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Assuming that the misprints are independently distributed across the book, and equally likely to be on any page, the number of misprints $X$ on a given page is binomially distributed with parameters $n = 1000$ ($=$ number of misprints) and $p = 1/1000$ ($=$ probability of a misprint ending up on the given page). Thus, the probability of there being at least ...

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No, your expression has a number of problems: First of all, even if you assume a binomial distribution applies to your situation, your expression only calculates the probability of exactly three errors, not at least three. Second, the binomial coefficient is incorrect: it should be $\binom{1000}{3}$ if you were to apply such a formula. Third--and this ...

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The number of ways 1000 mistakes can happen in 1000 pages is $1000^{1000}$. Assuming all of them are equally likely, we calculate the required probability as follows. The number of ways given page can have at most 2 mistakes is $999^{1000} + 1000(999)^{999}+\frac{1000*999}{2}999^{998}$. Therefore the required probability is 1-\frac{999^{1000} + ... 4 The claim is false, as shown by the following counterexample: let A and C be independent events with P(A) = P(C) = 1/3, and let B be the event "A or C", which has probability P(B) = 5/9. Then P(B \mid A) = 1 > P(B) and P(C \mid B) = P(C)/P(B) > P(C), but P(C \mid A) = 1/3 = P(C). EDIT: An even more concrete example (in the lines of ... 3 If D is a domain in \mathbb{R}^n, and x_0 \in D is a point, then the distribution of Brownian motion started at x_0 stopped at the time when it leaves D is the harmonic measure of \partial D with base point x_0. In the plane you can often use conformal invariance to calculate harmonic measure. Since you know that the harmonic measure of the ... 3 If we replace uniforms on (0,1) by uniforms on (0,w), the resulting random variable Y has the same distribution as wX, where X has Irwin-Hall distribution. In particular, \Pr(Y\le y)=\Pr(wX\le y)=\Pr(X\le \frac{y}{w}). It follows that if f_X is the density function of the Irwin-Hall, then Y has density f_Y(y)=\frac{1}{w}f_X(y/w). In a ... 3 The usual model here is a Poisson model: the number X of misprints in a randomly chosen page has roughly Poisson distrobution, mean (and therefore parameter) \lambda=\frac{1000}{1000}. We have \Pr(X\ge 3)=1-\Pr(X\le 2). And \Pr(X\le 2)\approx e^{-1}\frac{1^0}{0!}+e^{-1}\frac{1^1}{1!}+e^{-1}\frac{1^2}{2!}. 2 Use the fact that S=X_1+X_2 \sim \operatorname{Poisson}(a_1+a_2), then compute the conditional PMF: \begin{align} P(X_1=x_1|X_1+X_2=n) &= \frac{P(X_1+X_2=n|X_1=x_1) P(X_1=x_1)}{P(X_1+X_2=n)}\\ &=\frac{P(X_2=n-x_1)P(X_1=x_1)}{P(X_1+X_2=n)}\\ &=\frac{\frac{1}{(n-x_1)!}a_2^{n-x_1} e^{}-a_2 \frac{1}{x_1!}a_1^{x_1}e^{-a_1}}{\frac{1}{n!}(a_1+a_2)^n ... 2 The difference between the terms "probability measure" and "probability distribution" is in some ways more of a difference between terms rather than a difference between the things that the terms refer to. It's more about the way the terms are used. A probability distribution or a probability measure is a function assigning probabilities to measurable ... 2 Integral:\int_{\mathbb{R}}f(0)e^{px^2}dx=f(0)\int_{\mathbb{R}}e^{px^2}dx$$is Gaussian integral, see here how to calculate this integral, so:$$\int_{\mathbb{R}}e^{px^2}dx=\sqrt{\frac{\pi}{-p}}$$So p=\sqrt{\frac{-p}{\pi}}. 2 The variance of this random vector is \begin{bmatrix} 1 & \rho \\ \rho & 1 \end{bmatrix}. Level sets of the density are ellipses, which are circles only when the correlation \rho is 0. With a positive-definite symmetric matrix, one can always rotate the two coordinate axes to diagonalize the matrix; then one has two linear combinations of the ... 2 Alternatively, you can integrate by parts twice. First use u = x^2, v = e^{-x^2}. 2 I would say you can integrate by parts: \displaystyle I=\int_{0}^{\infty} x^{3}e^{-x^2}dx,\quad u = x^2, v' = xe^{-x^2}, u'=2x, v = -\frac{1}{2}e^{-x^2} \displaystyle I=\left[-x^2 \cdot \frac{1}{2}e^{-x^2}\right]_{0}^{\infty}+\int_{0}^{\infty}xe^{-x^2}dx=-\frac{1}{2}\left[e^{-x^2}\right]\int_{0}^{\infty}=\frac{1}{2} 2 Substitution u=x^2 leads to:$$\int_{0}^{\infty}x^{3}e^{-x^{2}}dx=\frac{1}{2}\int_{0}^{\infty}ue^{-u}du=\frac{1}{2}\mathbb{E}U$$where U has exponential distribution with parameter \lambda=1 (so \mathbb E(U)=\frac{1}{\lambda}=1) 2 Your answer is correct. The probability of the test passing is the probability that all oranges will pass the test, and since the oranges are independent, it is equal to the probability that the first orange will pass the test times the probability that the second orange will pass the test times etc, meaning it is equal to$$(1-\frac15)^{10}\approx 0.1074$$... 