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6

Arrange ZLGEBR, then replace Z with AA.


5

Step $I$: Pick $6$ balls and weigh them $3$ on each side. Step $II$: $1$. If the balls balance each other, then weigh the remaining two balls and the heavier one will tilt the balance on its side. $2$. If the balls do not balance each other, then the heavier ball should be one of the three balls on the side the balance tilts. From these three balls, ...


3

Pascal's triangle is based on an identity for choosing unordered samples taken without replacement (no item sampled more than once). Binomial coefficients. The number of unordered samples of size $k$ from a larger set of size $n$ is $$C(n, k) = {}_n C_k = {n \choose k} = \frac{n!}{k!(n-k)!},$$ for integers $0 \le k \le n.$ I have shown three common ...


3

Given that $Y$ is Bernoulli distributed, it will have probability mass function $P(Y=0)=(1-p)$ and $P(Y=1)=p$. As Frank has helpfully commented, we can condition on $Y$. When $Y=0$, we have $P(\frac{X}{Y-X}<0|Y=0)=P(-X<0|Y=0)=P(X>0|Y=0)$ When $Y=1$, we $P(\frac{X}{Y-X}<0|Y=1)=P(\frac{X}{1-X}<0|Y=1)=P(X>1|Y=1)$ Thus, we have the ...


3

Just a guess, but perhaps $f$ is the expectation operator and $x$ is a random variable? If you're trying to understand why $$E[(X-E(X))^2] = E(X^2)-E(X)^2\;,$$ then this post might help. Another derivation can be found here.


2

The "Note: $P(X=x^2)=P(X=x)$" is simply wrong. They should have stated: $P(X^2=x^2)=P(X=x)$   so that: $$\begin{align}E(X^2) & = \sum\limits_{x^2\in\{1,4, 9, 16\}} x^2 \;\mathsf P(X^2=x^2) & \color{silver}{= \sum\limits_{y\in\{1,4, 9, 16\}} y\; \mathsf P(X^2=y)} \\ & = \sum_{x=1}^4 x^2\; P(X=x)\end{align}$$ PS: More generally ...


2

$Y=0$ if and only if $X_1=X_2=X_3=X_4=0$. Since the four random variables are independent, the probability this happens is $$\Pr[X_1=0]\cdot\Pr[X_2=0]\cdot\Pr[X_3=0]\cdot\Pr[X_4=0]$$ which is equal to $$ (p^0(1-p)^{1-0})\cdot(p^0(1-p)^{1-0})\cdot(p^0(1-p)^{1-0})\cdot(p^0(1-p)^{1-0}) = p^0(1-p)^{4} $$


2

Let $X_i$ be a single draw from the uniform distribution. Then it follows that it has a CDF of $$F(x) = \left\{\begin{array}{lr} x, & 0\le x\le1,\\ 0, & \text{ Otherwise}. \end{array}\right.$$ For $N$ draws, then, what we are looking for is $Y=\max(X_1,X_2,\ldots X_N)$. The CDF of $Y$ is $$G(y)=\Pr(Y\le y)$$ Since $Y$ is the max, this means each ...


2

Label the men $1,2,\dots,50$. For $i=1$ to $50$, let $X_i=1$ if Man $i$ has a woman standing next to him, and let $X_i=0$ otherwise. The number $Y$ of men with a woman next to them is $X_1+\cdots+X_{50}$, so by the linearity of expectation $E(Y)=E(X_1)+\cdots+E(X_{50})$. This is $50E(X_1)$. It remains to find $E(X_i)$, which is $\Pr(X_i=1)$. We find ...


2

it rather depends on what information you have available! one simple approach might be to estimate the total size of the fleet, and the average number of years in service for a plane. when talking of number of planes the skewed distribution of plane size means that passenger numbers are not so helpful, though you could assume an approximately lognormal or ...


2

Not complicated, but doesn't fit nicely into a Comment: (a) OK (b) E(L) = \$160 [P(Ticket)] + \$0 [P(No Ticket)] = \$30 (c) Pay for 3 hours = \$6 < \$30, so paying is better.


2

Here's an explicit counterexample. Let $U=|X|$ and $V=|Y|$. Then $\text{Var}(X)\geq \text{Var}(Y) \Rightarrow E(|X|)\geq E(|Y|)$ is equivalent to $E(U^2)\geq E(V^2) \Rightarrow E(U)\geq E(V)$ for $U,V$ continuous in [0,1]. Consider a $\text{beta}(\alpha,\beta)$; it has mean $\mu=\frac{\alpha}{\alpha+\beta}$ and second raw moment $\mu_2'=\mu ...


2

I will set out a solution strategy and some hints, in keeping with the site's homework policy. It's quite an effective one in general - in this case, checking the back of my envelope, I used four simple lines of algebra and three rough sketch plots (two PMFs and an overlaid contour plot) - but other approaches are available. In particular it can make life ...


2

Linearity of expectation is (sometimes surprisingly) always valid. If I roll two dice then the expected value of the first is $3.5$ and the exopected value of the second is $3.5$ and the expected value of the sum is $7$. Now you'll say "Sure, that's cause they are independent." But now do this: I roll the first die and then manually place the second die so ...


