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Hint: Let $X$ and $Y$ be continuous and independent random variables and let $Z=X+Y$. Then $$f_Z(z)=\int_{-\infty}^\infty f_X(x)f_Y(z-x)\mathrm dx.$$

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Let $S=\{1,2, \dots ,100\}$, $A=\{x \in S: 4|x\}$, $B=\{x \in S: 5|x\}$, $A \cap B=\{x \in S: 20|x\}$. Then $|A|=25$, $|B|=20$ and $|A \cap B|=5$. Using the inclusion-exclusion formula $$|A \cup B|=|A|+|B|-|A \cap B|=25+20-5=40$$ Since $|S|=100$ that gives $p=40/100=2/5$.

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The event $A\cup B$ occurs if $A$ occurs, if $B$ occurs or if both $A$ and $B$ occur. The event $A\cap B$ occurs if both $A$ and $B$ occur.

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For every Bernoulli random variable $X$ with parameter $p$ there exists some Poisson random variable $Y$ with parameter $p$ such that $$P(X\ne Y)\leqslant p(1-\mathrm e^{-p})\leqslant p^2.$$ Thus, for every binomial random variable $S$ with parameter $(n,p)$ there exists some Poisson random variable $Z$ with parameter $np$ such that $$P(S\ne Z)\leqslant ... 0$$ P_0(t)=\phi(0,t) \qquad P_1(t)=\frac{\partial\phi}{\partial s}(0,t)  P_n(t)=\frac1{n!}\,\frac{\partial^n\phi}{\partial s^n}(0,t) $$0 the critical value can be looked up in the \chi^2-table. The number of degress of freedem is df=(r-1)\cdot (c-1) r=number of rows c=number of columns And here is \chi^2_{\alpha}=\chi^2_{0,01} The null hypothesis is: There is no connectedness between the "gender" and "speedness" The alernative hypothesis is: There is a connectedness between the ... 0 Since each ball can go into bin 1 or bin 2, mutually exclusively and exhaustively, it's a binomial distribution: \operatorname{P}(X=x) = \dbinom{10}{x} \dfrac{1}{2^{10}} \\ \quad = \dfrac{10!}{x!(10-x)! 2^{10}} The expected value is thus: \operatorname{E}[X] = \sum\limits_{x=0}^{10} x\cdot\operatorname{P}(X=x) \\ \quad = \sum\limits_{x=0}^{10} ... 0 Here is some elementary analysis. To follow your notation, let n be the number of balls in the bag, let i be the number of iterations of replacing the selected red balls with green balls, and let s (instead of \frac n{100}) be the number of balls you select on each iteration. Also, let's define p(i,g) to be the probability that after the ... 0 If exactly 3 of the eight coins are heads, then we don't need to worry about the bias of the coin flip. The other 5 coins flipped up tails. In order to assign an expected value to clustering we need a numerical measure. Let's assign it the count of tails between the first and last head. Thus the greatest cluster has a value of 0, the least a value of 5. ... 0 There are 4 equal spots, and 100 spins that each land on one spot This means that each spot should have 100/4 land on it, or 25 Since only 18 did, it should have landed on it 25-18 times or 7 times more than it did. 2 The random variable Z of your question has density function e^{-z/2} for z\gt 0 and 0 for z\lt 0. By your argument, we have$$\Pr(Y\le y)=\Pr(e^Z\le y).$$If y\le 1, then \Pr(e^Z\le y)=0, since Z\le 0 with probability 0. Thus the density function of Y is 0 for y\le 1. If y\gt 1, then$$\Pr(e^Z\le y)=\Pr(Z\le \ln y)=1-e^{-(\ln ...

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Randomly select an urn, draw a ball without replacement, then again randomly select an urn and draw a second ball. Let $D_n$ be the colour of the $n^{th}$ draw, and $U_n$ be the urn from which it is drawn. Using k for black, the urns are: $a = \{(w,4), (r,2)\}, b = \{(r,3), (k,3)\}$ Now, if we know a black ball is going to be removed in the second draw, ...

