Probability Examples and Explanations

Let $\mathrm{\Omega <!--\; \Omega \; -->}={\mathbb{N}}_0^2,\mathcal{A}=Pot(\mathrm{\Omega <!--\; \Omega \; -->})$ and P the product of two poisson distributions with paremeter ${\mathrm{\lambda <!--\; \lambda \; -->}}_1,{\mathrm{\lambda <!--\; \lambda \; -->}}_{>}0$, i.e
$$\mathbb{P}(\{(n_1,n_2)\})=\frac{{\mathrm{\lambda <!--\; \lambda \; -->}}_1^{n_1}{\mathrm{\lambda <!--\; \lambda \; -->}}_2^{n_2}}{n_1!n_2!}{e}^{-<!--\; -\; -->{\mathrm{\lambda <!--\; \lambda \; -->}}_1-<!--\; -\; -->{\mathrm{\lambda <!--\; \lambda \; -->}}_2}$$
Define then $X:\mathrm{\Omega <!--\; \Omega \; -->}\to <!--\; \to \; -->{\mathbb{N}}_0,(n_1,n_2)\mapsto <!--\; \mapsto \; -->n_1+n_2$. Prove that X is poisson distributed with parameter ${\mathrm{\lambda <!--\; \lambda \; -->}}_1+{\mathrm{\lambda <!--\; \lambda \; -->}}_2$

Is the likelihood function $L(\mathrm{\theta <!--\; \theta \; -->}|X)$ equal to or proportional to $p(X|\mathrm{\theta <!--\; \theta \; -->})$?
Sometimes I find that the likelihood function is written as $L(\mathrm{\theta <!--\; \theta \; -->}|X)=p(X|\mathrm{\theta <!--\; \theta \; -->})$ while other times $L(\mathrm{\theta <!--\; \theta \; -->}|X)\propto <!--\; \propto \; -->p(X|\mathrm{\theta <!--\; \theta \; -->})$.
which is correct?
If $L(\mathrm{\theta <!--\; \theta \; -->}|X)\propto <!--\; \propto \; -->p(X|\mathrm{\theta <!--\; \theta \; -->})$, then what is the constant?

Let g be a probability density function, what is the ratio $\frac{g(x)}{g(t)}$ for $x>t$?

Lottery chance (odds) calculation
We have two lottery game with 1:1000 odds to win. I have money to buy 20 tickets. Q1. If I spend my money on one of these games only then I have 1:50 chance to win. Am I right? Q2. If I spend my money to buy 10-10 tickets and play both of games then I have 1:100 chance to win on the first and the same chance to win on the second game. But what is my chance to win at least one game this time? I have only 1:100 or 1:50 like in question Q1? If I split my money I split (reduce) my chance too?

Marissa is playing a game where she throws darts at a dartboard until her most recent dart is further away from the bullseye than any dart she threw previously. What is the probability that this game ends in k darts thrown?
My solution: The probability is essentially equal to how many valid orderings of the darts there are over k!. Thus, label each dart from 1 to i,corresponding to when it was thrown. Similarly, label the positions, according to how close they are to the bullseye(with 1 being the position closest to the bullseye). We will call the first dart and first position $d_1$ and $p_1$ respectively. Now, we know that $d_k$ needs to be assigned to $p_k$, as the game ends at k darts. Thus, we need to order the first $k-<!--\; -\; -->1$ darts. It's clear that $d_1$ must be assigned to $p_{k-<!--\; -\; -->1}$, as otherwise, there will be a time where the most recent dart thrown will be the furthest from the bullseye, ending the game before k. But after we assign $d_1$ to $p_{k-<!--\; -\; -->1}$, the rest of the ordering can be any which way, right? So it's $\frac{(k-<!--\; -\; -->2)!}{k!}$?
However, the answer is actually $\frac{k-<!--\; -\; -->1}{k!}$, and I fail to see how.

