Get Ahead in Geometric Probability: Expert Guidance and Real-World Applications

Probability with elements of geometry
We have a wire of length 20. We bend this wire in random point. And then bend again to get rectangual frame. What's the probability that area of this rectangle is less than 21.

Geometric series with probability
Archeological evidence indicates that the ancient (pre-Roman) Etruscans played dice using a dodecahedral die having 12 pentagonal faces numbered 1 through 12 (figure above). One could simulate such a die by drawing a random card from a deck of cards numbered 1 through k. for your own personal value of k, begin with the largest digit in the sum of the digits in your student ID number. This is your value of k unless this digit is less than 5, in which case subtract it from 10 to get your own value of k. a.) John and Mary draw alternately from a deck of shuffled k cards. The first one to draw an ace-the card numbered one- wins. Assume that John draws first. Use the formula for the sum of a geometric series to calculate (both a rational number and a four place decimal) the probability J that John wins, and the probability M that Mary wins. Check that $M+J=1$. b.) Now John , Mary, Paul draw alternately from a deck of k cards. Calculate separately their respective probabilities of winning, given that John draws first and Mary draw second. Check that $J+M+P=1$

A Diminishing Geometric Distribution
A standard geometric distribution can be interpreted as the number of Bernoulli trials required to get one success. However, what if the probability of success if each trial diminishes by some factor with each failure?
Let p be probability of success of the first trial and d be the diminishing factor of each failure such that the probability of success of trial n is $p{d}^{n-<!--\; -\; -->1}$. I have been trying to calculate the expected value of this distribution without success. My main stumbling block is the probability of failing n times:
$P(n\text{ failures})=\underset{k=1}{\overset{n}{\prod <!--\; \prod \; -->}}(1-<!--\; -\; -->p{d}^{k-<!--\; -\; -->1})$
Given that, the expected value would be
$E(X)=\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}np{d}^{n-<!--\; -\; -->1}F(n-<!--\; -\; -->1)$
Is there a way to get a closed form for this?

Find the probability that a geometric random variable X is an even number
Let $\mathrm{\alpha <!--\; \alpha \; -->}$ be the probability that a geometric random variable X with parameter p is an even number
a) Find $\mathrm{\alpha <!--\; \alpha \; -->}$ using the identity $\mathrm{\alpha <!--\; \alpha \; -->}=\underset{i=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}P[X=2i]$.
b)Find $\mathrm{\alpha <!--\; \alpha \; -->}$ by conditioning on wether $X=1$ or $X>1$

Geometric Probability Distribution, Expected Values
Let $X\sim <!--\; \sim \; -->$ Geometric $(\mathrm{\theta <!--\; \theta \; -->})$, and let $Y=min(X,100)$. Compute (a) E(Y) and (b) $E(Y-<!--\; -\; -->X)$.
I know that the Geometric distribution is $(1-<!--\; -\; -->\mathrm{\theta <!--\; \theta \; -->}{)}^{k-<!--\; -\; -->1}\mathrm{\theta <!--\; \theta \; -->}$ and I also know how to calculate expected value but I'm confused about what it means that $Y=min(X,100)$?

I chose a random number from 1 to 6. Afterwards, I roll a die till I get a result that is even or higher than my chosen number. What is the E(x) of the number of times I throw the die?
So I thought it's Geometric distribution will "success" where "success" is to get my number. So first the probability of choosing a random number in the die is 1/6. Now I can't configure what is the probability to get something even or higher? Since I chose the number randomly. and afterthat, I just need to divide 1 by the probability I get (like the geometric E(x) formula)?

Geometric probabilities with rectangle
One side of rectangle is 1.2 other is 3.9. We randomly pick points on adjacent sides and then draw a stretch through them. What is the probability that the area of the received triangle is less than 1.15?

Probability That the Sides of an Isosceles Triangle is an Equilateral Triangle
The sides of an isosceles triangle are whole numbers, and its perimeter is 30 units. What is the probability that the triangle is equilateral?

Asymptotic behavior of the geometric sequence $U_{n+1}=Q{U}_n$ in a probability problem
We know the asymptotic behavior of the scalar geometric sequence $u_{n+1}=q{u}_n$ with respect to the values of q. That is, $|q|<1$ implies $\underset{n}{lim}{u}_n=0$, $q=1$ implies $u_n$ is constant, etc.
Now I am just asking myself the same question about the geometric sequence $U_{n+1}=Q{U}_n$, where Q is a $2\times <!--\; \times \; -->2$ matrix and $U_n$ is a two dimensional vector.

