# Tag Info

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To clarify the notation: The variable "D" in bullet point 2 is the "demand" that is referenced in bullet point 3?

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Let $C$ be the event a randomly chosen patient is critical, and $D$ be the event that the event that a randomly chosen patient dies. We want $\Pr(C|D)$. By the usual formula for conditional probability, we have $$\Pr(C|D)=\frac{\Pr(C\cap D)}{\Pr(D)}.\tag{1}$$ We want to find the probabilities on the right of (1). The probability $\Pr(C\cap D)$ is, as you ...

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The probability of A given B is equal to P(A|B) = $\frac {P(B | A)\, P(A)}{P(B)}$. In your problem, B = "Given that a patient dies", and A = "the probability that the patient was classified as critical". This is known as Baye's Theorem. I suggest you read up on it as it is very important in probability.

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Let $A=$ the event that the patient is in critical condition, $B=$ the event that the patient is in serious condition, and $C=$ the event that the patient is in stable condition. Let $D=$ the event that the patient dies. The we know $P(A)=0.2, P(B)=0.3, P(C)=0.5, P(D|A)=0.3, P(D|B)=0.1, P(D|C)=0.01$. Then ...

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Suppose we want to list all the possible ordered pairs of distinct elements from the set {1, 2, 3, 4, 5}. One way to do this is first to list all the possible permutations of the set {1, 2, 3, 4, 5}: 12345 12354 12435 12453 $\vdots$ 54321 and then throw away all but the first two entries in each: 12 12 12 12 $\vdots$ 54 Of course this produces each ...

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Use the law of total probability to break up $P(S_N=x)$ into the cases where $N$ is equal to every possibly integer: $$P(S_N=x)=\sum_{k=0}^\infty P(S_N=x|N=k)\cdot P(N=k)$$ Then each $P(S_N=x|N_k)$ is just $P(S_k=x)$, for a constant $k$, which you can compute.

