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From 10 socks you are choosing 2, so you have $\binom{10}{2}$ possibilities. And to two socks to be different color you have to pick one blue and one white so you have $\binom{5}{1}\binom{5}{1}$ possibilities. So the final result is $\frac{25}{45}=\frac{5}{9}$.

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We use a fairly crude counting approach, in order to rely minimally on intuition. There are $\binom{52}{4}$ equally likely ways to choose the positions of the $4$ Aces. We now count the "favourables." Maybe the first two Aces are in positions 1 and 2. That leaves $\binom{50}{2}$ for the rest. Maybe they are in positions 2 and 3. That leaves ...

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Let's call the set of all words $W$, the subsets of those containing $\sf ace$ $A$ and the set of words containing $\sf fd$ $F$. We are interested in $|W \setminus (A \cup F)|$, now $$|A \cup F| = |A| + |F| - |A \cap F|$$ Let's look at $|A|$. We have to make words out of the four "letters" - now $\sf ace$ counts as one letter! - $\sf (ace)$, $\sf b, d, ... 2 Your proof works quite all right! Arthur already gave the proper comment, this is just an alternative way to prove the statement - maybe a bit more straight forward, to elaborate a bit: We want to show for$X$positive, that $$(E(X) +E(\frac{1}{X})) \geq 2$$ holds, but this is due linearity of the expectation operator equivalent to $$E(X +\frac{1}{X}) ... 2 Your Error First of all Andre has posted a very nice answer. Already Upvoted it. I am trying to look at where all you have gone wrong. Too long for an comment, so posting it here. Let us look at the last case. You have 48! ways of the non-ace cards. Now the remaining cards are all aces anyways. So the question of multiplying this by 4 \cdot 3 does ... 2 Here is a bijection between set of all permutation with the first Ace followed by an Ace and set of all permutations with the first card being an Ace$$B^nAAX <-> AB^nAX$$where A stands for an Ace, B for not an Ace, 0\leq n \leq 48 and X contains 2 Aces and 48-n non-Aces. Hence the probability of an Ace following the first Ace is equail to ... 2 You are successful iff in both stages you throw the same number k of heads. If a\geq b this can be any number in the range [0\>..\>b]. Therefore the probability you are after amounts to$$p={1\over 2^{a+b}}\sum_{k=0}^b{a\choose k}{b\choose k}\ .$$2 The probability of going n times right and then n times left is just the product$$ {a\choose n}\left({1\over2}\right)^a{b\choose n}\left({1\over2}\right)^b= {a\choose n}{b\choose n}\left({1\over2}\right)^{a+b}. $$Of course you have to sum this over all possible n, that is for 0\le n\le\min(a,b). By using Vandermonde summation formula, if a\le b ... 2 The answers so far are correct, but I wanted to provide a direct combinatorial proof for your original correct answer. (2)^{(a+b)} is the total number of possible outcomes of all the flips, so we want to show that a+b\choose a is the total number of flips that result in you returning successfully. Assume, wlog, that a is the lesser number. Well, each ... 2 You wrote (.005^n). You should have written (.005^1). If you try solving it again, you should get the right answer. Based on the rest of your work, it seems like you know what you're doing, and this was just a typo. 2 You have a simple Markov chain on n+1 states (enumerated by the number of hats in the first drawer). Simple calculation shows that in steady state all states are equiprobable - hence they have probability \frac 1 {n+1}. A man doesn't wear a hat only in states 0 and n and in those states probability of non wearing a hat is \frac 1 2. Hence he ... 2 No. Given independently identically distributed Y_1, \ldots, Y_n, define {Y_1}' as Y_1 and {Y_2}' to Y'_n as the order statistics of Y_2 to Y_n. Then the order statistics of {Y_1}' to {Y_n}' have the given distribution, but {Y_1}' to {Y_n}' are not independently identically distributed. 1 Or to put it another way--there are 10^4 numbers 'between' 0000 and 9999 and you want ONE. So it's \cfrac{1}{10^{4}}. 