider the analogous question asking for how many multiples of $7$ there are in $\{700, \ldots, 999\}$), and the quotients aren't quite right as written, but we can repair them with floor notation, e.g., $\lfloor\frac{999}{7}\rfloor = 142$.


1

You are absolutely right, and that's the shortest way of doing it as far as I know.


1

If two events are independent, it means that, by definition $P(A\cap B) = P(A)P(B) $, such that if $P(A) > 0$ and $P(B) > 0$ then $P(A\cap B) >0$. Alternatively, $P(A|B)=P(A)$, where $P(B)>0$. So, basically, your diagram is incorrect. Assume for simplicity that $A\cup B = \Omega$ and $A\cap B = \emptyset$ and neither $B$ and $A$ are empty sets. ...


2

Isn't the frequency just once per 20 minutes? "Hello" is printed only when the number is 20. That is with 1/20 probability. So, even if the rate varies each time, over an average of 7 hours, by 5% error margin, you can say the rate is 1 "hello" per 20 minutes.


1

In every minute the probability that it will print hello is $\frac{1}{20}$ assuming that the "Random" is fair in JAVA Let $A$ denote the event that it will print hello per one minute so $\mathrm {P}[A]=\frac{|A|}{|\Omega|}=\frac{1}{20}$ Therfore the rate is one hello per $20$ minutes


1

Personally I would use a graph rather than a transition matrix. The first graph here corresponds to your transition matrix, with the blue numbers representing the number of coins to toss in the next round and the green numbers of probability of moving to the next state. But you are not interested in tosses which are all heads, i.e. do not change the ...


1

It might simplify slightly to say we just require the chain to reach state $4$ at any point, starting at $0$, and we can forget about the variable $T$. So re-define $u_i$ as $$u_i = P(X_n = 4 \text{ for some $n$}\mid X_0=i).$$ Our transition probability matrix: $$ \begin{matrix} \qquad 0 \qquad & 1 \qquad & 2 \qquad & 3 \quad & 4 ...


0

The best that you can expect is a convergent subsequence: $(X_n)$ bounded in $L^1\implies E(X_n)$ bounded $\implies$ $E(X_n)$ has a convergent subsequence.


0

No it's not true. Take the Lebesgue measure $\lambda$ and the sequence $\{1_{[0,n]}\}_{n \in \mathbb N}$ of indicator functions. These are obviously measureable as closed intervals are in the Borel sigma algebra. They are all bounded by the constant function $1: \Omega \to \mathbb R$ but $$lim_{n \to \infty}E(1_{[0,n]})=lim_{n \to \infty} ...


1

It is not true. Take, for example, $X_n=0$ almost surely when $n$ is odd and $X_n=1$ almost surely when $n$ is even.


0

Assuming everyone who can refund, does, and assuming you mean $x\le .9$ rather than $x=.9$, and $f(x)=e^{-x}$ rather than $e^x$, then $E(X)=-2P(X<.9)+3P(X>.9)=3-5\int_0^.9 f(x)=-2+5e^{-.9}.$


0

Yes, the desired formula holds without the assumption of normal distribution. Denote a probability space by $(\Omega,\mathcal{F},\mathbb{P})$. Assume that we know the joint probability density function $f_{X,Y,Z}(x,y,z)$. Write $$ g_{X,Y}^{z}(x,y) := \frac{f_{X,Y,Z}(x,y,z)}{f_{Z}(z)}$$ Note that $$g_{X,Y}^{z}(x,y)=f_{X,Y|Z}(x,y|z)$$ where the right-hand ...


3

a. Yes, since the density is not zero over $[0,3]$, then $$P(X\leq 1) = \int_0^1 f_X(x)\,dx.$$ b. In this case, it is asking to find the cdf $$P(X\leq x).$$ By Remember to include the values for $F_X$ for all real $x$ I believe it is saying to make sure you are explicit about the cases $$P(X\leq x)$$ when $x< 0 $ $0\leq x \leq 3$, and $x>3$. ...


1

That a sequence of members of $\{0,1\}$ approaches $0$ means that after some finite number of terms it is always $0$. For $\{X_n\}_{n=1}^\infty$ to converge almost surely to $0$, it would have to be the case that $$ P( \text{for some finite number } N, \text{ for all } n\ge N,\ X_n=0) = 1. $$ The second Borel–Cantelli lemma tells us that that happens only ...


4

You are choosing 4 people out of a set of 30 people, without replacement, and order does not matter. There are $30\choose4$ possible ways to do that, and Joe and his friends being picked is just 1 of those ways. Thus the probability of this occurring is $$\frac{1}{30\choose4}=\frac{1}{27405}\approx0.00003648969... $$


0

Let $q_i=P(Y_i=1)$. The variables $q_i$ are called nuisance parameters and to deal with them we have to put a prior on them and then average out over them. To be more precise, the model is the following: we have parameters $p_1,\dots,p_N$ and $q_1,\dots,q_N$. We know the $p_i$, but we don't know the $q_i$ and so must put a prior distribution on them, say ...