Probability mass function of sum of two independent geometric random variables
How could it be proved that the probability mass function of $X+Y$, where X and Y are independent random variables each geometrically distributed with parameter p; i.e. $p_X(n)=p_Y(n)=\{\begin{array}{cc}p(1-<!--\; -\; -->p{)}^{n-<!--\; -\; -->1}& n=1,2,...\\ 0& otherwise\end{array}$ equas to ${\mathbf{P}}_{\mathbf{X}\mathbf{+}\mathbf{Y}}\mathbf{(}\mathbf{n}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{n}\mathbf{-<!--\; -\; -->}\mathbf{1}\mathbf{)}{\mathbf{p}}^{\mathbf{2}}\mathbf{(}\mathbf{1}\mathbf{-<!--\; -\; -->}\mathbf{p}{\mathbf{)}}^{\mathbf{n}\mathbf{-<!--\; -\; -->}\mathbf{2}}$

A geometrically distributed random variable where the first success probability is different
Your first success probability $p_a$ is higher on the first trial, than on the remaining trials (where it is $p_b$, constantly). How does this impact the mean of the general experiment (1/p in the general, fixed geometric case) and the standard deviation $\sqrt{(1-<!--\; -\; -->p)/p^2}$ in the general, fixed geometric case)?
I think I got the mean:
$p_a+(1-<!--\; -\; -->p_a)(1/p_b+1)$

Negative Binomial and Geometric Distributions
An actuary has determined that the number of claims per month can take any number 0, 1, 2, 3,... and follows a negative binomial distribution with mean 3 and variance 12. Calculate the probability that the number of claims is at least three but less than six.
So by using some properties of the negative binomial we can derive that $p=0.25$, and $r=1$. It was my understanding that a geometric distribution is just a negative binomial distribution with $r=1$. Given this I tried using the pmf for geometric distribution and got the wrong answer. Can someone explain to me what is going on here.

Geometric distribution and the probability of getting success at "first try"?
If one knows the probability of success p,
Then how does one calculate the probability of getting success at "first try" using geometric distribution?
Is it simply the probability of success? Are the successive events independent?

Geometric probability of intersection of a square and a circle
In the unitary square we choose a point (X,Y) with iid coordinates U[0,1] and a radius R, independent of (X,Y) and U[0,1], and we draw the circle of radius R with center (X,Y). Find the probability that this circle intersects the circumference of the unit square.

Probability about a geometric distribution
Bill, Mary and Tom have coins with respective probabilities $p_1,p_2,p_3$ of turning up heads. They toss their coins independently at the same times.
What is the probability that neither Bill nor Tom get a head before Mary?

Let $X_1,X_2$ be a random variables with geometric distribution with p,
Find the probability function of the random variable $X_1+X_2$

Probability - Geometric Sequence Problem
I have been trying to solve a problem which states the following:
For every hour less than 8, an initial value depreciates by 7.9%. The initial value given is 87. This is what I have calculated so far:
8 hours = 87
7 hours = 80.127
6 hours = 73.797
5 hours = 67.967
4 hours = 62.598
3 hours = 57.652
2 hours = 53.098
1 hours = 48.903
From the above, it can be seen that the problem is a geometric sequence, having a constant ratio of 0.921.
The problem asks to find the number of hours it would take to reach 50. I know that multiplying a previous value by 0.921 gives you the next value for a whole hour (bottom to top). However, I am really struggling to find a way to calculate the hours for the value 50. How can I go about doing this?

Calculate the probability of circuit breakage.
A breakage of the electrical circuit occurs if either an element $K_1$ or both elements $K_2$ and $K_3$ break. The elements $K_1,K_2,K_3$ break independently of each other with probabilities respectively 0.3, 0.2, 0.1. Calculate the probability of circuit breakage.

Find probability function and E(X)
A bowl contains six marbles of which two are green and four are red. The marbles are removed one by one, at random and without replacement. Let X be the number of the drawing at which a red marble is removed for the first time. Find the probability function of X, and $\mathbb{E}(X)$.
For the probability function, I believe that this question is a geometric distribution, so I have $\mathbb{P}[X=x]=p{q}^{x-<!--\; -\; -->1}$
For $\mathbb{E}(X)=1/p$ since it's geometric distribution
Is it the Hypergeometric distribution?
So $E(Y)=nr/N$

Non-geometric way to calculate expected value of breaks?
"A bar is broken at random in two places. Find the average size of the smallest, of the middle-sized, and of the largest pieces."
The author gives what seems like a complicated geometric way of calculating the probabilities. He arrives at the solutions 1/9, 5/18, and 11/18. Is there a simpler, non-geometric way of calculating these probabilities?