Alice speaks the truth with probability 3/4 and Bob speaks the truth with probability 2/3. A die is thrown and both Alice and Bob observe the number. Afterwards, Alice asserts to Carl (who does not know the number) that the number is 3 while Bob says (to Carl) the number is 1. Find the probability that the number is actually 1.
UPDATE: To clear ambiguity, note that if person decides to lie, he/she will choose a false answer randomly from all the possible false answer ({1,2,⋯,6} - {The number that actually showed up}). Also, a die is thrown, and then both Alice and Bob will see the number. Then, they will lie/say truth accordingly.
My attempt:
Case 1: Number 1 showed up.
Chance of all this happening = $\frac{1}{6}\cdot <!--\; \cdot \; -->\frac{2}{3}\cdot <!--\; \cdot \; -->(\frac{1}{4}\cdot <!--\; \cdot \; -->\frac{1}{5})=\frac{1}{180}$
Case 2: Number 3 showed up
Chance of all this happening = $\frac{1}{6}\cdot <!--\; \cdot \; -->\frac{3}{4}\cdot <!--\; \cdot \; -->(\frac{1}{3}\cdot <!--\; \cdot \; -->\frac{1}{5})=\frac{1}{120}$
Case 3: Other number showed up
Chance of all this happening = $\frac{4}{6}\cdot <!--\; \cdot \; -->(\frac{1}{4}\cdot <!--\; \cdot \; -->\frac{1}{5})\cdot <!--\; \cdot \; -->(\frac{1}{3}\cdot <!--\; \cdot \; -->\frac{1}{5})=\frac{1}{450}$
So, total = $\boxed{{\displaystyle \frac{\frac{1}{180}}{\frac{29}{1800}}=\frac{10}{29}}}$
Is my attempt correct? If not, how to do this problem?

Confusion about sum of random variables conditional probabilities
Let X,Y be independent random variables, and $Z=X+Y.$. Then I want to calculate $Pr[X=x\mid <!--\; \mid \; -->Z=z].$. My confusion is on evaluating this expression. On the one hand, I have
$Pr[X=x\mid <!--\; \mid \; -->Z=z]=Pr[Z-<!--\; -\; -->Y=x\mid <!--\; \mid \; -->Z=z]=Pr[z-<!--\; -\; -->Y=x]=Pr[Y=z-<!--\; -\; -->x].$
But also, $Pr[Z=z\mid <!--\; \mid \; -->X=x]=Pr[X+Y=z\mid <!--\; \mid \; -->X=x]=Pr[Y=z-<!--\; -\; -->x].$ So these two probabilities are equal? But $Pr[Z=z\mid <!--\; \mid \; -->X=x]Pr[X=x]=Pr[X=x\mid <!--\; \mid \; -->Z=z]Pr[Z=z]$ and in general $Pr[X=x]$ and $Pr[Z=z]$ are not equal. I believe it should be $Pr[X=x\mid <!--\; \mid \; -->Z=z]=Pr[Y=z-<!--\; -\; -->x\mid <!--\; \mid \; -->Z=z]$ but I don't think the conditional $Z=z$ can be removed since Y and Z are not independent?
I'm not sure whether the first equation holds either. For example, if I roll a fair six sided die X (numbered 1 to 6) and roll a fair ten sided die Y and take the sum, then $Pr[X=1\mid <!--\; \mid \; -->Z=2]=1$ since the only possible outcome is $(x,y)=(1,1),$, and this is not equal to $Pr[Y=(2-<!--\; -\; -->1)]=1/10.$. On the other hand it is equal to $Pr[Y=(2-<!--\; -\; -->1)\mid <!--\; \mid \; -->Z=2]=1.$. I think I'm making a mistake in one of these but it's not clear to me in which step.
(The context of this was that X is a random variable with given distribution representing some unknown parameter and Y is a standard normal error. Then you observe $z=x+y$ and want to estimate the X.)

It's known that birth months are uniformly distributed. A class is divided into 10 groups of 5 students. A group that all five members were born in different months is our interest. What is the probability that there is one such group of interest among 10 groups?