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Without loss of generality, let's sort the points by $x$-coordinate, so that $p_1$ is the left-most point and $p_{n}$ is the rightmost point. Now, consider the following observation. If $p_{i}$ and $p_{j}$ both belong to $S$, with $i < j$, then $y_{j}$ (the $y$-coordinate of $p_{j}$) must be less than $y_{i}$; if not; then $p_{i}$ and $p_{j}$ for a line ...

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Let $R_1$ and $R_2$ be the event of picking a red ball on the first or second draw, respectively. Then: $$P(R_1 \mid R_2) = \frac{2}{8} = \frac{1}{4}$$ Reason: If you've picked 1 red ball then $2$ of the remaining $8$ balls are red. The order of doing this does not matter, so if you know the second ball is red then there is a $\frac14$ chance of the first ...

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There are two ways for the second ball to be red. Either blue then red, or red then red. If a blue is picked first, then the chance of a red is $\frac{3}{8}$. If a red is picked first, the odds are $\frac{2}{8}$. For the first pick, red happens $\frac{1}{3}$ of the time and blue $\frac{2}{3}$. The overall odds of blue then red is $\frac{2}{3}$ times ...

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To do this problem you need to calculate the Chi-Square test of independence. Its not horribly complicated, and its well detailed in the Wikipedia article Here. After that, the number you get should be compared to the ${X^2}_{0.010}$ column in This table of numbers. Now that you know the method, it should be a piece of cake.

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You can compute this number very quickly if the number of possible values $n$ ($n = 10$ in your case) is small. For a sequence of length $N$, note that the sequence is an admissable sequence if and only if the prefix of length $N-1$ is an admissable sequence and the last two digits are 2 or more apart. So you can just compute the number of admissable ...

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I did it in excel. The entries are the number of acceptable strings of length 1 through 4 ending in the number on the left. $$\begin {array} {c c c c c ... 0 This doesn't classify as an answer,though some of the info could be useful,maybe something about continuity or general solution for b(-t)=-b(t) for example.(note the last 2 expressions are quadratic formula if either of a(t),b(t) are fixed,and a(t),b(t) are real-valued functions) also clearly a(t)=1,b(t)=0 are solutions ... 0 Whenever you have and, you want to multiply. For this problem, all you have to do is think about it this way. You want the probability that the 1st driver wears a seat belt and the second wears a seat belt and the third wears a seat belt. The probability that each one of them is wearing it is 80\%=0.8. The probability that all three of them wear seat belts ... 0 This is a simple normal distribution. You are given two things:$$\mu=\$6050\sigma = \$1500$$From this, you can get the z-score. The z-score is modeled as:$$z=\frac{x-\mu}{\sigma}$$The z-score will tell you how many standard deviations you are away from the mean, at some value x. From the z-score, you can obtain the area that's under the normal ... 0 To do a scientific experiment truly objectively, the ideal process is: Describe the experiment to be performed. Enumerate all possible outcomes of the experiment. Declare what conclusion you would draw from each outcome. Perform the experiment. Publish the results and the conclusion. People often simplify this process in ways that can introduce minor (or ... 0 Hint: The normal distribution is described by a probability distribution function$$f_X(x) = \frac{1}{\sqrt{2\pi}\sigma}\exp \left( -\frac{(x-\mu)^2}{2\sigma^2}\right).$$The probability a randomly selected house spends less than \6350 is given by$$\int_{-\infty}^{6350} f_X(x)\ dx.$$Simply perform a rescaling of x to get it to standard normal form, ... 0 I should point out that testing the hypothesis that the coin is fair is most definitely not equivalent to testing the hypothesis that the sequence of heads/tails it generates is random. The sequence can be random but biased; conversely, the coin can be fair but in a deterministic (non-random) fashion. 0 A p-value is actually defined (in Statistical Inference by Casella and Berger, for instance) to be a statistic, not just a number, and it is called valid if it satisfies$$ P_\theta[p({\bf X}) \leq \alpha] \leq \alpha $$for \theta \in \Theta_0. So the probability of seeing small values is genuinely low under the null hypothesis. This makes sense. ... 0 Note that your expected gain given that you first rolled a 5 or 6 is the same as your expected gain initially… you just get to start over. Using linearity of expectation, then, you can write your expected gain as$$ E[G]=\sum_{i=1}^{6}E[G\;\vert\;X_1=i]\cdot P[X_1=1]=-\frac{1}{3}+\frac{2}{3}+\frac{1}{3}E[G]=\frac{1}{3}+\frac{1}{3}E[G], $$and then solve ... 0 If the order of dice counts, then first choose 2 dice that must take care of the sixes. That can be done on \binom{6}{2} ways. For each other 4 dice there are 5 faces acceptable (a six is not allowed) giving 5^{4} possibilities. That leads to a total of \binom{6}{2}\times5^{4} possibilities. 0 No. For each combination of {male, female}, {married, single}, {grad, not grad}, you can calculate the number who gave that combination of responses (lumping in "no answer" with "opposite answer" for simplicity). If you do so, you end up with a negative number of single female non-college-grads. There's a simple shortcut to calculate that by adding and ... 2 First choose which (in order) will be sixes. How many ways are there to choose them? Then the other four dice can be any of five numbers. 1 Actually the function p is not the PDF of a real random variable because the PDF of a real random variable is defined on \mathbb R hence one should rather consider the function q:\mathbb R\to\mathbb R defined by$$ q(x)=\left\{\begin{array}{ccl}0&\text{if}&x\leqslant0\\ x&\text{if}&0\lt x\leqslant1\\ 2-x&\text{if}&1\lt ...

