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I take the question to mean: We have $Y$ balls in a bucket and we sample $k$ balls with replacement; what is the expected value of the number $x$ of different balls sampled? (I assume that "are normally distributed and independent", which doesn't seem to make any sense here, is meant to express that the samples are independent and each sample has uniform ...

1

(a) Let r.v. $Y$ be the size of the team with player $A$. You already found $P(y=k)$, so $$E(Y) = \sum_{k=1}^{n-1}kP(Y=k) = \sum_{k=1}^{n-1} \dfrac{2k^2}{n(n-1)} = \dfrac{2n-1}{3} \\ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{using the identity \sum_{j=1}^m j^2=\dfrac{m(m+1)(2m+1)}{6}}.$$ (b) Define event $C =$ "Player $A$ is captain". Then, for ...

1

Hint: $P(Y \le y \mid X=x)= \dfrac{y-x}{1-x}$ provided that $x \le y$ So $\displaystyle P(Y \le y) = \int_{x=0}^{x=y} \frac{y-x}{1-x} \,dx$ Then differentiate with respect to $y$ to get the density

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$$f_Y(y)=\int f_{(X,Y)}(x,y)dx=\int f_{Y|X=x}(y)f_X(x)dx$$ You know $f_X$ and you know $f_{Y|X=x}$.

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I don't have enough reputation to comment, so I am attempting to answer: I don't think (a) is correct because, as you pointed out, we cannot assume the intervals are disjoint. Here is my attempt at a correct solution to (a). Note that I may very well be incorrect, but what I try to do is break the intervals into disjoint intervals and add back in the ...

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The question needs some interpretation. I will assume that we have a uniform distribution over a ball of radius $c$. If $P$ is a "randomly chosen" point in the ball, let $X$ be the distance of $P$ from the centre of the ball. We are interested in the various quantiles of $X$. For example, let us compute the $60$-th percentile of $X$. This is the number ...

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Your plan is good. Keep in mind that $Y=\sqrt{|X|}$ will be distributed on $[0,1]$, so we can start by calculating $$\mathbb P(Y\leq y)=\mathbb P(-y^2\leq X\leq y^2)$$ for $0\leq y\leq 1$. Since $X$ is uniform, the probability comes out to $y^2$. Thus $$F_Y(y)=\begin{cases} 0,& y\leq 0\\ y^2,& 0\leq y\leq 1\\ 1,& 1\leq y \end{cases}$$ As you ...

2

The variance of $Y$ is $\left(\frac{n+1}{n}\right)^2$ times the variance of $X$. Now we will find the variance of $X$. Recall that $X$ has cdf $\left(\frac{x}{\theta}\right)^n$ on the interval $(0,\theta)$. So $X$ has density function $\frac{n}{\theta^n}x^{n-1}$. Now we can compute $E(X^2)$ by using $$E(X^2)=\int_0^\theta x^2\cdot ... 0 Denote A = \max \{ U_1, U_2\} and B=U_1 + U_2. And I suppose you want to find the smallest such C for any p, thus the least information about U_1 + U_2. First, suppose C \le 0, then we have$$\mathbb{P}(A \ge p \mid B \ge C) = \mathbb{P}(A \ge p) = 1- p^2.$$If we want \mathbb{P}(A \ge p)=p, we must solve 1-p^2-p=0, which yields p = ... 1 Since F is a nondecreasing function, the preimage of every set of the form (a, \infty) has the form [b, \infty) or (b, \infty). This shows that F is a Borel-measurable function, which in turn shows that F(Y) is measurable. Your proof only works if F is actually injective, but in general this is not the case. You need to use the generalized ... 1 First, let's find the perpendicular bisector of the line p_0p_1. We'll call the midpoint of p_0p_1 M, and its slope s. Also, s' will be the slope of the bisector, s' = -1/s.$$y = s'(x - M_x) + M_y = -\frac{x - M_x}{s} + M_y$$Now, let's find the location of a p_2 that happens to be exactly the same distance from p_b as p_1 is (it'll ... 2 The area of the entire ellipse is about 1885, but only a quarter of the ellipse matters.$$\frac{\frac{1}{4} \pi \cdot 20 \cdot 30 + 800}{3200} \approx 0.3973. $$(I think there's a mistake, though; 39% is the probability that the rabbit is not safe, not the probability that it is safe.) 2 By symmetry, the probability that u_1\gt u_2 is equal to the probability that u_2\gt u_1. Since the distribution is continuous, u_1\ne u_2. Therefore, the probability that u_1\gt u_2=\frac{1}{2}. 0 (*) is correct, but you need to go one more step. Since, f_X(x) = \dfrac 1{R-L}\;\mathbf 1_{x\in[L;R]}$$\begin{align}f_Y(y) & = \frac {f_X(\frac{y-d}{c})}{c} \\[2ex] & = \frac{1}{c(R-L)}\;\mathbf 1_{y\in[cL+d;cR+d]}\end{align} That is all.

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The logical answer for this question is that if we suppose that there are 4 cars 1,2,3 and 4 and they are going in an order like 1 2 3 4 1 is the slowest car 2 is the second in matters of speed 3 is third in the matters of speed 4 is the fastest one when 2 aligns with 1 sod o 3 and 4 because they come closer to one faster than 2 ...

2

Let $c(n)$ denote the expected number of clusters. As written in the comments, I will assume that the cars have pairwise different speeds and then a random permutation is applied to rearrange the cars. So wlog the cars are all ordered in the initial distribution [before we apply our permutation]. This means car $1$ is the slowest, while car $n$ is the ...

2

Here's my thoughts. This isn't really my area, so please critique me if necessary. There will be a slowest car, and it will lie in the back as often as the front, so its expected position is the very center, forming a clump of $n/2$ cars behind it and $n/2$ cars driving in front of it. The cars in front will also have a slowest, which will, on average, split ...

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