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0

If there are N raffle tickets sold to N different people, then each ticket holder has the same probability of winning the prize. The distribution is uniform.


0

Here's a way to go : We know that the distribution is a gaussian vector and since we know the covariance matrix, we can compute the density function $f(x_1,x_2,..,x_k)$. Using this we can compute the cumulative distribution function for $ \max_{i=1,2,..,k} |X_i|$. $$ P(\max_{i=1,2,..,k} |X_i| < x )= \int_{]-x,x[^{k}} f(x_1,x_2,..,x_k)dx_1dx_2...dx_k ...


1

For the same sample space, simpler to compute favorable/total number of ways Without replacement: Ways of drawing two odd numbers = ${5\choose 2} = 10$ Ways of drawing two even numbers = ${4\choose 2} = 6$ P(draw 2 odd numbers | sum is even) = $\dfrac{10}{6+10} = \dfrac{5}{8}$ With replacement Ways of drawing two odd numbers = $5\cdot5 = 25$ Ways of ...


0

Here is the proof of Question 1: first arrow, left to right: \begin{eqnarray} P(X, A, B) & = & P(X, A\ |\ B)P(B)\\ & = & P(X\ |\ B)P(A\ |\ B)P(B)\ \ \ \ \ because\ X\ and\ A\ are\ conditionally\ independent\ on\ B\\ & = & P(X\ |\ B)P(AB)\\ & = & P(X)P(AB)\ \ \ \ \ because\ X\ and\ B\ are\ independent \end{eqnarray} first ...


3

There are five odd numbers and four even ones. Given that we drew either two odds or two evens, what is the probability that we drew two odds? Assuming we drew with replacement: Probability of two evens is $(4/9)^2$. Probability of two odds is $(5/9)^2$. Therefore the relative probability of two odds is $\frac{(5/9)^2}{(4/9)^2+(5/9)^2} = \frac{25}{16+25} = ...


1

Whenever you see $$\bar{X}\mid \mathbf{x}$$ or some other form of it, it means to fix the value of $\mathbf{x}$. So, in your case, we seek $$\text{Var}\left[\bar{X}\mid X_1, \cdots, X_n\right]\text{.}$$ But here's the thing. By definition, $$\bar{X} = \dfrac{1}{n}\sum\limits_{i=1}^{n}X_i$$ so thus $\bar{X}$ is constant because $X_1, \cdots, X_n$ are fixed ...


1

When you know all your $X_i$ then you know your $\bar{X}$, hence $E(\bar{X}|X_1,X_2,...,X_n)=\bar{X}$ as your sample mean is no longer a random variable but a constant now. Thus, $Var(\bar{X}|X_1,X_2,...,X_n)=E(\bar{X}^2|X_1,...,X_n)-(E(\bar{X}|X_1,...,X_n))^2=\bar{X}^2-\bar{X}^2=0$


1

You are essentially asking for the distribution of the maximum order statistic of $n$ iid exponential random variables. That is to say, what is $$\Pr\left[\max_{1\le k \le n} X_k \le x\right]?$$ Recall that $$\Pr[X_k \le x] = 1 - e^{-x}, \quad k = 1, 2, \ldots.$$ Since the $X_k$s are iid, it easily follows that $$\Pr\left[\max_{1\le k \le n} X_k \le ...


0

Maybe you could try to model this as a Negative Binomial. It describes how many trials we need to conduct, iterations in this case, before we have a certain number of successes, here errors. This distribution takes an $r$, a number of failures, in this problem $k$. It also takes a probability $p$ or success on each trial. We are asking ourselves how many ...


2

It is absolutely not correct that at least one should be infinite. It is true that for every real number $a$, infinitely many of the random variables will be more than $a$, but every one of them will be finite. But the fact that for every $a>0$, infinitely many of them are $>a$ is enough to imply that the limit is infinite. I'm going to follow up on ...


0

There are 25 coins and they consist of : 6 quarters, 4 dimes, 5 nickels, 10 pennies (since there are 25 total and we already have 15) The probability of getting a quarter is 6/25 : because there are 6 of 25 possibilities. The probability of getting a dime is 4/25 : because there are 4 of 25 possibilities. The probability of getting a nickel is 5/25 : ...


0

Hint: probability of event(s) is number of certain event divided by number of total event So, the probability to get a nickel is P(nickel) : P(total event) = 5 : 25 = 0.2


1

Assume that there are $k$ elements in $A_1$. There are $\binom{n}{k}$ ways of choosing such an $A_1$. Now for a given $A_1$ count the number of admissible $A_2$. Well, $k$ elements must not belong to $A_2$, so the others can be chosen or not in $2^{n-k}$ ways. Letting $k$ run from $0$ to $n$, we get the total number of possible combinations of $(A_1,A_2)$ ...