1

It should be $\sum_{n=0}^\infty P(X=2n)=\sum_{n=0}^\infty F(2n)-F(2n-1)=\sum_{n=1}^\infty1-2^{-2n-1}-1+2^{-2n}=\sum_{n=1}^\infty2^{-2n-1}=\frac{1}{2}\sum_{n=1}^\infty4^{-n}=\frac{1}{8}\sum_{n=0}^\infty4^{-n}=\frac{1}{6}$


1

Almost right, assuming of course that Zipf's law with the basic exponent of 1 applies to your corpus. But you have to divide by the approximated total of all the frequencies, which if your set contains $N$ words is $$T = c \sum_{k=1}^N \frac{1}{k} \approx c \ln N.$$ Thus if there are (say) 100000 words, $$ f_1 = \frac{90000}{1 \cdot 90000 \ln 100000} ...


1

$f_r = 90{,}000/r$ would mean the most frequent word occurs $90{,}000$ times, the second one $45{,}000$ times, the third $30{,}000$, and so on. If you're doing modeling then I would think you have the actual counts so you can see how good a fit this is. But you haven't given us any of your data.


1

Your first probability is incorrect. The probability of choosing an empty box on the first try is $\frac{3}{5}$; your $\frac{2}{5}$ is the probability of choosing a box that contains an item. Your second probability is good: with only 4 boxes left and 2 that contain items you want, the probability of getting an item is $\frac{1}{2}$. Mutliply them ...


1

Chebyshev's Inequality is true for any $c > 0$, but you are right that it only provides useful information for $c > 1$. This is actually surprisingly easy prove. Define $\mu = E(X)$ and $\sigma^2 = E((X-\mu)^2)$. Observe that for any $c \geq 0$ we have $\mathbb{1}\left\{\left|\frac{X-\mu}{\sigma}\right|\geq c\right\} \leq \frac{(X-\mu)^2}{\sigma^2 ...


1

You are confusing two different uses of "p" in this problem. First, p = 0.07 is the proportion of defective light bulbs (the stated defect rate). It simply means that the probability of a randomly selected bulb being defected is 0.07. Second, the p-value of the hypothesis test is given as 0.087. This means that the janitor has conducted a test. Typically ...


1

This density function is for an uncorrelated bivariate normal distribution. So it is the product of two normal density functions. Recognizing that is the easy path. Otherwise transform to polar coordinates to integrate. Of course, you need the integral over the plane to be unity.


1

Recall that $ \hat \beta_0$ and $ \hat \beta_1$ are determined by minimizing the term of the right-hand side. Thus, we have the pair of equalities $$\sum_{i=1}^n (y_i-\hat \beta_0-\hat \beta_1 x_i)=0$$ $$\sum_{i=1}^n (y_i-\hat \beta_0-\hat \beta_1 x_i)x_i=0$$ Using these relationships we see that $$\begin{align} \sum_{i=1}^n (y_i-\hat \beta_0-\hat ...


1

If $a=0$, then the correlation between $x$ and $y = ax + b$ is not defined. Computation of Pearson's correlation, would involve division by 0. In the computation of Spearman's correlation, all values of $y$ are equal, and there is no meaningful way to rank the $y$s. [Note: You give one formula for Spearman's correlation. Another valid computation is to ...


1

i.) Your first result is wrong because the question is related to the total waiting time, $X+Y$: $$P(X+Y<30)=\iint_{\{(x,y):x+y<30\}}f_{X,Y}(x,y)\ \ dxdy=$$ $$=\frac{1}{100}\int_0^{30} e^{-\frac{1}{10}x}\int_0^{30-x}e^{-\frac{1}{10}y}dydx=$$ $$=\frac{1}{10}\int_0^{30} e^{-\frac{1}{10}x}\left[1-e^{-\frac{1}{10}(30-x)} \right]dx=$$ ...


1

The sum of n independent Bernoulli random variables, all with the same p, is a binomial random variable with parameters n and p. For the case $n=4$, here are a few of the probabilities in the distribution: $P(Y = 0) = (1-p)^4,$ as you already know. $P(Y = 1) = 4p(1-p)^3,$ as in the hint from @Alex. You must take into account all of the sequences 1000, ...


1

It looks like you are given the correlation $\rho$ of the two random variables $X$ and $Y$. There is a formula connecting the correlation to the covariance, and it is $$\rho(X,Y) = \frac{\text{Cov}(X,Y)}{\sigma_X\sigma_Y}.$$ Do you see what to do from here?


1

I think by $p = 0.6$ you mean $\rho = 0.6,$ which is the correlation. If so, you have only to look at the relationship between covariance and correlation. If not, I wonder what $p$ might be.


1

Your density function matches the distribution $Beta(3, 1)$ which has mean $\mu = E(X) = 3/4$ and $\sigma^2 = 3/80 = 0.0375$, so $\sigma \approx 0.1936.$ You can verify all this with elementary calculus. Then $P(5/8 < X < 7/8) = P(|X = \mu| < 1/8),$ which is in a form convenient for the use of Chebyshev's inequality. (How many standard deviations ...


1

This question has been plaguing internet ranking sites for a while, e.g. reddit. Fortunately it has a nice theoretical solution. It involves a bit of statistics; the Ruby code is: require 'statistics2' def ci_lower_bound(pos, n, confidence) if n == 0 return 0 end z = Statistics2.pnormaldist(1-(1-confidence)/2) ...


1

You have $n(n-1)/2$ possible pairs of which $2(n-1)-1$ pairs are incorrect hence the probability to pick an incorrect pair is $(4(n-1)-2)/(n(n-1))$. Will making an invalid guess you gain information on where is, at least one, of the incorrect options. Thus the probability of getting an incorrect pair decrease. You have now $n(n-1)/2-1$ possible pairs (all ...



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