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7

Suppose you have $n$ men and the enemy has $m$. Let $W(n,m)$ be the probability that you win. There are three rules: $$ W(n,m)=(W(n-1,m)+W(n,m-1))/2\\ W(n,0)=1\\ W(0,m)=0 $$ The solution seems to be $$W(n,m) = \frac{1}{2^{n+m-1}}\sum_{k=0}^{n-1}{n+m-1\choose k}$$


4

We can also derive a summation in the following manner: When army X does not lose, there is only one way for the sequence of winning: a string of only W's, and the probability can be written as: \begin{align*} \binom{y-1}{0} p^{y} \end{align*} where $p$ is the probability of a player of army X winning in one battle. When army X loses once, fix the last ...


4

Your formulas are not really valid from a probabilistic viewpoint , since $E(X)$ is not a measurable event (i.e., it has no probability). $X=E(X)$ is a measurable event, but you are not asking that here. Intuitively, yes, you are correct that conditioning on useless information is not helpful, but the above formulation is not the way to express that - you ...


3

There are multiple defensible interpretations of the question, depending on how the "random arrangement" is conducted. As a multinomial problem The red balls determine $n=15$ spaces ("slots") between them. Suppose they are filled equiprobably and independently by the $k=10$ blue balls. $X$ counts the number of slots filled with exactly one blue ball. To ...


3

You succeed if and only if you win at least $Y$ battles out of $N = X + Y - 1$ attempts. Indeed, if you win less than $Y$, then you lose at least $X$, and you are dead before your enemy. Conversely, if you win at least $Y$, then your enemy will be dead before you. The number of battles won is given by the binomial distribution, so the probability of ...


2

Let $n$ be the size of army $X$ and $m$ be the size of army $Y$. Denote $f(n,m)$ the probability that army $X$ will win. It's easy to see that $f(n,0)=1$ and $f(0,m)=0$. Let's go through some examples. For the $n=m=1$ combination, we know that two outcomes could happen: 1) X wins 2) Y wins Which is 50/50. So $f(1,1)=0.5$. However, think of it this way: ...


2

To start with, your approach does not make much sense, because -as Eupraxis1981'says- you are not conditioning on a random event. Informally: what you know about the random variable occurrence (i.e., something about the value of $X$) is one thing, what you know about the probability law of $X$ (i.e., something about the density $f_X(\cdot)$) is another ...


2

Here is a suggestion for the case where the density function is $0$ outside the interval $[a,b]$, and is bounded by $c$ on $[a,b]$. Use a random number generator to generate random points om the rectangle with base $[a,b]$ and height $c$. If the pdf is $f(x)$, discard all points $(x,y)$ such that $y\gt f(x)$. For each point you keep, record $x$.


1

If he lives to 80, then he already lived to 50; therefore, one condition supersedes the other. In set-theory language, {People who live to 80} $\subseteq$ {People who live to 50} so " {People who live to 80} intersected with {People who live to 50} " is just {People who live to 80}.


1

We have something exciting to share: recently we have constructed a new class of estimators for entropy (and actually also for nearly every functional of distributions) that have provable optimal performance guarantees with linear complexity. In particular, our estimator for entropy can also achieve the optimal n/log n scaling achieved by Valiant and ...


1

There are $\binom{1392}{696}$ ways to choose the fixed points, then $$696!\sum_{k=0}^{696} \frac{(-1)^k}{k!}$$ derangements, so (a) is the correct answer. $$\binom{1392}{696}\cdot 696! = \frac{1392!}{696!}$$ so $(a)$ is correct.


1

To simplify the notations, introduce $Y=X_2+\cdots+X_n$ and note that $(X_1,Y)$ is independent with Poisson distributions of parameters $\lambda$ and $\mu=(n-1)\lambda$ respectively and that $T=X_1+Y$ is Poisson with parameter $\lambda+\mu$. For every nonnegative integer $t$, $c(t)=E(T'\mid T=t)$ is $$ c(t)=P(X_1\leqslant1\mid ...


1

"Near zero" is a very vague term and needs some context. Near compared to what? This is the problem with the covariance: $Cov(X,Y) = 1$ could be large or small, depending on the variance of X and Y. This is why we have correlation: it is a normalized covariance: $Corr(X,Y)=\frac{Cov(X,Y)}{\sqrt{Var(X)Var(Y)}}$, which corrects for the overall variability of ...


1

Your answer looks right to me. If you work out the probability that (strictly) fewer than $4$ cars are observed it comes to $26.5$%. If you are sure you have quoted the question correctly, then maybe the authors had a brain fade when they wrote the answer section ;-)


1

This formula may be derived from what we know about the variance of a sum of independent random variables.[4] If $X_1, X_2 , \ldots, X_n$ are $n$ independent observations from a population that has a mean $\mu$ and standard deviation $\sigma$, then the variance of the total $T = (X_1 + X_2 + \cdots + X_n)$ is $n\sigma^2$. The variance of $T/n$ ...


1

Assume constant timesteps of size $dt$. The increment of a Poisson process $dq\left(t\right)$ with constant rate $\lambda$ is described by $$ dq\left(t\right)=\begin{cases} 1 & \text{with probability }\lambda dt\\ 0 & \text{with probability }1-\lambda dt \end{cases}. $$ It should be obvious that the two events $dq\left(t\right)=1$ and ...


1

Question 2 is answered in this Wikipedia article about Bessel's correction. The meaning of the standard deviation is largely the same as the meaning of the mean absolute deviation: Both are translation invariant, i.e. if you add the same number to every data point, you don't change the measure of dispersion, whether it is the standard deviation or the ...


1

So first we convert the histogram to data to get a better feel for things: \begin{pmatrix} 23 & 24 & 25 & 26 & 27 & 28 & 29 & 30 & 31 \\ 3 & 7 & 13 & 18 & 23 & 17 & 8 & 6 & 5 \end{pmatrix} The definition of standard deviation is the square root of the ...


1

The Gamma distribution is supported on $[0,\infty)$, so $v(F) = 0$.


1

Shortcut for finding $P\left\{ Y>X\right\} $: As always we have: (1) $1=P\left\{ Y>X\right\} +P\left\{ X=Y\right\} +P\left\{ X>Y\right\} $ Next to that it can be shown in this case that: (2) $P\left\{ Y>X\right\} =P\left\{ X>Y\right\} $ (consequence of $f(x,y)=f(y,x)$ everywhere) (3) $P\left\{ X=Y\right\} =0$ These three facts lead to ...



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