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

I have two binary matrixes, of the same size (e.j. 5000x5000). Those matrixes represent the same area, divided in cells of the same size. Each cell of one matrix can be true or false, meaning some property is present or not in this cell. One matrix represents the presence of a property A, and the other one the property B.
Therefore, I can easily build a 2x2 contingency table using as variables the presence/absence of A and B:
A=1 A=0
B=1 a b
B=0 c d
a= number of cells where both A and B are present
b= number of cells where only B is present
etc.
And
I can apply a chi-square test on this table, building an "expected" contingency table, to assess the independency of both properties.
But I also need to assess if the number of cells that "overlap" (cells that are are true in both matrixes, i.e. where both A and B are present) is higher or lower than expected if both properties were independent. Of course I can compare real and expected value of a in the real and the expected contingency tables, but what I need is some thing like a probability or a measure of how much overlap is higher or lower than expected. In some way, it can also be seen as a measure of the "correlation" between both properties? I know if I had a smaller number of cells I could use Fisher's exact test, where obtained p-value will indicate the "direction" of the relationship between A and B. But as Fisher's exact test implies factorials, it is not possible.

Approximating the result of probability
10 dices are rolled simultaneously. What is the probability of getting the sum of numbers appearing on top of all the dice as 35?
If we calculate manually, I think it's pretty tedious with no calculators allowed. So, I thought of approximating the answer. We know that rolling 10 dice, has total number of ways as $6^{10}$ which is ~1e7 which makes it ideal for using a normal distribution. So, mean of 10 throws is 35 and SD of 10 throws is ~5.4 and so $Z={\displaystyle \frac{(X-<!--\; -\; -->35)}{5.4}}$ and so we want $P(\le <!--\; \le \; -->35)-<!--\; -\; -->P(\le <!--\; \le \; -->34)$ as my answer, $P(\le <!--\; \le \; -->35)=1/2$ and the other one can be obtained from the Normal graph and hence, we have the answer.
But to my surprise, the interviewer didn't seem convinced! Is there something wrong with my method? Please point out.

Relationship between binomial and negative binomial probabilities
Let X be a negative binomial random variable with parameters r and p, and let Y be a binomial random variable with parameters n and p. Show that $\mathbb{P}(X>n)=\mathbb{P}(Y<r).$
I would like to get an analytic solution. Basically I want to show the following equality mathematically:
$$\underset{i=n+1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}(\genfrac{}{}{0ex}{}{i-<!--\; -\; -->1}{r-<!--\; -\; -->1})p^r(1-<!--\; -\; -->p{)}^r=\underset{i=0}{\overset{r-<!--\; -\; -->1}{\sum <!--\; \sum \; -->}}(\genfrac{}{}{0ex}{}{n}{i})p^i(1-<!--\; -\; -->p{)}^{n-<!--\; -\; -->i}$$

Of a group of 50 students, 20 are freshmen, 10 are sophomores, 15 are juniors, and 5 are seniors. 5 members must be chosen. How many different committees can there be if there must be exactly one senior and exactly two freshmen on the committee?

People tend to attend a party on time with a probability of 0.7. For your birthday party, 45 people are invited. What is a probability that at least 30 arrive on time?

Domain of joint continuous random variables - Geometrically
Given the joint P.D.F of X and Y:
$f_{X,Y}(x,y)=6x^{2}y(x,y)\in <!--\; \in \; -->(0,1)\times <!--\; \times \; -->(0,1)$
$f_{X,Y}(x,y)=0$ elsewhere
The following double integral should be the probability of $P(Y>2X)$:
$P(Y>2X)={\int <!--\; \int \; -->}_0^1{\int <!--\; \int \; -->}_0^{y/2}6x^{2}y\mathrm{d}x\mathrm{d}y$
However what I don't understand is why X takes values from 0 to y/2.
I tried to sketch the corresponding area - upper left triangle of $1\times <!--\; \times \; -->1$ square above $Y=2X$ - and my best guess is that for each value of Y then $X=\frac{Y}{2}$
Finally how would one calculate the $P(Y<2X)$ area where X and Y move?
Right now what I struggle with is finding the corresponding area of $R^2$ regardless of the problem/question

Use the special addition rule to determine the probability of drawing either a spade OR a heart from a standard deck of cards, on one draw from the deck.

