Expert Help with High School Geometry

Geometric distribution where failure probability is not $1-<!--\; -\; -->p$
The typical geometric distribution is defined from the success probability p, i.e., a r.v. G~Geometric(p), would have PMF... $P[G=g]=(1-<!--\; -\; -->p{)}^{g-<!--\; -\; -->1}p$
Fischer and Spassky play a chess match in which the first played to win a game wins the match. After 10 successive draws. the match is declared drawn. Each game is won by Fischer with probability 0.4 is won by Spassky with probability 0.3. and is a draw with probability 0.3. independent of previous games.
a) What is the probability that Fischer wins the match?
For part a, the answer is obviously ${\mathrm{\Sigma <!--\; \Sigma \; -->}}_{k=1}^{10}(0.3{)}^{k-<!--\; -\; -->1}(0.4)\approx <!--\; \approx \; -->0.571$.
But we could also say that r.v. L~CustomGeometric($p=0.4$, $q=0.3$)...
$P[L=l]=q^{l-<!--\; -\; -->1}p=(0.3{)}^{l-<!--\; -\; -->1}(0.4)$
Where the answer to part a would just come from evaluating the CDF of L, i.e., $P[L\le <!--\; \le \; -->10]$. Is there a name for this more generic version of the Geometric distribution?

Finding the maximum volume of a cylinder.
The cylinder is open topped and is made from $200{\text{cm}}^2$ of material.
So what I have done is said $S=2\mathrm{\pi <!--\; \pi \; -->}{r}^{2}h+2\mathrm{\pi <!--\; \pi \; -->}{r}^2$.
$\begin{array}{rl}2\mathrm{\pi <!--\; \pi \; -->}{r}^{2}h+2\mathrm{\pi <!--\; \pi \; -->}{r}^2& =200\\ 2\mathrm{\pi <!--\; \pi \; -->}{r}^{2}h& =200-<!--\; -\; -->2\mathrm{\pi <!--\; \pi \; -->}{r}^2\\ h& =\frac{200-<!--\; -\; -->2\mathrm{\pi <!--\; \pi \; -->}{r}^2}{2\mathrm{\pi <!--\; \pi \; -->}{r}^2}\end{array}$
Subbing this into the volume equation I get $V=(\mathrm{\pi <!--\; \pi \; -->}{r}^2)\frac{200-<!--\; -\; -->2\mathrm{\pi <!--\; \pi \; -->}{r}^2}{2\mathrm{\pi <!--\; \pi \; -->}{r}^2}$.
Having trouble differentiating that equation.

Points A, B, C lie on a circle with center O. Show that $\mathrm{\measuredangle <!--\; \measuredangle \; -->}OAC=90-<!--\; -\; -->\mathrm{\measuredangle <!--\; \measuredangle \; -->}CBA$

Finding Geometric Probability
So there are 2 parallel lines 20 feet apart. A piece of pipe 20 feet long falls between the lines and one end is exactly 10 feet from one line. What is the probability that the pipe lies entirely between the lines?
I know how to do this if the pipes placement varies but not if it is fixed. Can anyone help me figure out the function and how I must integrate this?

Finding dimensions of a rectangular box to optimize volume
A rectangular box (base is not square) with an open top must have a length of 3ft, and a surface area of $16f{t}^2$. Compute the dimensions of the box that will maximize its volume.
I am going wrong somewhere but I can't see where. I have set $2(3h+3w+hw)=16$ and my optimization equation is $v=hwl$. I set $w=\frac{8-<!--\; -\; -->3h}{3+h}$ and plug into the optimization equation. Calculating v' gives me $\frac{9h^2-<!--\; -\; -->84h+72}{{(3+h)}^2}$,but setting that equal to zero does not give me the answer. I need for h, which is 1.

What is the distance between (31,-201) and (28,-209)?

Fair coin probability of Heads
I have a fair coin which I toss many times. What is the probability that it takes 12 rolls until I get two heads in a row?
I thought this was just standard geometric. But it doesn’t seem to work? Can someone please explain how to do this?

Distance between (-2,4,-13) and (-4,5,-22)?

Newton polygons
This question is primarily to clear up some confusion I have about Newton polygons.
Consider the polynomial $x^4+5x^2+25\in <!--\; \in \; -->{\mathbb{Q}}_5[x]$. I have to decide if this polynomial is irreducible over ${\mathbb{Q}}_5$
So, I compute its Newton polygon. On doing this I find that the vertices of the polygon are (0,2), (2,1) and (4,0). The segments joining (0,2) and (2,1), and (2,1) and (4,0) both have slope $-<!--\; -\; -->\frac{1}{2}$, and both segments have length 2 when we take their projections onto the horizontal axis.
Am I correct in concluding that the polynomial $x^4+5x^2+25$ factors into two quadratic polynomials over ${\mathbb{Q}}_5$, and so is not irreducible?
I am deducing this on the basis of the following definition of a pure polynomial given in Gouvea's P-adic Numbers, An Introduction (and the fact that irreducible polynomials are pure):
A polynomial is pure if its Newton polygon has one slope.
What I interpret this definition to mean is that a polynomial $f(x)=a_n{x}^n+...+a_0$ $\in <!--\; \in \; -->{\mathbb{Q}}_p[x]$ (with $a_n{a}_0\ne <!--\; \ne \; -->0$) is pure, iff the only vertices on its Newton polygon are $(0,v_p(a_0))$ and $(n,v_p(a_n))$. Am I right about this, or does the polynomial $x^4+5x^2+25$ also qualify as a pure polynomial?

What is the distance between (3,5) and (6,2)?

