Probability Examples and Explanations

An airline claims that there is a 0.10 probability that a coach-class ticket holder who flies frequently will be upgraded to first class on any flight. This outcome is independent from flight to flight. Sam is a frequent flier who always purchases coach-class tickets.
(a) What is the probability that Sam's first upgrade will occur after the third flight?
(b) What is the probability that Sam will be upgraded exactly 2 times in his next 20 flights?
(c) Sam will take 104 flights next year. Would you be surprised if Sam receives more than 20 upgrades to first class during the year?

To win a particular lottery game, a player chooses 4 numbers from 1 to 50. Each number can only be chosen once.
1. How many different number combinations are possible =
2. What is the probability that the chosen number is 2500=

What does a cross product of 0 mean?

Which of the following statements about probabilities and components is not true?
Choose the correc answer below:
a) $P(E)-<!--\; -\; -->P(E^{\prime })=1$
b) $P(E)=1-<!--\; -\; -->P(E^{\prime })$
c) $P(E)+P(E^{\prime })=1$
d) $P(E^{\prime })=1-<!--\; -\; -->P(E)$

If randomly picked from m symbols, each symbol being equally likely, how long on average would it take to pick m/2 different symbols?
Lets take the specific example of the English alphabet. In this case, I have 26 letters (symbols). Lets say these letters are plastic toys in a box. At random, I take one toy letter from the box, and write it down on a piece of paper. I then put that toy letter back into the box, and repeat the process. How many times, on average, would I have to repeat this process for my piece of paper to contain 13 of the 26 letters of the alphabet?
Is there a way to generalize the answer to m symbols instead of 26? What happens when m is odd?

Relationship among Exponential, Gamma, and Normal distribution
I am studying stochastic processes where I stumbled upon the theorem that says the sum of exponential distributions is gamma distribution. However, from the central limit theorem, we know that sum of a sufficiently large number of IID random variables converges to Normal distribution. So is the following relationship always true?

Let X be a separable Banach space. The p-Wasserstein space on X is defined as
$W_p(X)=\{\mathrm{\mu <!--\; \mu \; -->}\in <!--\; \in \; -->P(X):{\int <!--\; \int \; -->}_X||x|{|}^{p}d\mathrm{\mu <!--\; \mu \; -->}(x)<\mathrm{\infty <!--\; \infty \; -->}\},p\ge <!--\; \ge \; -->1$

Bayes' Theorem in Conditional Probability
We are testing for a disease D that we think is present, D+, with probability 0.4, and absent, D-, with probability 0.6. We believe that a test has sensitivity $P\{T+|D+\}=0.75$ and specificity $P\{T-<!--\; -\; -->|D-<!--\; -\; -->\}=0.8$.
Q1: What is our probability that the disease is present if we perform the test and it is positive, T+, and our probability the disease is absent if that test is negative, T-?
My ans:
We can apply Bayes' formula to calculate $P\{D+|T+\}$ and $P\{D-<!--\; -\; -->|T-<!--\; -\; -->\}$.
$P\{D+|T+\}=5/7$
$P\{D-<!--\; -\; -->|T-<!--\; -\; -->\}=24/29$
Q2: Suppose we perform three tests, conditionally independent given D. Given each possible number of positive test results, 0, 1, 2, or 3, what is our probability that the disease is present?
My ans:
Let k denote the number of positive test results.
We know that P{ [exactly] k successes in n trials | p } $=(nCk)x(p^k)\times <!--\; \times \; -->((1-<!--\; -\; -->p{)}^{n-k})$, hence we can calculate $P\{k=0|D+\}andP\{k=0|D-<!--\; -\; -->\}$.
We can then apply Bayes' formula to calculate $P\{D+|k=0\}$.
$P\{D+|k=0\}=0.692$
Using the same approach for $k=1:3$, we derive:
$P\{D+|k=1\}=0.5$
$P\{D+|k=2\}=0.308$
$P\{D+|k=3\}=0.165$
Would really appreciate it if someone can verify whether my reasoning and answers are correct.