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Because of the symmetry, once ties are excluded, the chance of winning must be $\frac 13\approx 33.33333\%$ and the chance of losing must be $\frac 23\approx 66.6667\%.$ You are showing $31-69$. This looks to be caused by the fact that you round to whole percents throughout the calculation.

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to find the critical value you have to use the table of the t-Distribution. What ist the value of $\alpha$ and how many degrees of freedom do you have here ? greetings, calculus

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$$P=\frac{\binom{1}{1}\binom{n-1}{k-1}}{\binom{n}{k}}$$ $$P=\frac{\frac{(n-1)!}{(k-1)!((n-1)-(k-1))!}}{\frac{n!}{k!(n-k)!}}$$ $$P=\frac{(n-1)!}{(k-1)!((n-1)-(k-1))!}\frac{k!(n-k)!}{n!}$$ $$P=\frac{(n-1)!}{(k-1)!(n-k)!}\frac{k!(n-k)!}{n!}$$ $$P=\frac{(n-1)!}{(k-1)!}\frac{k!}{n!}$$ $$P=\frac{k}{n}$$

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What you're looking for is called the Inclusion–exclusion principle.

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There are $k$ choices and $n$ people. First person will choose a random number. Second person will choose a random number and will have $1\over k$ probability to choose the number chosen by others. Third person will have $2\over k$ probability. $n$-th person will have ${n-1} \over k$ probability. Geting the probability that way will be way too hard, ...

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If we are careful, we can use cumulative distribution functions to solve this question. Let $X$ be a nonnegative (I assume $X$ to be nonnegative because a negative radius is a bit senseless) random variable whose cumulative distribution function is given by $F_X:[0,\infty)\to[0,1]$. If we define $Y=\pi X^2$, then $Y$ is the area of a circle of radius $X$, ...

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Transforming to standard normal is not necessary nor helpful. The distribution of $Y$ is actually given in the question! Look up how to compute $P(Y>c)$ when $Y$ follows that distribution, and you are done. If the distribution of $Y$ wasn't given, you could find the distribution of $Y$ and/or the probability of the event $Y>c$ based on the ...

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$$P(X_1+X_3=n,X_1+X_2+X_3+X_4=N)$$ $$=P(X_1+X_3=n,X_2+X_4=N-n)$$ $$=P(X_1+X_3=n)P(X_2+X_4=N-n)$$ $$=\sum_{i=1}^nP(X_1+i=n)P(X_3=i)+\sum_{i=1}^{N-n}P(X_2+i=N-n)P(X_4=i)$$ $$=\sum_{i=1}^nP(X_1=n-i)P(X_3=i)+\sum_{i=1}^{N-n}P(X_2=N-n-i)P(X_4=i)$$ Edit (proof independence of sum) $$P(X_1+X_3=n,X_2+X_4=m)=\sum_{i=1}^nP(X_1+i=n,X_2+X_4=m)P(X_3=i)$$ ...

