High school probability Uncovered: Discover the Secrets with Plainmath's Expert Help

How do I generate doubly-stochastic matrices uniform randomly?
A doubly-stochastic matrix is an n×n matrix P such that
$\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{p}_{ij}=\underset{j=1}{\overset{n}{\sum <!--\; \sum \; -->}}{p}_{ij}=1$
where $p_{ij}\ge <!--\; \ge \; -->0$. Can someone please suggest an algorithm for generating these matrices uniform randomly?

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.

Over a period of 100 days, weather forecasters issue forecasts for rain on a daily basis. Based on the forecasts and actual observations, you are supposed to help them finding out if the forecasts were any good. A forecast is simply one of the following three possible statements: “no rain”, “maybe rain”, or “certainly rain”. The forecasters record the number of days they issued each of these three forecasts, and also the corresponding number of days it rained. The results are collected below:
Forecast: "No rain" , Days with rain: 4, Days without rain: 18
Forecast: "Maybe rain" , Days with rain: 25, Days without rain: 22
Forecast: "Rain" , Days with rain: 25, Days without rain: 6
Show that $n_{ij}$ has a multinomial distribution with unknown probabilities $p_{ij}$, which are the unknowns parameters in this problem. Write down the likelihood of the data $n_{ij}$ as a function of $p_{ij}$ and N.
This is a past exam paper and I have no clue how to answer this in a way that could get me marks! I've written the multinomial distribution:
$\frac{n!}{X_1!...X_n!}{p}_1^{x_1}\ast <!--\; \ast \; -->p_2^{x_2}...\ast <!--\; \ast \; -->p_n^{x_n}$
But I can't work out how to do this question for the life of me! Can anyone help?

A card is drawn and replaced five times from an ordinary deck of 52 cards and the sequence of colors is observed. What is the probability that:
a) Five red cards were drawn?
b) Five black cards were drawn?
c) Three red and two black cards were drawn?
d) why is it necessary to replace the cards?
My thoughts:
a) ${}^5{P}_1{\left({\displaystyle \frac{26}{52}}\right)}^1(1-<!--\; -\; -->p{)}^4+...{+}^5{P}_5{\left({\displaystyle \frac{26}{52}}\right)}^5(1-<!--\; -\; -->p{)}^0$
b) isn't this the same as part (a) ?
c) isn't this the same as asking exactly 5 black or red cards were drawn ?
d) not sure about this one.

How many license plates can be made consisting of 2 letters followed by 3 digits (using the fundamental counting principle to solve)? What I know:
1. 26 letters in alphabet, so that means ${\displaystyle 2\times 26}$
2. 10 digits possible (0-9), so that means ${\displaystyle 3\times 10}$
3. FC principle says given m and n options gets you ${\displaystyle m\times n}$ varieties...
... However, the answer key says "676,000" when I got 1560...

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?

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.

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
$A_1,\dots <!--\; \dots \; -->,A_n\text{ are stochastically independent }⟹<!--\; ⟹\; -->P\left(\underset{j=1}{\overset{n}{\bigcap <!--\; \bigcap \; -->}}{A}_j^{k_j}\right)=\underset{j=1}{\overset{n}{\prod <!--\; \prod \; -->}}P(A_j^{k_j}).$

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?

A cricket club has 15 members, of whom only 5 can bowl. What is the probability that in a team of 11 members at least 3 bowlers are selected?

Differentiate between basic and binomial probability
So I saw a question under the topic for Binomial Distributions which asks that what is the probability of making 4 out of 7 free throws where the $P(makingafreethrow)=0.7$. Why can't the answer be a simple $(0.7{)}^4$? Why would it be $7C4\ast <!--\; \ast \; -->(0.7{)}^4\ast <!--\; \ast \; -->(0.3{)}^3$?

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?

Probability using binomial distribution
A probability that a manufactured device has 3% or more deffects is $p=0.02$. If a company buys 5 devices, what is the probability that at least one has 3% or more defects.
I am thinking using binomial distribution to find probabilities that 1, 2, 3, 4 or all 5 devices are defective. So
$P(A)=(\genfrac{}{}{0ex}{}{5}{1})p(1-<!--\; -\; -->p{)}^4+(\genfrac{}{}{0ex}{}{5}{2})p^2(1-<!--\; -\; -->p{)}^3+(\genfrac{}{}{0ex}{}{5}{3})p^3(1-<!--\; -\; -->p{)}^2+(\genfrac{}{}{0ex}{}{5}{4})p^4(1-<!--\; -\; -->p)+(\genfrac{}{}{0ex}{}{5}{5})p^5$
Is this a correct approach or there is an easier way to solve the problem?

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?

Mary is certainly one of medical doctor Brown's patients. She has carried out a domestic pregnancy take a look at which has given a high quality end result. what's the possibility that the pregnancy take a look at used by doctor Brown in his surgical procedure will say Mary is pregnant given that the home check became nice?
Home pregnancy test is 85% accurate
Doctor Brown pregnancy test is 95% accurate
20 females are pregnant and 80 females are not

Different 4 math books , 6 different physics books , 2 different chemistry books are to be arranged on a shelf. how many arrangement is possible if the books in particular subject must stand together ?

Find each of the following probabilities when n independent Bernoulli trials are carried out with probability of success p.
a) the probability of no successes
b) the probability of at least one success
c) the probability of at most one success
d) the probability of at least two successes
e) the probability of no failures
f) the probability of at least one failure
g) the probability of at most one failure
h) the probability of at least two failures

What is the probability an event occurs, given the event occurs exactly N times in the next M trials?
Assume the probably an event occurs are equal, but what if the probabilities were unequal for each trial?