Let X be a random variable with finite expectation, and suppose that $E[e^{-<!--\; -\; -->X}]\le <!--\; \le \; -->1$. Prove that $E[X]\ge <!--\; \ge \; -->P(X\ge <!--\; \ge \; -->2)$.<br<My first thought was to use Markov's inequality, but I can only use that if X is non-negative, and in any case, the resulting inequality doesn't seem very helpful. Even if I assume that $X\ge <!--\; \ge \; -->-<!--\; -\; -->c$ and apply it to $Y=X+c$, I'm left to show $P(X\ge <!--\; \ge \; -->2)\ge <!--\; \ge \; -->\frac{c}{c+1}$, which doesn't seem like it's always true.
I'm primarily unsure of how to use the $E[e^{-<!--\; -\; -->X}]$ condition; Jensen's inequality tells me that the expectation is non-negative but there's presumably more to it.

1. Is there evidence of diffetence in mean time spent exercising between males and females? In the sample, we have a mean of 12.4 hours a week for 20 males and a mean of 9.4 hours a week for 30 females.
2. Is there evidence of a negative correlation between blood pressure and heart rate? In a sample of 200 2. Is there patients, we found a sample correlation of -0.057.

About moment generating function
It's on P91, probability theory and examples, version 5, by Rick Durrett. Although it's a section about large deviation, I think my question is universal for moment generating functions.
Define $\mathrm{\phi <!--\; \phi \; -->}(\mathrm{\theta <!--\; \theta \; -->})=E(e^{\mathrm{\theta <!--\; \theta \; -->}X})$. Let ${\mathrm{\theta <!--\; \theta \; -->}}_{+}=sup\{\mathrm{\theta <!--\; \theta \; -->}:\mathrm{\phi <!--\; \phi \; -->}(\mathrm{\theta <!--\; \theta \; -->})<\mathrm{\infty <!--\; \infty \; -->}\}$ and ${\mathrm{\theta <!--\; \theta \; -->}}_{-<!--\; -\; -->}=inf\{\mathrm{\theta <!--\; \theta \; -->}:\mathrm{\phi <!--\; \phi \; -->}(\mathrm{\theta <!--\; \theta \; -->})<\mathrm{\infty <!--\; \infty \; -->}\}$.
1. Why $\mathrm{\phi <!--\; \phi \; -->}(\mathrm{\theta <!--\; \theta \; -->})<\mathrm{\infty <!--\; \infty \; -->}$ in $({\mathrm{\theta <!--\; \theta \; -->}}_{-<!--\; -\; -->},{\mathrm{\theta <!--\; \theta \; -->}}_{+})$? I think the reason is my proof in the second question. But I want to confirm it.
2. Assume ${\mathrm{\theta <!--\; \theta \; -->}}_{+}>0$, can I prove $(0,{\mathrm{\theta <!--\; \theta \; -->}}_{+})\subset <!--\; \subset \; -->({\mathrm{\theta <!--\; \theta \; -->}}_{-<!--\; -\; -->},{\mathrm{\theta <!--\; \theta \; -->}}_{+})$? My idea is the following. Let $max\{{\mathrm{\theta <!--\; \theta \; -->}}_{-<!--\; -\; -->},0\}<{\mathrm{\theta <!--\; \theta \; -->}}_0<{\mathrm{\theta <!--\; \theta \; -->}}_{+}$. Note that $e^{\mathrm{\theta <!--\; \theta \; -->}x}\le <!--\; \le \; -->1+e^{{\mathrm{\theta <!--\; \theta \; -->}}_{0}x}$, for all $0<\mathrm{\theta <!--\; \theta \; -->}<{\mathrm{\theta <!--\; \theta \; -->}}_0$. Thus $\mathrm{\phi <!--\; \phi \; -->}(\mathrm{\theta <!--\; \theta \; -->})\le <!--\; \le \; -->1+\mathrm{\phi <!--\; \phi \; -->}({\mathrm{\theta <!--\; \theta \; -->}}_0)<\mathrm{\infty <!--\; \infty \; -->}$.
If this is true, I am confused about the proof of Lemma 2.7.2, which discusses $0<\mathrm{\theta <!--\; \theta \; -->}<{\mathrm{\theta <!--\; \theta \; -->}}_{-<!--\; -\; -->}$. Because it's unnecessary.

