# Tag Info

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The error lies in the fact that your calculator's invNorm function finds the $z$-value for which $75$ percent of the normal distribution lies between $-\infty$ and $+z$, not between $-z$ and $+z$. In particular, $50$ percent lies between $-\infty$ and $0$ (obviously, by symmetry) and then another $25$ percent lies between $0$ and $+0.674$. By symmetry, ...

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A cumulative distribution function must be non-decreasing and right-continuous with a to-negative-infinite limit of $0$ and a to-positive-infinite limit of $1$. IE $\;\lim\limits_{x\to-\infty} F(x)=0\;$ and $\;\lim\limits_{x\to+\infty} F(x)=1\;$. If the CDF is a continuous function, then the random variable is called a continuous random variable.   ...

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I've never seen $(k, \lambda)$ to parametrize a Gamma distribution before, but I have seen $\lambda$ to parametrize an exponential distribution (with mean $1/\lambda$). I am not going to provide a complete solution here, but some hints. Hint: Assuming you've defined a Gamma ($X$) distribution to be the sum of $k$ independent exponential random variables ...

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$$P=1-({\frac{13999}{14000}})^{14000}$$ let $a=14000$ then $$x\equiv ({\frac{13999}{14000}})^{14000}=(1-1/a)^a$$ taking logs and using $\ln (1-x )\sim -x$ $$ln(x) = a \ln(1-1/a) \sim a(-1/a)=-1$$ so $P=1-e^{-1}\sim 63.2\%$ is good to about 5 decimal places.

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Let $y$ be a positive integer. Then $$\Pr(Y=y)=\Pr(y-1\lt X\le y)=\int_{y-1}^y \lambda e^{-\lambda t}\,dt.$$ When we evaluate the integral, we obtain $e^{-\lambda(y-1)}-e^{-\lambda y}$. This can be "simplified" to $$(e^{-\lambda})^{y-1}(1-e^{-\lambda}).$$ We leave it to you to identify the distribution. Once you have done that, mean and variance will ...

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Q: A dice is thrown and the result is summed to the previous throw until the sum equals or exceeds $11$. Call each attempt to reach $11$ a run of dice. How many times can we expect to roll the dice until we roll a run which sums to exactly $11$? The minimum length of a run is $2$: $(5, 6)$ or $(6,5)$. The maximum length of a run is $11$: Rolling $1$ eleven ...

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Let $E(s)$ be the expected number of throws with current sum $s$. $E(11) = 0$ by definition, and $E(s) = E(0)$ for $s>11$. Then: $$E(s) = 1 + \frac{1}{6}\sum_{k=1}^{6}E(s+k)$$ This yields a system of equations. Solve for $E(0)$. For example: $$E(7) = 1 + \frac{1}{6}(E(8) + E(9) + E(10) + E(11) + E(12) + E(13))$$ can be framed as $$-6E(7) + E(8) + ... 0 You should try setting this situation up as a Markov chain. Your states can correspond to the total score achieved. Eg x_8 is "score = 8." Your starting state is x_0. The transition probabilities are given by the result of the die throw. For example, from x_4 you could go to x_5 with probability \frac 16, to x_6 with probability \frac 16, ... 0 Hint: Because the maximum of several quantities is less than a number if and only if each quantity is less than the number, you have \Bbb P[M_n\le x]=[1-\exp(-\lambda x)]^n. 0 Since$$ \mathrm{Pr}\left(M_n-\frac{\ln n}{\lambda}>t\right)=\mathrm{Pr}\left(X_1>t+\frac{\ln n}{\lambda}\right)^n=\mathrm{exp}\left(-\lambda n(t+\ln n/\lambda)\right) $$it follows that$$ \lim_n\mathrm{Pr}\left(M_n-\frac{\ln n}{\lambda}\le t\right)=1- \mathrm{exp}\left(-\lim_n (\lambda nt + n\ln n)\right)=1 $$whenever t>0. Are you sure about ... 0 Calculate the number of possibilities for everything that doesn't make the cut. Here's what doesn't make the cut: All birthdays different Exactly 1 pair the same; all others different Exactly 2 distinct pairs the same; all others different ... Exactly 19 distinct pairs the same; all others different Exactly 20 distinct pairs the same Line the ... 2 Find \mathsf P(Y_{(3)}>2Y_{(1)}) , the probability that the largest of three iid rv is more than twice the least. That is the probability that one of the values is more than twice another, and that the third value lays somewhere between both of them; for any given selection of the three r.v..$$\begin{align}\mathsf P(Y_{(3)}>2Y_{(1)}) & = ...