2

Consider an $1>\epsilon>0$. For all $n$, $|X_n-0|=0$ with probability $1-1/n$, and $|X_n-0|=1$ with probability $1/n$. So $$Pr(|X_n-0|>\epsilon)=\frac{1}{n}$$ since $|X_n-0|>\epsilon$ only when $X_n=1$. $1/n$ converges to 0 as $n \to \infty$. If $\epsilon \geq 1$, then $Pr(|X_n-0|>\epsilon)=0$ for all $n$.


2

For simplicity, consider the case where $X = 0$ and $X_n$ is the indicator function of an event $E_n$. "$X_n$ converges almost surely to $0$" says that with probability $1$, only finitely many of the events $E_n$ occur. "$X_n$ converges in probability to $0$" says that the probability of event $E_n$ goes to $0$ as $n \to \infty$. Consider a case where ...


1

Literally solving the first integral without being clever at all, the result is $$ \left.\frac{n-1}{2}\cdot\frac{x^2}{2} \right\rvert_{-\frac{1}{n}}^{\frac 1n}. $$ You rewrote this as $$ \left.\frac{n-1}{2}\cdot 2 \frac{x^2}{2} \right\rvert_{0}^{\frac 1n} $$ which is not correct, because $$ \left.\frac{n-1}{2} \cdot\frac{x^2}{2} \right\rvert_{-\frac 1n}^{0} ...


3

You're looking for the waiting time distribution for either bus, which is exponential. This is because you know the bus has equal probability of arriving at any given time in an hour, and that the distribution is memoryless. So the waiting time for a bus has density $f(t)=\lambda e^{-\lambda t}$, where $\lambda$ is the rate. To understand the rate, you know ...


0

The concept of uniform integrability gives a generalization of the conditions of the dominated convergence theorem. Uniform integrability of a sequence $\left(Y_n\right)_{n\geqslant 1}$ means that $$\lim_{x\to +\infty}\sup_{n\geqslant 1}\mathbb E\left[\left|Y_n\right|\mathbf 1\left\{|Y_n|\gt x\right\}\right]=0.$$ We can prove that a uniformly integrable ...


0

Note that the support iof $X$ is dependent on the unknown parameter $\theta$, so $X_{(n)} \le \theta$, and $\hat{\theta}_n = X_{(n)}$. For consistency you have to show that $$ \lim_{n\to\infty}P(|X_{(n)} - \theta| \ge \epsilon) = 0, $$ due to the fact that $X_{(n)} \le \theta$, you have \begin{align} \lim_{n\to\infty}P(|X_{(n)} - \theta|\ge\epsilon) &= ...


1

For brevity, I'm going to say "we win" if the game stops with a sum of $5$, and "we lose" if the game stops with a sum of $7$. Let $p_k$ be the probability that we win, given that the first roll is $k$. Suppose we roll a $1$ and then a $2$. After the $2$, the probability that we will win is $p_2$, that is, all that matters after; that roll is that we last ...


2

(First number the 30 cards to distinguish them.) Let S be the set of all possible selections of 10 cards, and let $E_i$ be the set of selections which do not include either card from pair $i$, for $1\le i\le5$. Using Inclusion-Exclusion, $|E_1^c\cap\cdots\cap E_5^c|=|S|-\sum|E_i|+\sum|E_i\cap E_j|-\sum|E_i\cap E_j\cap E_k|+\cdots$ $\hspace{1 ...


0

Your solution is correct, though it may be incomplete. I'll start with what you have, including the $\frac{4n^2}{4n^2}$ term: $$\frac{(4n^2)(4n^2-2)(4n^2-4)\cdots(4n^2-2(2n-1))}{(4n^2)(4n^2-1)(4n^2-2)\cdots(4n^2-(2n-1))}$$ Removing a factor of $2$ from each term in the numerator gives: ...


2

The table you've linked is a pretty nonstandard format for a z-score table, but it seems to be referring to a situation like this: The area under the curve between $0$ and $t$ is the probability of a normally distributed variable falling between $0$ and $t$. Using the fact that the curve is symmetric about $0$, you can deduce the probability of a normally ...


0

You need to know more. For problems like these, it is always good to take particular examples, even extreme examples. In what follows assume that, in Reality, $1\%$ of the population has the disease. Example A: A person without the disease NEVER tests positive. In that case, a positive test result is absolute proof that the person actually has the ...



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