How are draws without replacement formalized?
Suppose I have some i.i.d. random variables $X_1,\dots <!--\; \dots \; -->,X_n$ which represent the nth balls drawn uniformly from a bin without replacement. Is sampling without replacement a condition that is put on the set I define my probability measure on? In that all elements $w\in <!--\; \in \; -->\mathrm{\Omega <!--\; \Omega \; -->}$ are such that they do not have repeated draws? Or is it a property of the random variable with which I use to sample?

Expectation of an average-centralised sum
Let $X_1,\dots <!--\; \dots \; -->,X_n$ be independently identically distributed random variables with $EX=\mathrm{\mu <!--\; \mu \; -->}$ and $\text{Var}X={\mathrm{\sigma <!--\; \sigma \; -->}}^2$. Let:
${\stackrel{¯<!--\; ¯\; -->}{X}}_n=\frac{1}{n}\underset{k=1}{\overset{n}{\sum <!--\; \sum \; -->}}{X}_k,$
$m_n^2=\frac{1}{n}\underset{k=1}{\overset{n}{\sum <!--\; \sum \; -->}}(X_k-<!--\; -\; -->{\stackrel{¯<!--\; ¯\; -->}{X}}_n{)}^2.$
Prove that $E(m_n^2)={\mathrm{\sigma <!--\; \sigma \; -->}}^2\frac{n-<!--\; -\; -->1}{n}$
This is a problem from Allan Gut's Probability - A Graduate Course, and after messing around with the sums for a while I have no idea what to do.

Measurable and non-singular equivalence relation in a one-sided Markov Shift
In a one-sided Markov shift $(X,\mathcal{B},\mathrm{\mu <!--\; \mu \; -->},T)$ where $X=\{1,2,\cdots <!--\; \cdots \; -->,n{\}}^{\mathbb{N}}$ is the sample space, B the $\mathrm{\sigma <!--\; \sigma \; -->}$-algebra generated by finite cylinder sets, $\mathrm{\mu <!--\; \mu \; -->}$ a product measure and T the shift transformation, first, we define a metric $d(x,y)=2^{-<!--\; -\; -->n}$ where n is the least integer such that $x[n]\ne <!--\; \ne \; -->y[n]$. Then, we define the following relation:
Given $x,y\in <!--\; \in \; -->X$, $x{\sim <!--\; \sim \; -->}_{r}y$ iff there exists $n,m\in <!--\; \in \; -->{\mathbb{Z}}_{\ge <!--\; \ge \; -->0}$ such that $T^{n}x=T^{m}y$.
One can easily check ${\sim <!--\; \sim \; -->}_r$ is an equivalent relation. We set $R=\{(x,y)\in <!--\; \in \; -->X\times <!--\; \times \; -->X|x{\sim <!--\; \sim \; -->}_{r}y\}$ and, for each subset $A\subseteq <!--\; \subseteq \; -->X$, we set $R(A)=\underset{a\in <!--\; \in \; -->A}{\bigcup <!--\; \bigcup \; -->}[a{]}_{{\sim <!--\; \sim \; -->}_r}$ to be the union of equivalent classes of each element in A.
My question is:
1. Is $R\in <!--\; \in \; -->\mathcal{B}\times <!--\; \times \; -->\mathcal{B}=\{A\times <!--\; \times \; -->B|A,B\in <!--\; \in \; -->\mathcal{B}\}$?
2. For each $B\in <!--\; \in \; -->\mathcal{B}$, is it true that $\mathrm{\mu <!--\; \mu \; -->}(B)=0$ iff $\mathrm{\mu <!--\; \mu \; -->}[R(B)]=0$?
The first question is asking if R is measurable and the second is asking if R is non-singular. For the first one, it suffices to show that, for each $B\in <!--\; \in \; -->\mathcal{B},R(B)\in <!--\; \in \; -->\mathcal{B}$. However, R(B) could be properly contained in $\underset{n\in <!--\; \in \; -->\mathbb{N}}{\bigcup <!--\; \bigcup \; -->}{T}^n(B)$ and then I do not know how to proceed. I also try to consider the function $(x,y)\mapsto <!--\; \mapsto \; -->\underset{n,m}{inf}d(T^{n}x,T^{m}y)$. This function is continuous but it may obtain zero outside of R.
For the second question, I suppose whenever $\mathrm{\mu <!--\; \mu \; -->}(B)>0$, B will contain a finite cylinder set and hence so does R(B). However, I do not how to deal with the other direction.