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$S_n$ is binomial for a given n: $S_n\sim \mathcal{B}(n,p)$ $N$ is poisson: $N\sim \mathcal P(\lambda)$ Use: \begin{align} \mathsf P(S_N=x) & = \mathsf P(\bigcap_{n=x}^\infty (S_N=x\cap N=n)) & \text{since n\geqslant x} \\[1ex] &= \sum_{n=x}^{\infty} \mathsf P(S_n=x)\mathsf P(N=n) & \text{by law of total probability} \\[1ex] & = ... 1\begin{align*} \frac12 x_1 + 7 x_2 &= 60 \\ x_1-2x_2 &= 8 \end{align*}\frac12x_1 = 4+x_2.4+x_2+7x_2 = 60.8x_2 = 56x_2 = 7 \implies x_1 = 22.$$I used no words. Probably. 1 You can use algebra to solve this problem. Represent the first number by x and the second number by y. Your first statement is thus just \frac12x+7y=60 and your second statement is x-2y=8. Do you know how to solve this system of equations? 1 There are two possible interpretations of the question: (1) what is the probability that all birthdays are concentrated in just two months, (2) what is the probability that all birthdays take place in exactly two months. You're trying to answer (2). Your reasoning assumes that person 1 and person 2 have a birthday in two different months, but that is not ... 1 Well it doesn't matter what order you pick them in, which your working implies. Hmmm this is rough working out and may be completely wrong, so be forewarned. Choose the 2 months \binom{12}{2}=66 ways. Now every person can be in either of the 2 months. As there are 5 people there are 2^5=32 possibilities. We then subtract 2 cases (where their birthdays ... 2 We give an alternate approach that (barely) involves cases. It will give a number to check your answer against. The box that gets 3 balls can be chosen in 7 ways, and for each way the balls it gets can be chosen in \binom{7}{3} ways. What about the rest? The remaining 4 balls can, almost, be assigned in 6^4 ways. Except that we do not want any ... 0 What you've calculated in Method 1 is the change in entropy from all five dice showing not 1 to all five dice showing 1. You haven't included the possibility that one or more die starts out showing 1. Hence you've counted only 5^5 microstates. What you've calculated in Method 2 is the change in entropy from some of the five dice show not 1 ... 1 Your first way of counting is perfectly good. If you want to count another way, let us invent 2 identical ghosts. The seats for them can be chosen in \binom{5}{2} ways, since ghosts aren't allowed to drive. (There is a problem with taking their picture for the licence.) The rest of the seats can be filled in 4! ways. 0 Hints: 52\choose 5 different ways to make a 5 card hand. 4\choose 1 different ways to choose a suit. 13\choose 3 different ways to pick 3 from that suit. 13\choose 4 for 4 of that suit. 13\choose 5 for 5 of that suit. That leaves you with 39 cards left in which you must choose 2, 1, or none of them to fill out the 5 cards ... 1 Hint. To count the number of five-card hands with exactly three spades, choose 3 spades from 13. . . . . . C(13,3) ways; choose 2 cards from the other 39. . . . . . C(39,2) ways. So the number of such hands is C(13,3)C(39,2). Now you need to allow for the fact that the question said at least three spades; divide by the total possible ... 0 For (a), consider Alice, who owns two motorcycles, and Bob, who owns one. What was the prior probability that Alice would be selected for the survey? What was Bob's probability? For (b), one might think that the department could determine the total number of motorcycle owners with exactly one registered motorcycle each, and multiply the estimated fraction ... 0 Presumably, the opinion R of someone doesn't change when their asked multiple times, but it may depend on the number of motorcycles owned O. If so, then the expected value of your sample proportion p is \sum P(O=i)E(X|O=i). Therefore, your estimator is biased if there is substantial stratification of opinions based on number of motorcycles owned. In ... 0 WithOUT replacement. For the 6th bait to be the 5th worm drawn, then 4 worms and 1 leech must be drawn first, and 5 worms and 9 leeches drawn after, in some order. So you need to count the ways to select 5 worms and 1 leech, the ways to arrange 4 worms and 1 leech, and the ways to arrange the remaining 5 worms and 9 leech. Divide the product of that ... 0 If you are drawing at a time then \sum_{i=2}^3 \frac{\binom{3}{i} \times \binom{15-i}{5-i}}{\binom{15}{5}}. 2 Why guess randomly? First off, your first three digits are nearly uniquely determined by your state of birth. This information is not hard to deduce, as most people's current state is also their state of birth. The group numbers are generated not consecutively, but in a defined order based on birthdate, so if you can get an estimate of the time of birth, you ... 0 I would say that you did work this "normally", but if you wanted to work this without using the complement principle you could break it into cases: 1) If all 4 dice are the same, there are 6 possibilities. 2) If exactly 3 of the dice are the same, there are \binom{4}{3} ways to choose the 3 that will be the same, 6 ways to choose their number, and 5 ways ... 1 Knowing Fatou's lemma, which applies for non-negative sequences, we want to use it with such sequences related to X_n and Y. Since Y bounds the absolute value of X_n, it is natural to use this with Y_n:=Y\pm X_n. Then in the bound we will be able to put the integral of Y outside the \liminf. 1  n= 100 ,  \mu = 8 \times 10^5 ,  \sigma = 3 \times 10^5 . Since sample size is more than  30  we can use CLT and use normal approximation. a)  P(\bar{X} > 825000)  =  P((\bar{X} - \mu)/ \sigma > (825000-800000)/300000)  =  P(Z>1/12) . Where  Z  follows standard normal distribution. b)  P((7900000-800000)/300000 < Z < ... 1 I got a different answer. May be I made mistake somewhere. OK we randomly choose M pair. The way how you choose the fragments is equivalent to M independent drawing of two numbers from the set of n (and putting them back). So always the probability the pair be equal (1,1) is p^2 like you wrote. Now lets denote N_w(M) as a function of M ... 0 If P is a probability measure, then it satisfies countable additivity and nonnegativity. So consider X_1=B, X_2=A\backslash B, and X_n=\emptyset for all n\geq 3. Because B\subset A, \{X_n\} constitutes a countable collection of disjoint set. So countable additivity implies$$ P(A)=P(\cup_nX_n)=\sum_n P(X_n)=P(B)+P(A\backslash ...