Computing volume using shell resulted in negative value?
The question I have solved (not sure if correctly), as I ended up with a negative volume to which I am confused whether or not I can have a negative Volume is...
Use cylindrical shells to compute the volume of $y=x^2$ and $y=2-<!--\; -\; -->x^2$, revolved about $x=2$.
$x^2=2-<!--\; -\; -->x^2\to 2x^2=2\to \frac{2x^2}{2}=\frac{2}{2}\to x^2=1$
$x=1,x=-<!--\; -\; -->1$
Radius is $r=2-<!--\; -\; -->x$
Height is $x^2-<!--\; -\; -->(2-<!--\; -\; -->x^2)$
Finding the Integral, ${\int <!--\; \int \; -->}_{\mathrm{\neg <!--\; \neg \; -->}1}^{1}2\pi (2-<!--\; -\; -->x)(x^2-<!--\; -\; -->(2-<!--\; -\; -->x^2))dx=2\pi (\frac{4x^3}{3}-<!--\; -\; -->4x-<!--\; -\; -->\frac{2x^4}{3}+2x^2-<!--\; -\; -->x^2+\frac{x^4}{6})+C$
Finding the limits, $\underset{x\to -<!--\; -\; -->1+}{lim}2\pi (\frac{4x^3}{3}-<!--\; -\; -->4x-<!--\; -\; -->\frac{2x^4}{3}+2x^2-<!--\; -\; -->x^2+\frac{x^4}{6})=2\pi (\frac{4(-<!--\; -\; -->1{)}^3}{3}-<!--\; -\; -->4(-<!--\; -\; -->1)-<!--\; -\; -->\frac{2(-<!--\; -\; -->1{)}^4}{3}+2(-<!--\; -\; -->1{)}^2-<!--\; -\; -->(-<!--\; -\; -->1{)}^2+\frac{(-<!--\; -\; -->1{)}^4}{6})=19.897$
$\underset{x\to 1-<!--\; -\; -->}{lim}2\pi (\frac{4x^3}{3}-<!--\; -\; -->4x-<!--\; -\; -->\frac{2x^4}{3}+2x^2-<!--\; -\; -->x^2+\frac{x^4}{6})=2\pi (\frac{4(1{)}^3}{3}-<!--\; -\; -->4(1)-<!--\; -\; -->\frac{2(1{)}^4}{3}+2(1{)}^2-<!--\; -\; -->(1{)}^2+\frac{(1{)}^4}{6})=-<!--\; -\; -->13.614$
And finally, computing the volume from the obtained limits $V=-<!--\; -\; -->13.614-<!--\; -\; -->19.897=-<!--\; -\; -->33.510$.
So, if someone could let me know if (or where) I went wrong, that would be great.

What is the distance between (5,-21,14) and (18,3,-1)?

Let segment gh be a chord of a circle $\mathrm{\omega <!--\; \omega \; -->}$ which is not a diameter, and let n be a fixed point on gh. For which point b on arc gh is the length n minimized?

Let $S^1$ be a circle of perimeter equal to 1 (not radius 1) .Let $f:S^1\to <!--\; \to \; -->S^1$ be an arbitrary homeomorphism. By uniform continuity, it always possible to find an $\mathrm{ϵ<!--\; ϵ\; -->}>0$ such that $d(x,y)<\mathrm{ϵ<!--\; ϵ\; -->}$ implies
$d(f(x),f(y))<\mathrm{ϵ<!--\; ϵ\; -->},$
where d denote the shortest distance between two points on the circle.
My question is: Is it always possible to find an $\mathrm{ϵ<!--\; ϵ\; -->}>0$ such that for any arc $I$ (or call it interval if you like) on $S^1$ such that $length(I)<\mathrm{ϵ<!--\; ϵ\; -->}$ implies that
$length(f(I))<\frac{1}{4}?$
In other words, do homeomorphisms always send very short arcs to short arcs, rather than long arcs?
Since a homeomorphism can be expanding or contracting, I am really puzzled by this. There must be some tricks I don't know.

Angle between chord AB and tangent at A is the same as subtended by segment AB at any point on the circumference.
How to prove this theorem?

What is the midpoint of (-3, 5) and (2, -1)?

A, B, and C Roll Dice
Three players A, B, and C take turns rolling a pair of dice. The winner is the first player who obtains the sum of 7 $(P[7]=1/6)$ on a given roll of the dice. If A rolls 1st, then B, then C, then back to A (etc. etc. etc.). What is the P[A wins], P[B wins], P[C wins]?

Geometric Random Variable Problem
A fair coin is tossed repeatedly and independently until two consecutive heads or two consecutive tails appear. Find the PMF, the expected value, and the variance of the number of tosses.

Mathematically prove that a Beta prior distribution is conjugate to a Geometric likelihood function
I have to prove with a simple example and a plot how prior beta distribution is conjugate to the geometric likelihood function. I know the basic definition as
'In Bayesian probability theory, a class of distribution of prior distribution θ is said to be the conjugate to a class of likelihood function $f(x|\mathrm{\theta <!--\; \theta \; -->})$ if the resulting posterior distribution is of the same class as of $f(\mathrm{\theta <!--\; \theta \; -->})$.'
But I don't know how to prove it mathematically.

Posterior Probability Distribution from Geometric Distribution
Suppose an experimenter believes that $p\sim <!--\; \sim \; -->Unif([0,1])$ is the chance of success on any given experiment. The experimenter then tries the experiment over and over until the experiment is a success. This number of trials will be $[G\mid <!--\; \mid \; -->p]\sim <!--\; \sim \; -->Geo(p)$.
How can you find the distribution of [p∣G]? Do you use Bayes' Theorem?