Probability that a 5 cards hand from a standard deck of 52 contains at least 1 ace $\frac{C(4,1)C(51,4)}{C(52,5)}$?
I understand the direct and indirect methods for solving this problem, but I do not understand why $\frac{C(4,1)C(51,4)}{C(52,5)}$ is wrong. The answer that $\frac{C(4,1)C(51,4)}{C(52,5)}$ results is about 4% bigger than the answer I got with $1-<!--\; -\; -->\frac{C(4,0)C(48,5)}{C(52,5)}$ or the direct approach. What are the extra scenarios that $\frac{C(4,1)C(51,4)}{C(52,5)}$ incorrectly cover that causes the 4% difference?

Given $(X_i{)}_{i=1}^{\mathrm{\infty <!--\; \infty \; -->}}$ a series of independent random variables, $X_i\sim <!--\; \sim \; -->G(\frac{1}{2})$ (Geometric distribution) for all $i\ge <!--\; \ge \; -->1$.
For $n\ge <!--\; \ge \; -->2$, we mark the following:
${\stackrel{¯<!--\; ¯\; -->}{X}}_n=\frac{1}{n}\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{X}_i{Y}_n=\ln <!--\; \; -->(1+({\stackrel{¯<!--\; ¯\; -->}{X}}_n{)}^2)T_n=(Y_n-<!--\; -\; -->\ln <!--\; \; -->5{)}^2$
And we need to find the value $C_n$ and the distribution of T, which satisfies:
$\frac{1}{C_n}\underset{n=1}{\overset{12}{\sum <!--\; \sum \; -->}}{T}_n\stackrel{d}{\to <!--\; \to \; -->}T$
Well, i'll save you the trouble. The final and correct answer is
$C_n=\frac{32}{25n},T_n\sim <!--\; \sim \; -->{\mathrm{\chi <!--\; \chi \; -->}}_{(12)}^2\text{ (chi distribution with 12 degrees of freedom)}$
But i can't understand why. This is all I know so far:
Let $(Z_i{)}_{i=1}^n$ be a series of independent random variable with the same distribution $Z_i\sim <!--\; \sim \; -->N(\mathrm{\mu <!--\; \mu \; -->},{\mathrm{\sigma <!--\; \sigma \; -->}}^2)$, then we know that $\frac{Z_i-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->}}{\mathrm{\sigma <!--\; \sigma \; -->}}\sim <!--\; \sim \; -->N(0,1)$ for all i, and we know that the distribution of the following sum is: $\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}(\frac{Z_i-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->}}{\mathrm{\sigma <!--\; \sigma \; -->}}{)}^2\sim <!--\; \sim \; -->{\mathrm{\chi <!--\; \chi \; -->}}_{(n)}^2$
My guess is that in my question:
$Z_i:=Y_n=\ln <!--\; \; -->(1+({\stackrel{¯<!--\; ¯\; -->}{X}}_n{)}^2)$
$\mathrm{\mu <!--\; \mu \; -->}:=E(Y_n)=\ln <!--\; \; -->5$
${\mathrm{\sigma <!--\; \sigma \; -->}}^2:=V(Z_i)=C_n$
and thus:
$\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}(\frac{Z_i-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->}}{\mathrm{\sigma <!--\; \sigma \; -->}}{)}^2=\frac{\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}(Z_i-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->}{)}^2}{{\mathrm{\sigma <!--\; \sigma \; -->}}^2}:=\frac{\underset{n=1}{\overset{12}{\sum <!--\; \sum \; -->}}{T}_n}{C_n}\stackrel{d}{\to <!--\; \to \; -->}T$
But i can't understand how is that $E(Y_n)=\ln <!--\; \; -->5$, or even what is the distribution of $Y_n$?
Am I on the right track? What are your thoughts about this question? How would you solve it?