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I don't see any easier way than going over all options and summing the probabilities: Probability $\frac{3!}{6^3}=\frac{1}{36}$: 4,5,6 Probability $\frac{3}{6^3}=\frac{1}{72}$: 3,6,6 Probability $\frac{1}{6^3}=\frac{1}{216}$: 5,5,5 All together, you get $\frac{5}{108}$.

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You say that '...90% of the mails marked is indeed spam'. That should probably be changed into '...90% of the mails that is spam is indeed marked'. On base of$P\left[\text{spam}\right]=0.1$, $P\left[\text{marked}\mid\text{spam}\right]=0.9$ and $P\left[\text{marked}\mid\text{no spam}\right]=0.1$ we find: ...

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There seems to be an error either in the question or your understanding of it. If the question indeed says 90% of the mails it marks as spam are indeed spam and 90% of spam mails are correctly labelled as spam, this means $$P(\textrm{spam} \mid \textrm{marked}) =0.9, P(\textrm{marked} \mid \textrm{spam}) = 0.9,$$ and the ...

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Why did you decide ${P(marked\ as\ spam)} = 0.1$ is incorrect? In fact, this makes perfect sense.. Out of 100 mails, the classifier is expected to mark 10 as spam, 9 of which are actual spam and there's another spam mail which successfully bypassed the filter. In your question you wrote "and 9 of 90 non-spam as spam." Which is not the expected ...

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Hint: Show and use that the likelihood function is of the form $$L(\theta)=\mathrm{e}^{\theta n}\mathrm{e}^{-\sum x_i}\mathbf{1}_{\min(x_1,\ldots,x_n)\geq \theta}.$$

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Here is one way of doing this. To make things simpler, let's call the label the colors $1$ through $n$, and let's say you have $c_{i}$ stones of color $i$ (so in your example, $n = 9$). Let $C = \sum_{i} c_{i}$ be the total number of stones, and let's say you choose $m$ stones from this set. For question 1, without loss of generality let's say the desirable ...

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If we have log of the two ratios P(z=1|L)/P(z=0|L) how can be compute P(z=1|L) from this equation? Note that $$P(z=1|L)+P(z=0|L)=1$$ hence $$P(z=1|L)=\frac{ \frac{P(z=1|L)}{P(z=0|L)}}{1+ \frac{P(z=1|L)}{P(z=0|L)}}$$ that is, $$P(z=1|L)=\frac{\mathrm e^\ell}{1+\mathrm e^\ell},$$ where $$\ell=\log\left(\frac{P(z=1|L)}{P(z=0|L)}\right).$$ Likewise, $$... 2 The best you can hope for is \left\lceil\operatorname{lg}(152)\right\rceil=8 guesses in the worst case. If you are limited to 6 questions, you won't be able to guarantee a correct answer, in which case the game becomes maximizing your chance of winning. You could, for example, split off the 32 most likely cases (accounting for 161 of the 400 possible ... 1 From the multiset, S = \{\mathrm{Blue}^3, \mathrm{Red}^3, \mathrm{Green}^3, \mathrm{White}^3, \mathrm{Yellow}^3, \mathrm{Orange}^3, \mathrm{blacK}^1, \mathrm{grEy}^{18}\} Select 6 stones, find P(B \geq 1, R \geq 1). P(B \geq 1, R \geq 1) = P(B = 1, R \geq 1) + P(B = 2, R \geq 1) + P(B = 3, R \geq 1) \\ \quad = P(B=1)P(R\geq1 \mid B=1) +P(B=2)P(R\geq1 ... 0 The expected value of 5d8-20 is 5\cdot 4.5-20=2.5 if negative results are allowed. If the roll totals 10 before the deduction, do you report -10 or 0?. If you replace all negative results by zero I get an expected value of about 3.54 I would have expected much higher. 0 I'm assuming you're drawing cards from a shuffled standard deck (of 52) without repetition.$$1 - \frac{\,^{48}P_9}{\,^{52}P_9} = 1- \frac{48!\;43!}{52!\;39!} = \frac{4209}{7735} That is, the complement of the probability of not drawing an ace in the first nine cards.

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The chance of not drawing an ace is to have the first nine cards come from the non-aces. How many ways are there to select nine cards out of the non-aces? Divide by the number of ways to select nine cards. Subtract from 1.

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