3

Take an element $i$. Let $p_i$ denote the probability that $i$ is not in the intersection of your two randomly chosen sets. It is in exactly $2^{n-1}$ subsets, hence the probability that it is in a randomly chosen subset is $\frac 12$. It follows that the probability that it is in both of two randomly chosen subsets is $\frac 14$. Therefore $p_i =\frac ...


2

The keyword is "fat-tail" distributions. So anything that falls off like $1/x^p$ as $x\rightarrow\infty$ will have all moments up to and including $p-2$. These come up more often than people expect, especially in finance and actuarial sciences and are absolutely catastrophic when it comes to (underestimating) rare events [read something like Black Swan by ...


1

According to MathWorld, the density for $d$ is $$ \frac{3d^2}{r^3}-\frac{9d^3}{4r^4}+\frac{3d^5}{16r^6}\;. $$ Thus the cumulative distribution function that you're looking for is $$ \int_0^d\left(\frac{3d'^2}{r^3}-\frac{9d'^3}{4r^4}+\frac{3d'^5}{16r^6}\right)\mathrm dd'=\left(\frac dr\right)^3-\frac9{16}\left(\frac dr\right)^4+\frac1{32}\left(\frac ...


1

Sure...it's not complicated but it is badly error prone. In particular, it is prone to the so-called off-by-one, or "fencepost" error (so-called because of the error one tends to make in counting the number of fenceposts needed to border $N$ yards if you want to separate each by $1$ yard). I think that what follows is correct, but I advise testing it. ...


-1

My question would be "why do you think this problem can be solved?" We are told that $5$ children got $100\%$ , another $10$ got $90\%$. Okay, that is a total of $$5 * 100 + 10 * 90= 1400$$ If the other $35$ children got an average $x\%$ then the total would be $$1400+ 35x$$ and, since we are told that the average for all students is $45\%$, that ...


0

Hint: Assume that the other $35$ children have lower scores than $90\%$. Then these $35$ children are below the $70$th percentile (since we know the top $15/50 = 30\%$ scores). Let $X$ be the sum of the $35$ scores. Using the class average, we can solve for $X$. The desired answer is $X/35$.


2

If all the exponential random variables have the same parameter $\lambda$, the answer is $\Gamma(n,\lambda)$ (Gamma distribution), whose density is: $$\frac{\lambda^n}{\Gamma(n)}e^{-\lambda x}x^{n-1},\, x>0$$ First prove it with $n=2$ using a convolution, this is pretty easy. Then you can prove the general case by induction. Note that ...


0

In order to calculate $P(B^c\mid C)$ observe that: $$ P(B^c\mid C)+P(B|C)=\frac{P(B^c\cap C)}{P(C)}+\frac{P(B\cap C)}{P(C)}=\frac{P((B^c\cap C)\cup(B\cap C)}{P(C)}=\frac{P(C)}{P(C)}=1 $$ So that $P(B^c\mid C)=1-P(B\mid C)$ and $P(B\mid C)$ is known. The same applies to $P(B^c\mid C^c)=1-P(B\mid C^c)$, which is used to obtain $P(A|C^c)$.


1

The number of sequences in which the minimal integer $i$ appears exactly $e$ times is $$ \binom ne (k-i)^{n-e}, $$ since there are $\binom ne$ ways to decide where the $i$s go and $k-i$ choices of larger integers for each other spot. The quantity you want is then the sum of this over all $e\ge j$. Equivalently, $$ p_{ij} = \frac1{k^n} \sum_{j\le e\le n} ...


0

Your syntax is fine, although it is more typical to consider conditional probabilities of the form P(M | X) rather than the way you've phrased it. However, you would need some extra information to solve your problem (i.e. your problem is under-constrained). Consider a simpler case where we only have two conditions – gender and location, both of which only ...


0

There is insufficient information to answer the question. What is missing is the relationship between the statement "the probability of catching a fish" and "the fisherman cast his rod three times and caught only one fish." Are we to assume, for example, that the probability of $0.4$ for pond $A$ is the per-casting probability, or the probability of ...