How to show that $\int <!--\; \int \; -->\Vert <!--\; \Vert \; -->y-<!--\; -\; -->y_0\Vert <!--\; \Vert \; -->\mathrm{\nu <!--\; \nu \; -->}(dy)<\mathrm{\infty <!--\; \infty \; -->}?$
Let X be a uniformly random variable on [0,1]. Since Y is supported on [cos(2),1], then we can define a probability measure on R so that $Y\sim <!--\; \sim \; -->\mathrm{\nu <!--\; \nu \; -->}$
How to show that for any $y_0\in <!--\; \in \; -->R$ we have
$$\int <!--\; \int \; -->\Vert <!--\; \Vert \; -->y-<!--\; -\; -->y_0\Vert <!--\; \Vert \; -->\mathrm{\nu <!--\; \nu \; -->}(dy)<\mathrm{\infty <!--\; \infty \; -->}?$$

If 99% of all new applicants tell the truth on their applications, then are submitted to a polygraph test which is 90% accurate what is the probability that:
for an applicant who did not lie his test will confirm this
for an applicant who did lie, the test will confirm this?
for an applicant who did lie he will pass the polygraph
for an applicant who is truthful will pass the polygraph
for an applicant who failed the polygraph,
lied on the application for an applicant who passed the polygraph,
was fully truthful on his application
How do I set up a 2-way contingency table or a tree diagram for these?

What is the formula to find the permutation of r objects given that there are n total object and out of them n1 and n2 are indistinguishable?
The question is I've n the total number of objects, out of them I've to find the permutation of r objects given that $r<n$. It's also given that there m and t are indistinguishable objects in n such that $m+t=n$. What will be the formula to find the permutation in such a case?
As an example, we've 8 marbles out of which 5 are red and 3 and blue, and the number of objects we have to choose is 2. The permutation we know would be equal to 4. But I don't don't the formula to evaluate this value of 4.

There is 16 white, 9 black and 7 yellow tulips. What's the probability of randomly picking 9 flowers where 2 are white, 3 are black, and 4 are yellow?

i am seeking to solve a query regarding contingency table as a ways as I know contingency desk display matter no longer densities,and i am having difficult time comprehending this simple desk. My tries turned into basically calculating the marginal distribution however the possibilities failed to sum to one. for example I tried solving the first question via:
$$P(A,B)=\int <!--\; \int \; -->(P(A,B,C)dC$$
but I'm missing something is dC probabilities or just count?

Probability: What is probability of a student solving at least x questions from n randomly chosen questions?
Out of a pool of 50 questions, 7 random questions are selected for the exam. If a student prepares a total of 16 questions perfectly, what is the probability that he can solve at least one of the 7 random questions. My solution is :
- Number of possible question combinations:C(50,7)
- Combinations where no prepared question occurs:C(34,7)
P(no questions solved) : C(34,7)/C(50,7)
P(atleast one question): $1-<!--\; -\; -->(C(34,7)/C(50,7))$
But I am not sure if this approach is correct. In addition, I have to also prepare a script with monte carlo method to verify it empirically. Which I am not sure how to proceed with.

Complete the following probability table:
$\begin{array}{|cccc|}\hline \text{Prior Probability}& \text{Conditional Probability}& \text{Join Probability}& \text{Posterior Probability}\\ P(B)0.70& P(A|B)0.10& P(A\cap <!--\; \cap \; -->B)0.07& P(B|A)\\ P(B^C)0.30& P(A|B^C)0.65& P(A\cap <!--\; \cap \; -->B^C)& P(B^C|A)\\ \text{Total}1& & P(A)& \text{Total}0.9999\\ \hline\end{array}$

Flipping a coin we get either head or tail can't get both so events are mutually exclusive. i.e $P(A\text{and}B)=0$. but flipping the same coin twice may result in either head or tail and result of flipping a coin twice is independent of what appeared the first time. So we can say events are independent?

Conditional expectation of X given (X, Y, Z)
Suppose there are three random variables X, Y, Z (may be dependent on each other).
Then, $E[X|X,Y,Z]=X$, regardless of their inter-relationship?

Can a contingent argument be proved invalid with natural deduction?
I know that when it is a non-contingent contradiction it can be refuted, but for example in this reasoning:
P⊢
P∧Q
I don't know how to refute it.
EDIT: What I mean is how someone can figure out that for example the fallacy of Affirming the consequent P→Q,Q⊢P is invalid reasoning without making truth tables. A mathematician could use a contingency in his reasoning inadvertently, so how could he be aware of his error?

