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

Geometric Probability in Multiple Dimensions
I understand how Geometric Probability works in 1, 2, and 3 dimensions, but is it possible to do these problems in, say, 5 dimensions? For example,
Five friends are to show up at a party from 1:00 to 2:00 and are to stay for 6 minutes each. What is the probability that there exists a point in time such that all of the 5 friends meet?

Mathematical expectation of the average distance of some set of points
In cartesian plane, N points are randomly generated within a unit square defined by points (0,0)(0,1)(1,1)(1,0), with uniform distribution. What is the mathematical expectation of the average distance of all distances between all points (distances between the point and the same point (which are obviously 0) are excluded from the calculation of the average).
There are actually two versions of this question:
- Distance is defined as Euclidian.
- Distance is defined as Manhattan.
The context of this question is some performance benchmarking in computer science, which are of practical nature and far from pure mathemathics, so I am not going the bother the reader with these details.

Geometric Sequence with Normal Distribution Problem
The running time (in seconds) of an algorithm on a data set is approximately normally distributed with mean 3 and variance 0.25.
a. What is the probability that the running time of a run selected at random will exceed 2.6 seconds?
Answer for (a): I've computed this and found it to be $p=\mathrm{0.7881.}$.
b. What is the probability that the running time of exactly one of four randomly selected runs will exceed 2.6 seconds?
Answer for (b): Not sure, I know it's a geometric sequence and I believe the formula that I need to use to be $p(1-<!--\; -\; -->p{)}^3$, thus giving me $0.7881(1-<!--\; -\; -->0.7881{)}^3$ however this was marked incorrect. I'm trying to figure out why, can someone explain my error and how to arrive at the correct solution?

How can you find $P(\frac{X}{Y-<!--\; -\; -->X}<0)$ if $X\sim <!--\; \sim \; -->Geometric(p)$ and $Y\sim <!--\; \sim \; -->Bernoulli(p)$
Let the independent random variables $X\sim <!--\; \sim \; -->Geometric(p)$ and $Y\sim <!--\; \sim \; -->Bernoulli(p)$, I want to prove that $P({\displaystyle \frac{X}{Y-<!--\; -\; -->X}}<0)=(p-<!--\; -\; -->1{)}^2(p+1)$
Do I need the joint probability mass function for this or should it be proven in some other way?

Calculating the probability of an event occurring in a specific time period
I am confused at how to approach the following question, i.e. what probability formula I am supposed to use.
If the probability of a flood is 0.12 during a year, what is the probability of two floods over the next 10 years...?
I have thought perhaps trying Geometric distribution at first, but it didn't seem to work out properly. I also tried Poisson, but it turned out to be quite a small number... which doesn't seem viable.
So my question is, how can I go about solving this and which probability distribution am I supposed to use?

X and Y are geometric RV's with parameter p.
A) $P\{X+Y=n\}(n=1,2,...)$?

Probability of picking 2 numbers between 0 and 1 to be within 1/2 distance of each other?
What's the probability of picking 2 numbers, x and y, between 0 and 1 such that they will be within the distance of 12 of each other?
In other words, $Pr(\text{distance between x and y}\le <!--\; \le \; -->\frac{1}{2})=?$
I solved the problem through a geometric approach by rewriting the probability as $Pr(|x-<!--\; -\; -->y|\le <!--\; \le \; -->\frac{1}{2})$ and graphing $|x-<!--\; -\; -->y|\le <!--\; \le \; -->\frac{1}{2}$
From the graph, I calculated the red area to be 75%.
Question: What would be a non-geometric solution to this problem?

Probability of multiple variables, geometric distribution?
You are on a basketball team, and at the end of every practice, you shoot half-court shots until you make one. Once you make a shot, you go home. Each half-court shot, independent of all other shots, has a 0.1 probability of going in. Your team has 100 practices per season. Estimate the probability that you shoot more than 1111 half-court shots after practices this season.
So I set X equal to the number of shots until finished per practice, and since $X\sim <!--\; \sim \; -->\mathrm{G}\mathrm{e}\mathrm{o}(0.1)$, I know that the P(X) is $(1-<!--\; -\; -->p{)}^{x-<!--\; -\; -->1}p$ but I don't know what to set x as.
I also set Y equal to the number of shots until finished per season so that $Y=100X$. I'm not sure how to go about finding the probability though? I understand that $P(Y)=P(X{)}^{100}$, but how can I find P(X)?

Non-zero probability of hitting a convex hull of $d+1$ i.i.d. points in ${\mathbb{R}}^d$.
Let $X_1,\dots <!--\; \dots \; -->,X_{d+1}$ be $d+1$ i.i.d. random points in ${\mathbb{R}}^d$ sampled from a continuous probability μ of density f.
Let $x_0\in <!--\; \in \; -->{\mathbb{R}}^d$. Is it true that almost surely with respect to $\mathrm{\mu <!--\; \mu \; -->}$, $\mathbb{P}(x_0\in <!--\; \in \; -->\mathrm{C}\mathrm{o}\mathrm{n}\mathrm{v}(X_1,\dots <!--\; \dots \; -->,X_{d+1}))>0$ where Conv denotes the convex hull of the points?

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 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?

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?

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?

The probability of the dart hitting the smaller ring is?
A dart is randomly thrown at a circular board on which two concentric rings of radii R and 2R having the same width (width much less than R) are marked. The probability of the dart hitting the smaller ring is:
(1) Twice the probability that it hits the larger ring.
(2) Half of the probability that it hits the larger ring.
(3) Four times the probability that it hits the larger ring.
(4) One-fourth the probability that it hits the larger ring.

Let $X_n$ be a geometric random variable with parameter $p=\mathrm{\lambda <!--\; \lambda \; -->}/n$. Compute $P(X_n/n>x)$
$x>0$ and show that as n approaches infinity this probability converges to $P(Y>x)$, where Y is an exponential random variable with parameter $\mathrm{\lambda <!--\; \lambda \; -->}$. This shows that $X_n/n$ is approximately an exponential random variable.

What is the expected volume of the simplex formed by $n+1$ points independently uniformly distributed on ${\mathbb{S}}^{n-<!--\; -\; -->1}$?

When A and B flip coins, the one coming closest to a given line wins 1 penny from the other. If A starts with 3 and B with 7 pennies, what is the probability that A winds up with all of the money if both players are equally skilled? What if A were a better player who won 60% percent of the time?