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If $U$ and $V$ are independent standard normal (that is, $(N(0,1)$) random variables, then $\sigma_1U \sim N(0,\sigma_1^2)$ and $\sigma_2V\sim N(0,\sigma_2^2)$, and this answer contains a short proof (no explicit integrations, no convolutions, no MGFs or characteristic functions; only the left half of the hint given in the problem) of the fact that ...

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The conditional expectation is $$\mathbb{E}[Y \mid X \in C] = \dfrac{\displaystyle \int_{y\in \mathbb{R}}\int_{x \in C} y f(x,y) \, dx\, dy}{\displaystyle \int_{y\in \mathbb{R}}\int_{x \in C} f(x,y) \, dx\, dy}$$ though as Ian says there is an issue if the denominator is zero.

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An easy way to get an arrangement by hand is to group the marbles into two groups of twelve, say the yellow+green and everything else. We can make an arrangement of the two groups that meets your requirement a by doing a checkerboard. There are twelve light squares and twelve dark squares. Now distribute the yellow+green any way you like on the (say) ...

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The k'th success occurs at the i'th trial exactly when there is k-1 successes in the first i-1 trials, $B(i-1,p) = k-1$, and the i'th trial is a success. $$P(X=i) = p \cdot P(B(i-1,p)=k-1) = p \left(\begin{matrix}i-1\\k-1\end{matrix}\right)p^{k-1}(1-p)^{i-k}$$

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No. I think you mean $\mathbb E$ instead of $\mathbb P$ It does not make sense to take the intersection of a collection of events (e.g. $\mathcal F_{t-1}$) and a collection of sample points (e.g. $(B_t = 1)$). Perhaps you meant $\sigma((B_t = 1))$. Careful about the extension you're trying to make here. You seem to be thinking we can do something like: ...

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We have the fact that: $$P(A)P(B)=P(A\cap B)\iff P(B|A)=P(B)\iff P(A|B)=P(A)$$ if $A, B$ are independent events. So we can check if $P(B|A)=P(B)$ or $P(A|B)=P(A)$ to see if they are independent. In this case, this is as simple as considering all card pairs that belong to $B$ say, namely those pairs with no figures $(JQKA)$ and showing that the proportion ...