1

One interpretation of the problem is that every time she casts her rod say at Pond A, she has probability $0.4$ of catching a fish. We will assume, unreasonably, that the results of the casts are independent. Then given that she cast $3$ times in Pond A, the probability she caught exactly $1$ fish is $\binom{3}{1}(0.4)^1(0.6)^2$. Compute similar ...


0

As has already been pointed out, the precise and careful choice of notation is paramount. The event $C$ of choosing the same color is the disjoint union of two separate events, so it is better to choose the following notation: Let $(b_1, b_2)$ be the random outcome of the two ball draws in order, where $b_i \in \{W, B\}$ for $i = 1, 2$. Thus, we want: ...


4

A picture is worth a thousand words


0

Note quite right. The correct version is $$\sum_{c}p(a|c\cap b)p(c|b)=p(a|b).$$ Proof: \begin{eqnarray*} \sum_{c}p(a|c\cap b)p(c|b) &=& \sum_{c}\dfrac{p(a\cap c\cap b)}{p(c\cap b)} \dfrac{p(c\cap b)}{p(b)} \\ &=& \sum_{c}\dfrac{p(a\cap c\cap b)}{p(b)} \\ &=& \dfrac{p(a\cap b)}{p(b)} \\ &=& p(a|b). \end{eqnarray*}


3

Directly, Pr = $\dfrac{4\choose3}{12\choose3} = \dfrac{1}{55}$ In the tortuous way you were attempting, Pr = 1 - $\dfrac{{8\choose3}{4\choose0} +{8\choose2}{4\choose1} +{8\choose 1}{4\choose2}}{12\choose3} = \dfrac{1}{55} $


1

The probability of randomly drawing 3 balls not red would mean selecting green balls such that there is a combination of 4 green balls from which we can choose 3 divided by the combination of 12 balls, green and red, choosing 3 green balls at random therefore: 4C3 / 12C3 = 1/55 or 0.0182


1

The total no. of ways in which green balls can be chosen $={4\choose 3}$. The total no. of ways in which balls can be chosen $={12\choose 3}$ Probability that the balls will not be red(i.e., they will be green)$= \frac{4\choose 3}{12\choose 3}=\frac{1}{55}$.


1

Well you are summing events prob which are not independent, see there : https://en.wikipedia.org/wiki/Law_of_total_probability and the approach given in comments by @true blue and using the very same theorem could lead you to the correct probability, with a three steps process, first draw : you get a green, same for second, etc, updating the odds with the ...


2

Here's a mathematically rigorous proof of the statement: Let's assume the board consists of $n$ fields of which $k$ are mines and the field is created by randomly choosing $k$ numbers from $1$ to $n$ without repetitions. This is a reasonable assumption, since the Fisher-Yates-Shuffles would yield an effective and easy way to create a random field and every ...


0

If your random permutation polynomials are of degree one, they have the form $T(x) = ax+b$ for some $a, b \in \mathbb{Z}_{p}$, $a \neq 0$. I'm not sure how many fixed inputs there are. For a single input $x_{1}$, $T(x_{1}) = ax_{1}+b = y_{1}$ whenever $b = y_{1}-ax_{1}$; there is precisely one $b$ for any choice of $a$, and $p$ choices for $b$, so the ...


3

How are you at three-dimensional integrals? Let the first point be at $(0,0,z)$. There are two spheres: One of radius $r$, centered at the origin, and one of radius $d$, centered at $(0,0,z)$. I would let $r=1$ to remove one letter from your calculations. The spheres' intersection is symmetric about the $z$ axis, which should help you calculate the volume ...


0

Hint: Have you learned about expected value in a probability or statistics course? Let's make $H$ the random variable that takes on the value of the number of heads in 10 flips. Then $E(H)$ is the expected number of heads in 10 flips. You calculate this using the formula $$E(H) = \sum\limits_{i=0}^{10}P(H = i)\times i.$$ Verify that if you are flipping a ...


17

You've correctly inferred that there are only two possibilities: The two bombs are at $A$ and $D$. The two bombs are at $B$ and $C$. Common sense now dictates that these two options are equally likely. The only way to prove that the two options are equally likely requires knowledge of how the minesweeper board was generated. It would be in the spirit of ...


1

They are all equally likely to be safe. The restrictions given by the numbers tell you that (Exactly) one of A & C is a mine. One of A & B is a mine. One of B & D is a mine. There are exactly 2 configurations that satisfy these restrictions: A & D are mines or B & C are mines. Assuming these are equally likely (probably a fair ...