2 The following, quoted from the question, is wrong:$$ \Pr[N=n] = \Pr[X_1,\ldots,X_{n-1} \le X_0] = \Pr[X_1 \le X_0] \cdots \Pr[X_{n-1} \le X_0] = F(X_0)^{n-1}. $$This would be correct if it said$$ \Pr[N=n\mid X_0] = \Pr[X_1,\ldots,X_{n-1} \le X_0\mid X_0] = \Pr[X_1 \le X_0\mid X_0] \cdots \Pr[X_{n-1} \le X_0\mid X_0] = G(X_0)^{n-1} $$where G is the ... 2 First off, clearly it doesn't matter if we assume the boys and girls are all different or all the same, so we assume all boys are identical, and all girls are identical. We now create a pictorial representation for each arrangement: In this arrangement we place a bar for each boy and a star for each girl. The following is a valid configuration: ... 2 Hint: Note that  A\cup \emptyset = A, thus: P(A\cup \emptyset) = P(A)+P(\emptyset)+P(A\cap\emptyset) = P(A), now P(A)>0 \implies ? 2 If X is uniformly distributed in (0, 10), then f_{X}(x) =  ?. If Y is uniformly distributed in (0, 20), then f_{Y}(y) =  ?. Since X and Y are independent, then f_{X, Y}(x, y) =  ?. 1 This is a balls and urns problem solvable using a combination of inclusion-exclusion and stars&bars |a|\leq 10 \Rightarrow -10\leq a \leq 10. Make a change of variable x_1 = a+10\Rightarrow 0\leq x_1 \leq 20 Do similarly to the other letters, so we are at x_1 + x_2 + x_3 + x_4 = (a+10) + (b+10) + (c+10) + (d+10) = (a+b+c+d)+40 = 18+40 = 58 ... 1 For any measure space (X,\mathcal M,\mu), \mu(\varnothing)=0 by definition. Since a probability measure is just a special case where \mu(X)=1, we still have \mu(\varnothing)=0. 1 A_n is increasing in n because the event$$ \bigg\{ \delta \sum_{i=1}^{n-1} X_i > b \bigg\} $$is a subset of the event$$ \bigg\{ \delta \sum_{i=1}^n X_i > b \bigg\}, $$simply because X_n is nonnegative. 1 Hint:$$P(\emptyset)=P(\emptyset\cup\emptyset)=P(\emptyset)+P(\emptyset)$$This because \emptyset\cap\emptyset=\emptyset, i.e. the sets are disjoint. 1$$Ε[Y]=Ε|Χ|=\int_{-1}^{1}|x|f_X(x)dx=\int_{-1}^{0}-x\frac12dx+\int_{0}^{1}x\frac12dx=\not 2\int_{0}^{1}x\cdot\frac1{\not2}dx=\left.\frac{x^2}{2}\right|_{0}^{1}=\frac12$$and since P(Z=0)=P(X=0)=0:$$E[Z]=E\left[\frac{X}{|X|}\right]=\int_{-1}^{0}\frac{x}{-x}\frac12dx+\int_{0}^{1}\frac{x}{x}\frac12dx=-\int_{-1}^{0}\frac12dx+\int_{0}^{1}\frac12dx=0$$and ... 1 The expression for the probability is gotten in this way: What is the probability that some specific j elements are all less than or equal to t, but that the rest are greater than t? That is obviously$$t^j(1-t)^{n-j}$$Now how many ways can I choose a that specific set of j elements? That answer is \binom{n}{j}. So the probaility that exactly ... 1 The distribution is discrete. Since the given probabilities add up to 1, the "missing" entries are all 0, so are not missing at all. To compute the correlation coefficient, you will need the covariance and the two variances. Let us start. From the table, E(XY)=(1)(1)(0.25)+(1)(2)(0.25)+(2)(1)(0.25)+(0)(0)(0.25). Now let us compute E(X). This can be ... 1 You can have a look at the following paper http://www.dm.fct.unl.pt/sites/www.dm.fct.unl.pt/files/preprints/2012/7_12.pdf which tackles your problem 1 There are several paper for this problem. See for example: On the Distribution of the Product of Independent Beta Random Variables – Applications Products of Beta distributed random variables BETA PRODUCTS WITH COMPLEX PARAMETERS 1 Note the convolution of density functions is the density of the sum:$$g^+ = f \ast f$$Depending on the given g^+ and knowing about the symmetry and support of f you can try to solve the equation$$g^+(s) = \int_{a}^{b} f(t) f(s-t) \ \mathrm dt = (f \ast f)(s)

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Maybe you are looking for a statement like this: The higher the probability of high values of $X$, the more likely it is that the firm defaults. That is, you are comparing probability distributions of $X$. Mathematically, you could use the notion of first order stochastic dominance (FOSD). Suppose you have two different distributions for $X$. Denote the ...

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