We place $n$ points on the unit circle uniformly and independently . What is the probability that their convex hull does not contain the origin?  I came up with this:      If the convex hull does not contain the origin then there is one point $x$ such that the next point $y$ in clockwise order forms an angle $xOy$ larger than $\pi$. Therefore we just need to calculate the probaility that this happens at least once, but notice that this cannot happen more than once . Suppose that $x_1,x_2,\dots x_n$ are the points that are to be placed independently, then the probability that the next point after $x_i$ forms an angle larger than $\pi$ is $\frac{1}{2^{n-1}}$. Therefore the probability the hull does not contain the origin is $\frac{n}{2^{n-1}}$   I would greatly appreciate a proof verification, alternative proofs and insight as to whether this can be generalized to higher dimensions, when there are $d+2$ points it is possible, as is shown .
123141
What's the probability that a n-sided polygon made from n distinct random points on circle contain the centre?
79721
Take the square $[0,1]^2$ we take $n$ random lines through the square. You can choose random lines by choosing a point on one of the sides of the squares and randomly choosing a different point on any of the other three sides and then drawing a line connecting them.   What is the probability that after $n$ of these random lines the largest region is greater than or equal to $0.5$?  Any tips or cases help, thanks.
43056
I want to prove the following statement:   If the unit sphere in a normed space contains a segment, then there exists two vectors $x,y$ such that $\Vert x + y \Vert = \Vert x \Vert + \Vert y \Vert$ with $x$ and $y$ linearly independent.   This is what I have done: Let $x$ and $y$ be the end point of the segment contained in the unit sphere. Then $\forall \lambda \in (0,1)$ I can write:   $$ \Vert \lambda x + (1-\lambda)y \Vert \leq 1 $$  From here I am struggling on getting the relationship on $x+y$. Any pointers?
3587
Find the recurrence relation satisfied by $R_n$, where $R_n$ is the number of regions into which the surface of a sphere is divided by n great circles (which are the intersections of the sphere and planes passing through the center of the sphere), if no three of the great circles go through the same point   From: Rosen Discrete Mathematics textbook  My attempt: I was unable to visualize these greater circles for values of $n \geq 4$. So, I launched GeoGebra and made the :     My initial planes were $x=y$, $x=-y$, and the $z=0$ plane. Now, I added a fourth yellow plane $y=z$. However, I still find it hard to understand how to calculate what number of new sections are formed. I am able to calculate the number of new sections in this specific case, but, I am unable to arrive at a general formula for the new number of sections, through recurrence.  I visited the following website -  - which happens to have a solution, where they mention the addition of $2(n-1)$ regions to the existing regions, after the addition of the $n$-th plane. However, I am unable to understand why exactly that is true.
106467
Good evening, I need to compute $\pi_1(\mathbb{R}^2 \setminus (\mathbb{Z} \times \{0\}))$.  I see that the set $\mathbb{R}^2 \setminus \mathbb{Z} \times \{0\}$ has a countable set of points less than whole $\mathbb{R}^2$, so at each point we can retract the space into circles, and finally the space can retract to a wedge of infinite countable circles. Aplying Seifert-van Kampen theorem we can conclude that the fundamental group is free product of $\mathbb{Z}$ with itself countable times. Is this idea correct? Thanks a lot.
95159
Given a random permutation $\sigma \in S_n$ from $[n] \to [n]$ in a uniform probability space, what is the probability that $\sigma $ has exactly $k$ fixed points for a given $k$ between $1$ and $n$?  In other words: what is the probability that $\exists x_1 ,...,x_k \in [n] : \sigma (x_i) = x_i  $ for $\ i\in \{1,...,k\}$ and for every $y \notin \{x_1 , ... , x_k\}$ we get $\sigma(y) \neq y$.  I saw that $\lim_{n \to \infty } prob(A_0) = e^{-1}$ using Inclusion–exclusion principle and i belive that for a given k : $\lim_{n \to \infty} prob(A_k) = \frac{e^{-1}}{k!}$ but I am not sure how to show it.  *$A_k$ stands for the event "k".
86065
Two numbers are chosen randomly and independently from the unit interval $(0,1)$.  What is the probability that they differ by more than $\frac{1}{2}$? Hint: Think of the $2$ numbers chosen as coordinates in the Cartesian plane. Draw a square with vertices at $(0,0)$,$(1,0)$,$(1,1)$, and $(0,1)$. Allowable choices for the $2$ points lie below the line $y=x-\frac{1}{2}$ or the above $y=x+\frac{1}{2}$. Find the combined area of these regions.  I need help on this. I know what $y=x-\frac{1}{2}$ and $y=x+\frac{1}{2}$ looks like but I am not understanding the problem as a whole
82240
We have $n$ distinct balls ($n&gt;7$) and want to randomly (and independently) distribute them into $N$ distinct boxes ($N&gt;n$) which are placed one next to the other. a) What is the probability that all balls are placed in consecutive boxes? b) What is the probability that all balls are placed in consecutive boxes and balls with number $1$, $4$ and $7$ are also placed in consecutive boxes?  I am not sure but I will try: a) There is a block of n balls which must be placed in one of the $N-n+1$ gaps formed if we arrange the boxes and fill $n$ consecutive of them. So there are $N-n+1$ places place this block. The $n$ balls in the block can be arranged in $n!$ ways. Also the N boxes can be arranged in $N!$ ways. So all in all we have $n!N!(N-n+1)$. b) Again we have $N!$ ways to arrange the boxes but the balls can be arranged in $(n-3)!$ ways, right? I don't know what is the total number of ways to place all balls in the boxes - plus that I see no restriction regarding the number of balls per box. Any help?