Show lemma on probability of n independent sets
Let be $(\mathrm{\Omega <!--\; \Omega \; -->},\mathcal{A},P)$ a probability space and $A_1,\dots <!--\; \dots \; -->,A_n\in <!--\; \in \; -->\mathcal{A}$. For sake of convenience we define $A^1:=A$ and $A^c:=\mathrm{\Omega <!--\; \Omega \; -->}\setminus <!--\; \setminus \; -->A$. Let be $(k_1,\dots <!--\; \dots \; -->,k_n)\in <!--\; \in \; -->\{1,c{\}}^n$. Show that
$A_1,\dots <!--\; \dots \; -->,A_n\text{ are stochastically independent }⟹<!--\; ⟹\; -->P\left(\underset{j=1}{\overset{n}{\bigcap <!--\; \bigcap \; -->}}{A}_j^{k_j}\right)=\underset{j=1}{\overset{n}{\prod <!--\; \prod \; -->}}P(A_j^{k_j}).$

Should the probability of an event set only be monotonic?
So our server had this question posted.
What are the chances of an event X to happen by the year 2025, by 2028 and by 2030?
Everyone but one of the user predicted non-monotonically.
i.e. these were his predictions :
- By 2025 : 40%
- By 2028 : 50%
- By 2030 : 45%
Now this is where our server is divided. We can't seem to agree on whether the predictions here should be monotonic or not. Is it logically incorrect to have the third probability less than second probability? If so, why?

Interpreting probability question in terms of measure theory
Let X and Y be real valued random variables with joint pdf
$$f_{X,Y}(x,y)=\{\begin{array}{ll}\frac{1}{4}(x+y),& 0\le <!--\; \le \; -->x\le <!--\; \le \; -->y\le <!--\; \le \; -->2\\ 0,& \text{ otherwise }\end{array}$$
Calculate the probability $\mathbb{P}\{Y<2X\}$
I am trying to view this problem in a measure-theoretic perspective and I am wondering if I am thinking about this properly.
Let $(\mathrm{\Omega <!--\; \Omega \; -->},\mathcal{F},\mathbb{P})$ be a probability space on which we define two random variables (measurable functions) X and Y. We then consider their joint distribution, the pushforward measure of the random variable $T(\mathrm{\omega <!--\; \omega \; -->})=(X(\mathrm{\omega <!--\; \omega \; -->}),Y(\mathrm{\omega <!--\; \omega \; -->}))$ on $({\mathbb{R}}^2,\mathcal{B}\times <!--\; \times \; -->\mathcal{B})$ defined by
$$T_{\star <!--\; \star \; -->}\mathbb{P}(A)=\mathbb{P}(T^{-<!--\; -\; -->1}(A))=\mathbb{P}((X(\mathrm{\omega <!--\; \omega \; -->}),Y(\mathrm{\omega <!--\; \omega \; -->}))\in <!--\; \in \; -->A).$$
Then by the question, we have that ${\displaystyle \frac{d{T}_{\star <!--\; \star \; -->}\mathbb{P}}{d\mathrm{\lambda <!--\; \lambda \; -->}}}=f_{X,Y}$ i.e. the Radon-Nikodym derivative of the pushforward of T with respect to the Lebesgue measure is the pdf of (X,Y).
How can I then calculate $\mathbb{P}\{Y<2X\}$? Somehow I have to relate the probability of this set to the pushforward measuere of which I know the density. What is the theorem that allows me to relate these two measures?
Essentially, im looking for the measure of the set $\{\mathrm{\omega <!--\; \omega \; -->}:Y(\mathrm{\omega <!--\; \omega \; -->})<2X(\mathrm{\omega <!--\; \omega \; -->})\}\subset <!--\; \subset \; -->\mathrm{\Omega <!--\; \Omega \; -->}$. I know the density of a measure on ${\mathbb{R}}^2$ which is a different set than $\mathrm{\Omega <!--\; \Omega \; -->}$. How do I know that $\{(x,y):y<2x\}\subset <!--\; \subset \; -->{\mathbb{R}}^2$ is the subset of R2 with which I need to integrate over?