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If $X$ is a standard Gaussian random variable, then we have $$\mathbb P\{X>\sqrt 2t\} \overset{t\to +\infty}{\sim}\frac Ct\exp\left(-t^2\right),$$ where $C$ is independent of $t$. In the context of the question, one can define $X_n:=X-\sqrt 2c/n$ and $a_n:=\sqrt 2n$, where $c\in\mathbb R$. We have $$P(X_n > a_{n} ) \, / \, P(X> ... 0 If p is the probability that a senator in the group C is voting for the bill, then the probability that at least 24 of them votes in this way is$$P_1=\sum_{k=24}^{39} \binom{39}k p^k(1-p)^{34-k}.$$You want to calculate this for p=1/2. So in this case this simplifies to$$P_1=\frac{\sum\limits_{k=24}^{39} \binom{39}k}{2^{39}}.$$If you do not want to ... 1 Hint. Firstly, P(A)=1-\frac{39 \choose 2}{52 \choose 2}, P(B)=\frac{36 \choose 2}{52 \choose 2}. Secondly, use the following expression for easier calculation:$${n \choose 2}=\frac{n(n-1)}{2}$$Hope it helps. 0 All 48 even values of n in the range qualify, as values of (n+1) from 8\times 1 thru 8\times 12, thus there are 60 favorable cases, and  Pr = \dfrac{60}{96} = \dfrac{5}{8} 1 The simplest way is perhaps to use the Moment Generating Function or the Characteristic Function of an \mathcal{N}(\mu,\sigma) random variable.$$\mathbb{E}[e^{s(X+Y)}] = \mathbb{E}[e^{sX}]\mathbb{E}[e^{sY}]$$by independence of X and Y. Since \mathbb{E}[e^{sX}]\mathbb{E}[e^{sY}] = e^{s\mu_1 + \tfrac{s^2}{2}\sigma_1^2}\bigg[e^{s\mu_2 + ... 1 HINT: n\equiv0\pmod8 \implies n\cdot(n+1)\cdot(n+2)\equiv0\cdot(0+1)\cdot(0+2)\equiv\color\green0\pmod8 n\equiv1\pmod8 \implies n\cdot(n+1)\cdot(n+2)\equiv1\cdot(1+1)\cdot(1+2)\equiv\color\red 6\pmod8 n\equiv2\pmod8 \implies n\cdot(n+1)\cdot(n+2)\equiv2\cdot(2+1)\cdot(2+2)\equiv\color\green0\pmod8 n\equiv3\pmod8 \implies ... 0 I don't have enough reputation to comment, but here is a pretty strong hint: For independent random variables X and Y, the distribution of Z = X+Y equals the convolution of X and Y. Use the definition of the normal distribution. 3 Looking at the equation$$n(n+1)(n+2)\equiv 0\pmod{8},$$we see that this happens when$$n\equiv 0,2,4,6,7,8\pmod{8},$$and so we get a probability of \frac{5}{8}. 1 n=8m n=8m-1 n=8m-2. n=4m n=4m-2 1) is a subset of 4) so we can remove 1). 3) is a subset of 5) so we can remove 3). So, n=8m-1 \rightarrow P=\frac18 n=4m \rightarrow P=\frac14 n=4m-2 \rightarrow P=\frac14$$\frac18+\frac14+\frac14=\frac58$$0 the total numbers are 96. for an odd number to be multiple of 8 using expression n(n+1)(n+2) the number n should be a number preceding to a multiple of 8 starting from 8 itself so first number is 7 and such is an AP whose last term is 95. So total terms which are 1 less than a multiple of 8 ie odd are 12 thus the probability is ... 0 Clearly if K \gt n_1 or K \gt n_2, the outcome that the two samples share exactly K elements is impossible. So let's assume K \le n_1,n_2. Further it is required, if the outcome is to be possible, that (accounting for the samples' overlap) N \ge n_1 + n_2 - K. Computing a probability here probably assumes all subsets of a given size are equally ... 1 The standard deviation of the sampling distribution of \bar{X} is \sigma_\bar{X}=\frac{\sigma_X}{\sqrt{n}}, where \sigma_X is the standard deviation of X and n is the sample size. Since the variance of X is 25kg^2, the standard deviation of X is \sigma_X=5kg. Then the standard deviation of the sampling distribution of \bar{X} is ... 0 (a) First of all: there is no essential difference between "selecting 1 out of 3 that are on their turn selected out of 6" and "selecting 1 out of 6". If B denotes the event that the biased coin was selected and E the event that 3 heads show up by 3 tosses then:$$P(B\mid E)P(E)=P(B\cap E)=P(E\mid ...

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First part is ok. For the second part, there is no restriction on the first choice. And once you have chosen, you can only choose from $8$ non-siblings, then from $6$ non-siblings and so on, thus $Pr = \dfrac{10\cdot8\cdot6\cdot4}{10\cdot9\cdot8\cdot7}$

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The first answer is right, but the second is not. If you apply the same reasoning to the second problem that you used in the first, you can argue that the probability is $$\frac{10}{10}\cdot\frac89\cdot\frac68\cdot\frac47=\frac8{21}\;.$$ There is no restriction on the first pick. Once you’ve picked one person, $9$ remain, and $8$ of them are not siblings ...

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Use https://en.wikipedia.org/wiki/Conditional_probability#Kolmogorov_definition With A = coin is biased, B = three coin flips are all heads. $P(A \cap B) = 1/6 * 27/64$ $P(B) = 5/6 * 1/8 + 1/6 * 27/64$ So $P(A|B) = 27/67$ (b) can be done in a similar way.

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Let $p$ be the probability that the first player wins. You can calculate $p$ without having to deal with an infinite sequence or series as follows. First, $p$ is clearly the probability that the first player wins on the first toss plus the probability that he wins after first getting a tail. The probability that he wins on the first toss is $\frac12$. ...

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I think the given answer is double-counting. Take a simple example with \begin{align} k &= \text{$2$ colors} \\ N_1 &= 2 \\ N_2 &= 1 \\ N &= N_1+N_2 = 3 \\ m &= 3. \end{align} Of course, with $m=N$, there is only $1$ way to select the balls. However, $$N_1\cdot N_2\cdot\binom{N-k}{m-k} = 2\cdot 1\cdot\binom{3-2}{3-2} = 2.$$ The ...