8

You know exactly one of A and C is a bomb exactly one of B and D is a bomb exactly one of A and B is a bomb So, as you say, there are two possibilities A and D are bombs while B and C are not B and C are bombs while A and D are not As far as I can tell, these have the same likelihood


1

Split it into disjoint events, and add up their probabilities: The probability of choosing white from the first box and then from the second box is: $$\frac{4}{9}\cdot\frac{6}{10}=\frac{24}{90}$$ The probability of choosing black from the first box and then from the second box is: $$\frac{5}{9}\cdot\frac{5}{10}=\frac{25}{90}$$ So the probability ...


3

The calculation on the right-hand side is correct; just the notation is bad, because as you say the right-hand factors in both terms are conditional probabilities. A better way to write this would be $$ P(\text{C})=P(\text{W})+P(\text{Bl})= ...


3

Sub $u=x+y$, $v=x-y$ with Jacobian $J=1/2$: $$\int_0^{\infty} dx \, \int_0^{\infty} dy \frac{f(x+y)}{x+y} = \frac12 \int_0^{\infty} du \int_{-u}^u dv \frac{f(u)}{u} = \int_0^{\infty} du \, f(u) = 1$$


-1

Actually, it is higher: $13,749,422,954,239,300,000,000,000,000,000$. As you have UP TO $20$ characters, you need to take into account all the possibilities prior to the max, i.e. $$36 \\ 1,296 \\ 46,656 \\ 1,679,616 \\ 60,466,176 \\ 2,176,782,336 \\ 78,364,164,096 \\ 2,821,109,907,456 \\ 101,559,956,668,416 \\ 3,656,158,440,062,980 \\ ...


1

I think for any realization of random variable : $ X_{1},...,X_{n}$, the empirical distribution $F_{n}(x)=\frac{1}{n}\sum_{i=1}^{n}I_{\{X_{i}\le x\}}$ is just a discrete distribution function concentrated on these n value and attached weight $\frac{1}{n}$ to each of them. Thus, the integral $\int g(x)dF_{n}(x)$ is same as expected value of $g(x)$ with ...


0

A little logic and computation will help you decide between your answers: (1) Your answer has to be less than $(1/2)^{10} = 0.0009765625$. That would be the probability all 10 rolls showed 3 or fewer spots. (2) Your first answer computes to 0.00006946032, which $is$ smaller than the answer in 1. (3) Your second answer computes to 0.04501029, which is ...


0

Start defining $$A=(\frac{x}{x+1})^{(x/z)}$$ Take logarithms $$\log(A)=\frac xz \log\frac{x}{x+1}=-\frac xz \log\frac{x+1}{x}=-\frac xz \log(1+\frac{1}{x})$$ Now remember that, for small values of $y$, $\log(1+y)=y-\frac{y^2}{2}+\frac{y^3}{3}+O\left(y^4\right)$; replace $y$ by $\frac{1}{x}$. So, $$\log(A)=-\frac xz \times \Big( \frac{1}{x}-\frac{1}{2 ...


1

In analog-to-digital conversion a quantization error occurs. This error is either due to rounding or truncation. When the original signal is much larger than one least significant bit (LSB), the quantization error is not significantly correlated with the signal, and has an approximately uniform distribution. The RMS error therefore follows from the variance ...


2

Consider a "spinner": an object like an unmagnetized compass needle that can pivots freely around an axis, and is stable pointing in any direction. You give it a spin and see where it comes to rest, measuring the resulting angle (divided by $2\pi$) as a number from $0$ to $1$.


2

It is well-known that $(\frac{x+1}{x})^{x} =(1+\frac{1}{x})^{x} \to e \approx 2.71828... $ as $x \to \infty $. Therefore $(\frac{x}{x+1})^{x} \to \frac1{e} $ as $x \to \infty $. Similarly, $(\frac{x}{x+1})^{(x/z)} =\left((\frac{x}{x+1})^{x}\right)^{1/z} \to \left(\frac1{e}\right)^{1/z} =\frac1{e^{1/z}} $ as $x \to \infty $. This assumes that $z$ is ...


2

$$\left(\frac{x}{x+1}\right)^x=\left(1-\frac{1}{x+1}\right)^x=\left(1-\frac{1}{x+1}\right)^{x+1}\left(1-\frac{1}{x+1}\right)^{-1}$$ This tends to $$\frac{1}{e}$$ https://en.wikipedia.org/wiki/E_%28mathematical_constant%29 For the second one, note that $\left(\frac{x}{x+1}\right)^{x/z}=\left(\left(\frac{x}{x+1}\right)^{x}\right)^{1/z}$ so this tends to ...



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