Sigma-algebra on the image of a random variable
Usually a random variable is considered to be a function X:$X:(\mathrm{\Omega <!--\; \Omega \; -->},\mathrm{\sigma <!--\; \sigma \; -->},P)\to <!--\; \to \; -->(\mathbb{R},B(\mathbb{R}))$ but I wonder what advantage choosing B(R) as a sigma algebra on R has over using the pushforward of $\mathrm{\sigma <!--\; \sigma \; -->}$ under X.

Let X be a random variable with $\mathbb{V}(X)=25$
It also satisfies $\mathbb{P}(X>20)>\frac{1}{4}$
Which of the following is true?
- $\mathbb{E}(X)>10$
- $5<\mathbb{E}(X)\le <!--\; \le \; -->10$
- $0<\mathbb{E}(X)\le <!--\; \le \; -->5$
- $\mathbb{E}(X)\le <!--\; \le \; -->0$
I tried using Chebyshev's inequality and found out that $\mathbb{E}(X)>5$ but wasn't sure how to go on about it.

How to find the average number of iterations for a variable to reach a number.
Suppose I have an infinite loop with a variable x that starts at 0. Every iteration of the loop, x has a 10% chance of being increased by 1 and a 90% chance of being decreased by 1, but it cannot go below 0. How can I calculate the average number of iterations of the loop for x to reach a certain number?

How to read and interpret the following mathematical notation? ${\mathrm{\Gamma <!--\; \Gamma \; -->}}_{su}=\frac{P_s{G}_{su}}{\underset{i\in <!--\; \in \; -->S\{\mathrm{\setminus <!--\; \setminus \; -->}s\}}{\sum <!--\; \sum \; -->}{P}_i{G}_{iu}+P_{AWGN}}$
I came across the following expression in a research paper:
$${\mathrm{\Gamma <!--\; \Gamma \; -->}}_{su}=\frac{P_s{G}_{su}}{\underset{i\in <!--\; \in \; -->S\{\mathrm{\setminus <!--\; \setminus \; -->}s\}}{\sum <!--\; \sum \; -->}{P}_i{G}_{iu}+P_{AWGN}}$$
My query is how to read that small s in denominator which has left slash with it.
Any help in this regard will be highly appreciated.

MLE in a Gaussian sequence model
Consider the Gaussian sequence model
$$y_i={\mathrm{\theta <!--\; \theta \; -->}}_i+\frac{w_i}{\sqrt{n}},\text{for}1\le <!--\; \le \; -->i\le <!--\; \le \; -->n$$
and $w_i\sim <!--\; \sim \; -->N(0,1)$ are i.i.d. My goal is to estimate $\{{\mathrm{\theta <!--\; \theta \; -->}}_i{\}}_{i=1}^n$ based on $y_1,...,y_n$.
Assume that there exists an $S\subset <!--\; \subset \; -->\{1,2,...,n\}$ such that $|S|=s$ and ${\mathrm{\theta <!--\; \theta \; -->}}_i\in <!--\; \in \; -->\{-<!--\; -\; -->1,+1\}$ for $i\in <!--\; \in \; -->S$ and ${\mathrm{\theta <!--\; \theta \; -->}}_i=0$ for $i\in <!--\; \in \; -->S^c$. What should be the MLE of $\mathrm{\theta <!--\; \theta \; -->}=({\mathrm{\theta <!--\; \theta \; -->}}_1,...,{\mathrm{\theta <!--\; \theta \; -->}}_n)$?
The likelihood in this case becomes:
$$L(\mathrm{\theta <!--\; \theta \; -->})={\left(\frac{\sqrt{n}}{\sqrt{2\mathrm{\pi <!--\; \pi \; -->}}}\right)}^n\exp <!--\; \; -->\{-<!--\; -\; -->\frac{n}{2}(\underset{i\in <!--\; \in \; -->S^c}{\sum <!--\; \sum \; -->}{y}_i^2+\underset{i\in <!--\; \in \; -->S}{\sum <!--\; \sum \; -->}(y_i-<!--\; -\; -->{\mathrm{\theta <!--\; \theta \; -->}}_i{)}^2)\}.$$
We need to minimize the exponent and thus we need ${\mathrm{\theta <!--\; \theta \; -->}}_i=y_i$ for $i\in <!--\; \in \; -->S$, but in order to respect the fact that ${\mathrm{\theta <!--\; \theta \; -->}}_i=\pm <!--\; \pm \; -->1$, we need $y_i=\pm <!--\; \pm \; -->1$ for $i\in <!--\; \in \; -->S$, but it is not always the case. So, how should I go about this?