1 Common sense answer: On a day he wore a hat, he flips a coin to decide which drawer to put it in. He has a fifty fifty chance of putting it in the same drawer as the other hat and a fifty fifty chance of putting it in the other drawer. So on half the days the hats are in the same drawer and on half the days they are in different drawers. On the days they ... 1 The problem is that it's simply not true that "then the value of each X_j should be the same". Hmm. Example: Say \Omega=[0,1] with P equal to Lebesgue measure. Define X,Y:\Omega\to\Bbb R by X(t)=t and Y(t)=1-t. Then X and Y are iid, but X(\omega)\ne Y(\omega). 1 It can help to look at some sample paths. From Bernt Øksendal's Stochastic Differential Equations: The image shows five sample paths of a geometric Brownian motion process \{X_t\}_{t\ge0}. The paths are different functions of t. Each one of them shows the values of X under different outcomes \omega_1,\dots,\omega_5 in the sample space \Omega. ... 1 The Cauchy distribution with location parameter \mu and scale \theta has PDF \frac{1}{\pi \theta \left( 1 + \left( \frac{x-\mu}{\theta} \right)^2 \right)} (in the variable x) and characteristic function \mathrm{e}^{\mathrm{i} \mu t - \theta |t|} (in the variable t). Since \mathrm{e}^z \neq 0 for every z \in \mathbb{C}, this characteristic ... 1 Fixing my original calculation Thanks to Shailesh, we now know that the expression$$ p_2 = \frac{48! \cdot 4 \cdot 3}{ {_{52}P_{50}} } $$Should be replaced with:$$ p_2 = \frac{48!}{ {_{52}P_{48}} } $$Because that after extracting 48 non-Ace cards, we'll necessarily have the second card as Ace. Moreover, I had a tiny mistake in the denominator. Instead: ... 1 \mathsf P(\bigcap_{v\in V}\psi_v) = 1-\mathsf P(\bigcup_{v\in V}\psi_v^\complement)\geq 1 - \sum_{v\in V} \mathsf P(\psi_v^\complement) Because it is likely that the complements of the events \psi_v are not disjoint. The event could not-occur at more than one point in the set V. 1 Did you get {5\choose 3} = 10 ? {5\choose 3} = \frac{5!}{3!2!}=10 1 (1) Yes, the Poisson point process is homogeneous (constant intensity \lambda throughout \mathcal{B}) so the distribution of n is Poisson with parameter: \lambda\times(\text{Area of \mathcal{B}(0,R)}) = \lambda\pi R^2. (2) Given n,\; the n points are uniformly distributed throughout \mathcal{B}. Since a circle's circumference is proportional ... 1 You concluded correctly that F_W(w)=F_V(e^w). Also we know the distribution of V so:$$f_W(w)=f_V(e^w)e^w=e^{-e^w}e^w=e^{-we^w}$$More directly you can derive this by differentiating the RHS of F_W(w)=F_V(e^w)=1-e^{-e^w} 1 This is not the case. The same probabilities also hold for Y_\lt=\min(Y_1,Y_2) and Y_\gt=\max(Y_1,Y_2) (where Y_1 and Y_2 are independently uniformly distributed), yet those are not independently uniformly distributed. 1 You first have$$\log T_n = \sum_{i=1}^n \log C_i$$(by definition of T_n and the C_i's, taking the logarithm on both sides of T_n = \prod_{i=1}^n C_i). From this, you "artificially" divide by \frac{1}{n} on both sides, to make it look like an average over the C_i's:$$ \frac{1}{n}\log T_n = \frac{1}{n}\sum_{i=1}^n \log C_i $$Since all choices ... 1 We are given that the number of correctly answered questions X is a binomial random variable with parameters n = 40 and p = 0.5. We are also given that$$\Pr[X > N] > 0.1, \quad \Pr[X > N+1] < 0.1.$$Now approximate X as$$Y \sim \operatorname{Normal}(\mu = np = 20, \sigma^2 = np(1-p) = 10),$$we have$$\Pr[X > N] \approx \Pr[Y > ... 1 The events are independent and you have a 1/2 probability of choosing a white ball each time due to the symmetry. So the answer is$\frac{1}{2^k}$for the first question. Second answer is$[\sum_{i=0}^n(\frac{i}{n})^k]/(n+1)$1 You should treat "ace" as a token and then you are permuting this token with b,f,d, totally$4!$and we have$6!-4!$left. Then you treat fd as a token and we have$6!-4!-5!$left. But you have minused those have both "ace" and "fd" twice so we need to add once back, giving us the final answer$6!-4!-5!+3!=582\$

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