Binomial probabilities in a production process
Screws are made in a production process where the probability of any one screw being defective is constant at $p=0.1$ i.e. 10% of screws produced are defective.
Screws are placed in bags of 15 at the end of the process. At various intervals a bag is checked and the process is stopped if more than 3 screws in that bag are found to be defective.
I am trying to answer the question "What is the probability that there will be sufficient defective screws to stop the process?".
I am using a sum of binomial probabilities and have calculated the answer to be 0.0555. Does this appear correct?

Theorem 16.13. If f is nonnegative, then
$${\int <!--\; \int \; -->}_{\mathrm{\Omega <!--\; \Omega \; -->}}f(T\mathrm{\omega <!--\; \omega \; -->})\mathrm{\mu <!--\; \mu \; -->}(d\mathrm{\omega <!--\; \omega \; -->})={\int <!--\; \int \; -->}_{{\mathrm{\Omega <!--\; \Omega \; -->}}^{\mathrm{\prime <!--\; \prime \; -->}}}f\left({\mathrm{\omega <!--\; \omega \; -->}}^{\mathrm{\prime <!--\; \prime \; -->}}\right)\mathrm{\mu <!--\; \mu \; -->}{T}^{-<!--\; -\; -->1}\left(d{\mathrm{\omega <!--\; \omega \; -->}}^{\mathrm{\prime <!--\; \prime \; -->}}\right).$$
A function f (not necessarily nonnegative) is integrable with respect to $\mathrm{\mu <!--\; \mu \; -->}{T}^{-<!--\; -\; -->1}$ if and only if fT is integrable with respect to $\mathrm{\mu <!--\; \mu \; -->}$, in which case (16.17) and
$${\int <!--\; \int \; -->}_{T^{-<!--\; -\; -->1}{A}^{\mathrm{\prime <!--\; \prime \; -->}}}f(T\mathrm{\omega <!--\; \omega \; -->})\mathrm{\mu <!--\; \mu \; -->}(d\mathrm{\omega <!--\; \omega \; -->})={\int <!--\; \int \; -->}_{A^{\mathrm{\prime <!--\; \prime \; -->}}}f\left({\mathrm{\omega <!--\; \omega \; -->}}^{\mathrm{\prime <!--\; \prime \; -->}}\right)\mathrm{\mu <!--\; \mu \; -->}{T}^{-<!--\; -\; -->1}\left(d{\mathrm{\omega <!--\; \omega \; -->}}^{\mathrm{\prime <!--\; \prime \; -->}}\right)$$
hold. For nonnegative f, (16.18) always holds.
(2.47) Theorem. Suppose $\mathrm{\Omega <!--\; \Omega \; -->}$ is an open set in ${\mathbf{R}}^n$ and $G:\mathrm{\Omega <!--\; \Omega \; -->}\to <!--\; \to \; -->{\mathbf{R}}^n$ is a $C^1$ diffeomorphism. (a) If f is a Lebesgue measurable function on $G(\mathrm{\Omega <!--\; \Omega \; -->})$, then $f\circ <!--\; \circ \; -->G$ is Lebesgue measurable on $\mathrm{\Omega <!--\; \Omega \; -->}$. If $f\ge <!--\; \ge \; -->0$ or $f\in <!--\; \in \; -->L^1(G(\mathrm{\Omega <!--\; \Omega \; -->}),m)$, then
$${\int <!--\; \int \; -->}_{G(\mathrm{\Omega <!--\; \Omega \; -->})}f(x)dx={\int <!--\; \int \; -->}_{\mathrm{\Omega <!--\; \Omega \; -->}}f\circ <!--\; \circ \; -->G(x)|\det <!--\; \; -->D_{x}G|dx$$
(b) If $E\subset <!--\; \subset \; -->\mathrm{\Omega <!--\; \Omega \; -->}$ and $E\in <!--\; \in \; -->{\mathcal{L}}^n$, then $G(E)\in <!--\; \in \; -->{\mathcal{L}}^n$ and $m(G(E))=$ ${\int <!--\; \int \; -->}_E|\det <!--\; \; -->D_{x}G|dx$.

Fair die roll 8 times, number of of perfect squares
You have a fair die you roll 8 times. Let X be the number of times you roll a perfect square. What is $E(X^2)$?
I tried doing summation from 1 to 8 of $(n/8)\cdot <!--\; \cdot \; -->2.5)+(8-<!--\; -\; -->n)/8\cdot <!--\; \cdot \; -->(3.5)$ and divide by 8 but that was wrong. G