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Hint: For the second tallest boy to have rank $r$, you'd need $r-2$ taller girls, and $1$ taller boy. Choose these taller children, order them, and then choose the second tallest boy. For instance, $P(X=4)=\frac{12\cdot {4\choose 2}\cdot 6\cdot 11}{16\cdot 15\cdot 14\cdot 13}=\frac{99}{910}$

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suppose the second player places matching card, the probability is: $$\frac{\binom{50}1\binom41}{\binom{50}1\binom{49}1}\approx0.0816=8.16\%$$ Select one card of 50, then 1 out of 4 remaining matching same type. At the first turn only.The probabilities for other later turns depend on location of cards and is a little-little bit lengthy.Hope you got the ...

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This is just a straightforward computation. Let $Y_i = 1_{\{X_i=1\}}$ \begin{align*} \text{Var}(N_W) &= \text{Var}\left( \sum_{i=1}^M Y_{j_i}Y_{j_i+1} \right) \\ &= \sum_{i=1}^M \text{Var}(Y_{j_1}Y_{j_1+1}) + \sum_{i \neq k} \text{Cov}(Y_{j_i}Y_{j_i+1}, Y_{j_k}Y_{j_k+1}) \\ &= M\text{Var}(Y_{j_1}Y_{j_1+1}) + 2\sum_{i=1}^{M-1} ...

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This is because we use multiplication theorem in case of dependent events and addition theorem in independent event. Clearly some students selected for a group cannot be selected for another group. But suppose selecting three flavours of icecream for yourself, even if you take one flavour,you can sleect any flavours for other two choices, here choices are ...

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Add up the following: The probability that it lands on the red side exactly $4$ times is $\dbinom{6}{4}\cdot\dfrac{5^2}{6^6}=\dfrac{375}{46656}$ The probability that it lands on the red side exactly $5$ times is $\dbinom{6}{5}\cdot\dfrac{5^1}{6^6}=\dfrac{30}{46656}$ The probability that it lands on the red side exactly $6$ times is ...

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Actually the second convergence can be almost surely Fix $\epsilon >0$. Let $\tau$ be the first moment when the total probability of elements in $D_n$ is greater than $1 - \frac{\epsilon}{2}$. By the fact that every element will eventually fall into $D_n$, we know $\tau$ is finite almost surely. To strictly prove it, begin with the element with biggest ...

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If you intend the chain lengths to averaged over all slots, including the empty slots, then the answer is just the average number of elements per hash slot. This is obviously $\dfrac{n}{m}$ no matter what the probability distribution is. If you intend the chain lengths to be averaged over only the non-empty slots, then we want the expected number of ...

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$G(x)=Pr(X\leq x|a<X\leq b) = Pr(X\leq x, a<X\leq b) /Pr(a<X\leq b) = Pr(a<X\leq x) / Pr(a<X\leq b)$ Now numerator is simplify as $F(x)-F(a)$. Denominator is $F(b)-F(a)$. That's the answer.

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The probability that the maximum $Y$ is $\le k$ is the probability that all the $X_i$ are $\le k$. This is $\left(\frac{k}{6}\right)^n$. The probability that $Y=k$ is the probability that $Y$ is $\le k$ minus the probability that $Y$ is $\le k-1$.

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Your guess is true, at least if the measure spaces $\Omega'$ and $\Omega''$ are $\sigma$-finite. Since measures on the product of two $\sigma$-finite measure spaces are uniquely determined by their values on measurable rectangles and both side of the equation you guessed define measures on such a product space, it suffices to check the equality for ...