Many people one try vs one person many tries - is there a name for this?
Assuming we have two situations:
- (A) 500 People throw a six-sided dice, the result of each throw is the points given to each person
- (B) One person throws a six-sided dice 500 times, the points of that person is the average of the throws
The expected value in A over all throws is the same as for B (3.5), but those are in my opinion obviously very different situations for the statistical considerations of the single person.
Is there a name to differentiate between those scenarios?
For background:
I had a heated dispute over whether its better-as in aiming for higher points-for a single person to throw the dice once or 500 times (No choice in-between, though that would be interesting, too..)
I was of the opinion that the 500 throws as a single person are quantitatively beneficial, as you are guaranteed 3.5 points, while a single throw depends strongly on luck.
I guess there is a the topic of max-min and min-max. In any case, I did notice that I lack a formal description/understanding for this particular situation, would love some help on that front!

Can we say $X_n$ converge to 0 in probability?
For a sequence of random variables $X_i$ for $i=1,\dots <!--\; \dots \; -->,n$, we have known the result
$$X_n=O_p(n^{-<!--\; -\; -->1/3}).$$
Can we say $X_n$ converge to 0 in probability?
It seems that $o_p$ means the convergence in probability. That means if $X_n=o_p(n)$, then,
$$\frac{X_n}{n}\to <!--\; \to \; -->0$$
in probability.

Probability of first increase in ordering of iid random variables
What is the probability that the first $n-<!--\; -\; -->1$ terms of iid Unif(0,1) random draws are in decreasing order, but the first n terms are not?
I know that due to exchangability, $\mathbb{P}(X_1<X_2<\cdots <!--\; \cdots \; --><X_{n-<!--\; -\; -->1})=\frac{1}{n-<!--\; -\; -->1!}$. Why can't a similar exchangability argument hold to show that $\mathbb{P}(X_1>X_2>\cdots <!--\; \cdots \; -->>X_{n-<!--\; -\; -->1}<X_n)=\frac{1}{n!}$? My reasoning is that the distribution of $(X_1,X_2,\dots <!--\; \dots \; -->,X_n)$ is symmetric, so no matter which permutation of the random variables exist in the random vector, they will all have the same probability.

Derivatives and Integrals of Stochastic Processes?
I am trying to wrap my head around this idea and am trying to understand why this might be useful. For instance, suppose I have some stochastic process like the Brownian Motion. Why might I be interested in knowing "how quickly the Brownian Motion might change" at some point (i.e. the derivative) and the "area that the Brownian Motion might cover over two time intervals" (i.e. the integral)?
I understand this is more complicated than evaluating derivatives and integrals on deterministic functions. In a deterministic function, you only need to take one derivative and one integral to answer your question. On the other hand, each time I simulate a Stochastic Process on a computer, each realization of this Stochastic Process will look different. Thus, it is likely more challenging to take the integral and derivative of a function that can take many forms, compared to a function that can only take a single form.
But this point aside - why is it useful to know the derivative and the integral of a stochastic process? What do I gain from knowing this - how might I be able to apply this information to some real world application? For example, in financial models such as the Black-Scholes Model, are we using Stochastic Calculus to infer the amount of "risk" or "volatility" (i.e. the area under the stochastic process) that the stochastic process corresponds to over some period of time?

For a binomial distribution, which probability is not equal to the probability of 1 success in 7 trials where the probability of success is 0.277?
Select one:
a) Probability of 6.000 failures in 7 trials where the probability of success is 0.723
b) Probability of 1 success in 7 trials where the probability of failure is 0.723
c) Probability of 6.000 failures in 7 trials where the probability of failure is 0.723
d) Probability of 6.000 failures in 7 trials where the probability of success is 0.277