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The total number of ways to choose $13$ out of $52$ cards is: $$\binom{52}{13}$$ The number of ways to choose $13$ out of $52$ cards with exactly $1$ king is: $$\binom{4}{1}\cdot\binom{52-4}{13-1}$$ Hence the probability of choosing $13$ out of $52$ cards with exactly $1$ king is: ...

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you have two disjoints subsets: 'kings' and 'everything else'. $\binom{4}{1}\binom{48}{12}$ simply means 'any 1 out of 4' AND 'any 12 out of 48'. There's no order involved. Any form of order would be if, for example, you would need to get 3 kings. Then you would need to divide by $3!$ because the order doesn't matter. Does this answer your question?

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(a) If by $\mu$ you mean $E(Y)$ then \begin{align} E(y) &= \int_{y=0}^{1} yf_Y(y)\;dy \\ &= \int_0^1 2y^2\;dy \\ &= 2/3. \end{align} You don't need the median at all for this. (b) The formula you have for $g_{(k)}(y)$ is the way to go. Here, we have $n=5, k=1$ and $F_Y(y) = \int_0^y 2y\; dy = y^2$. If my calculation is correct, you should end ...

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You want to draw 6 balls and the remaining ball should be black. Exactly one of your seven balls is black. So the six balls you draw in 1. are white and the remaining ball is black If you draw 7 balls (all balls) then 2. is the same as if you draw 6 white balls and at last a black ball. So the balls are drawn in the following order: w w w w w w b The ...

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If you have $n$ balls, you can, in the beginning, fix the black ball (i.e. it'll be present in all of your combinations). Then, you can choose $n-2$ balls from the remaining $n-1$ balls. You can do it in ${n-1\choose n-2}=n-1$. If you were to choose any $n-1$ balls from $n$ balls, you could do it in ${n \choose n-1}=n$ ways. So, the probability of getting ...

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Hint: Extracting $n-1$ balls from $n$ balls without replacement (1) comes to the same as extracting $1$ ball from $n$ balls without replacement (2). Just think of this extracted ball in (2) as the unique ball wich is not extracted in (1).

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I think you're meant to realise that the probability of drawing $n-1$ balls is the same as the probability of leaving $1$ ball in the bag at the end. Therefore probability of six white draws $= \frac{1}{7}$ and probability of five white and one black is the same as leaving a white behind $=\frac{6}{7}$

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Another thinking of your answer would be $1-{1\over{7\choose 6}}$ but is essentially the same thing as the above technique. Other than that you would use ${6\choose 5}\over{7\choose 6}$ by fixing the black ball to be chosen and choose $5$ from the remaining $6$ balls.

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The probability of drawing a black ball is the complement of not drawing a black ball: $$P (\mbox {drawing black})=1-P (\mbox {not drawing black})$$ Not drawing the single black ball when drawing $n-1$ balls from $n$ balls is simply $\frac {1}{n}$, so the probability you're after is: $$1-\frac {1}{n}=\frac {n-1}{n}$$

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So your problem is basically, we pick five numbers from $0$ to $99$ two times and what is the probability that four numbers end up to be the same both times. It is simple: $\large{5 \choose 4}{5\choose4}100^4\cdot100^2\over100^{10}$

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There are $\binom{52}5$ ways to select $5$ cards form $52$ cards. There are $\binom43$ ways to select $3$ queens from $4$ queens. There are $\binom42$ ways to select $2$ tens from $4$ tens. There is $1$ way to select heart king, heart queen, heart ten, heart nine and heart jack from a deck. So: P(A)=\frac{\binom43\binom42}{\binom{52}5}\text{ and ...

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Split the allocations into disjoint sets:both 1 and 2 are in slots 3-10, thats $8 \cdot 7 \cdot 8!$ and 2 is slot 1 and 1 is in slots 3-10: $8 \cdot 8!$

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http://cseweb.ucsd.edu/classes/fa15/cse21-abc/HW8_F15.pdf Maybe it is better to ask the TA in tomorrow discussions?

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To figure out the probability $P(A_{1} \mbox{ and } A_{2})$: Count the number of permutations satisfying both, divide by total number of permutations. To address independence: Does $A_{1}$ being true affect $P(A_{2})$?

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