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The distribution in each bin is binomial. You have $n$ "trials", which are the $n$ elements. For any given bin, you have a probability $p = \frac{1}{m}$ of any given element winding up in the bin. The mean of the binomial distribution is given by $\mu = np$, which in your case is $$\mu = \frac{n}{m}$$

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There is no known distribution function but you can create by your own. Order the biased coin $1,2 \ldots ,M$. Now in the binomial coefficient we have $\binom Mk \times p^k \times q^{M-k}$ . Here for a fixed $k$ it will be $(p_{i_1}p_{i_2}p_{i_3} \ldots p_{i_k})(q_{i_1}q_{i_2}q_{i_3} \ldots q_{i_k})$ product will run over all possible choosing of $k$ in ...

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Here's a hint. Assume that at some stage $n$, you know that that the maximal value you have observed is $k$. At the next step $n+1$, what can happen? Either next $\xi_{n+1}$ is smaller than your observed maximum, in which case your chain remains at the same place. (Can you find the probability of $\xi_{n+1} \leq k$?). Or $\xi_{n+1}$ can be higher than ...

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You may want to look for "Poisson binomial distribution", e.g., http://en.wikipedia.org/wiki/Poisson_binomial_distribution

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They are probably using one symbol to denote the risk-adverse measure and the other to denote the risk-neutral measure. Often authors use $\mathbb{P}$ and $\mathbb{Q}$, respectively, but sometimes $\tilde{\mathbb{P}}$ is used for the risk-neutral measure. However, there's no reason why the author can't use the notation in your question. See Girsonov's ...

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In the theory of Markov chains, the transition matrix is often denoted $P$ so there it is clearer to use some other symbol like $\mathbb P$ for probability. But such decisions are up to the author, and will vary from person to person.

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No difference--one can use either $P$ or $\mathbb P$ or $\mathbf P$ or $\Pr$ or...

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Gamma(a) is nothing but a chi-square distribution. And you know t-distribution is ratio of Standard normal and chi-square distribution upon number of degree of freedom. From this relation you will get answer.

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What's the chance that C wins? In his first game he plays someone who has already won a game, either A or B. C must win that game to have any chance (50%). He then plays the player who lost that first game (either B or A). If C loses, then both A and B have one win, and they play each other in the next round. One of them will have two wins and win the ...

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Answer on (i): $\left\{ D_{n}\rightarrow\infty\right\} ^{c}=\bigcup_{k=1}^{\infty}\bigcap_{n=1}^{\infty}\left\{ X_{n}\leq k\right\}$ so that $P\left(\left\{ D_{n}\rightarrow\infty\right\} ^{c}\right)\leq\sum_{k=1}^{\infty}P\left(\bigcap_{n=1}^{\infty}\left\{ X_{n}\leq k\right\} \right)$. For a fixed $k$ we find $P\left(\bigcap_{n=1}^{\infty}\left\{ ... 0 The function is undefined on$y_2\ge y_1$and$y_2\le0$and integration assumes the function is continuous on the given domain, and if the function is not defined on some of the domain it is definitely not continuous at those points. 1 The function you are trying to integrate does not exist on the whole$\mathbb R^+$, because the radicand will sooner or later become negative. This implies that the integral is meaningless. 1 Define$P$the chance a player will win the tournament when he gets a new opportunity to, playing against last game's victor. Now what can happen? If he loses, two games in a row are won by the same player, and the tournament is over , ie.$P_{L}=0$If he wins twice, he wins the tournament:$P_{WW} = \frac1{2}\frac1{2} = \frac1{4}$If he wins first, and ... 1 Clearly$A$and$B$have the same chance of winning, and$C$has whatever probability remains. So let's look at$A$. There are unique sequences for$A$to win in rounds$2$,$4$,$5$,$7$, ... (any nonmultiple of$3$, except for$1$). These sequences (of match winners) are:$AABCAAACBAABCABCAAACBACBAA$etc. So the probability that$A\$ ...

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