30709
If $n$ distinct objects are arranged in a circle, I need to prove that the number of ways of selecting three of these $n$ things so that no two of them are next to each other is $\frac{1}{6}n(n-4)(n-5)$.  Initially I can select $1$ object in $n$ ways. Then its neighbours cannot be selected. So I will have $n-3$ objects to choose $2$ objects from. Again, I can select the second object in $n-3$ ways. The neighbours of this object cannot be selected. However from here onwards, I am unable to extend this argument as the selection of the third object is dependent on the position of the first and the second object. Is there any simpler method to prove the result?
34648
Consider the triangle formed by randomly distributing three points on a circle. What is the probability of the center of the circle be contained within the triangle?
6437
How many points can one can place in $\mathbb{R}^n$, with the requirement that no $n+1$ points lie in the same $\mathbb{R}^{n-1}$-plane, and the euclidean distance between every two points is an integer?
64423
What is the probability that a particular set of integer edge lengths selected from an interval $[1, N]$, can form a triangle?  How might this extend to the case where one selects real number edge lengths from the unit interval?   I'd also be very interested in any comments on what the probability distribution would look like for the distances between the triangles vertices.
29365
"Three points are chosen uniformly and at random from a unit square. What is the expected value of the area of the resulting triangle?"  I need to do a research about that problem and i found this suggested solution: .  Now, I understand almost everything except anecdote (2) when he computes the expected value of $b$ and $v$.  I can't understand how he reaches those calculations.  If someone can explain to me this that would be great.  Thanks.
22494
Let $C$ be a convex subset of $\mathbb{R}^n$. I've been trying for hours to prove that $\dot{\overline{C}}=\dot{C}$. Somehow my intuition completely fails me. I found a proof in a textbook, but just got stuck on another statement the author considered obvious. Could someone please give a proof that uses little more than elementary linear algebra, topology, and the definition of a convex set?  Edit:  The proof mentioned above is from : Let $y\in\dot{\overline{C}}$ and $T$ be a ball around $y$ contained in $\overline{C}$. Then $C\cap T$ has an inner point, as otherwise $C\cap T$ would be contained in a hyperplane and $\overline{C\cap T}=\overline{T}$ would be contained in the same hyperplane. The problematic statement is "as otherwise $C\cap T$ would be contained in a hyperplane".  Another thing: I would be interested in a proof that doesn't use the theorem about separating a convex set from a point by a hyperplane, as I came across this problem in a proof of that very theorem (in the appendix of . To be more precise, it occurs in the case of the point in question being in the boundary of $C$, when it is tacitly assumed, that it is also in the boundary of $\overline{C}$. I know this isn't strictly necessary, as I could use another proof, e.g. the one referred to by Mike, but now I'm curious.
68938
Here is a professor's solution to the exercise which states, " Prove that if $x$ is an isolated point of a set $S \subseteq \mathbb{R}$, then $x$ is a boundary point of $S$."    The professor derived a contradiction upon assuming that for an arbitrary neighborhood $N$ centered at an isolated point $x$ of $S$, we have $N \subseteq S$, so he then proceeds in saying that every $N$ must have a noenmpty intersection with $S^c$. But I feel that there is something wrong with his argument: Using mathematical logic, we can represent his assumption that every $N$ is a subset of $S$ as $\forall \, N : p(N)$ where $p(N)$ means, "$N$ is contained in $S$." He then showed that this assumption is false, so $\sim \forall \, N : p(N) \equiv \exists \, N : \, \sim p(N)$ is true, which means there are some $N$'s that are not contained in $S$. But this information is not sufficient in concluding as to whether or not there is at least one $N$ contained in $S$; it just means not every $N$ is contained in $S$. But if I am right, how can he proceed with his proof since whatever result he deduces about $N$ after the contradiction does not necessarily apply to all $N$?  I hope what I am trying to convey is clear, and I thank you for your help
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Can somebody explain to me graphically or intuitively why this is true? Because if we think in terms of the base of the triangle formed inside a unit circle it seems like it should not be negative.
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It's known that .  Five random points can go right into the $6\times6$ matrix, and then the $A x^2 + B xy + C y^2$ part can be looked at.  If $B^2-4AC&lt;0$, it's an ellipse.  Five random points will almost never produce circles or parabolas, so the results will be ellipses and hyperbolas.  What are the odds of an ellipse?    In a random run of 100000 trials, I got 27974 ellipses. "It's less than $e/10$," seems like a solid answer.  Anyone have anything more specific?  EDIT:  As Oscar points out, I should have said "It's more than $e/10$."  In my trial, real-values points were randomly picked from a unit square.   methods might be applicable.  EDIT2: Aretino points out that  are $49/144≈0.34$. So how can points making a convex pentagon give a non-ellipse?  Here's a picture. With the red points fixed, the black points are outside of the convex hull yet still yield a non-ellipse.     EDIT3: That spray of points above goes back to Newton, Philosophiae naturalis principia mathematica, 1687, where he solved the  (). If a point is between one of the two parabolas and the degenerate lines, then it gives a hyperbola.