Calculate Statistic False Positives with Our Examples

Given the problem of a patient taking a test for a disease where having the disease is denoted by X and the a positive test is denoted by Y,
- the rate of occurrence of the disease in the general population is 1%
- the probability of a false positive is 3%
- the odds of getting tested positive is 90% if you have the disease
is it appropriate to solve the following through rearranging the total probability
$P(A)=P(A|B)P(B)+P(A|NotB)P(NotB)$
Into
$P(A|B)=(P(A)-<!--\; -\; -->P(A|B)P(B))/P(NotB)$
and then getting P(B) from the total probability
$P(B)=P(B|A)P(A)+P(B|NotA)P(NotA)$
then given B is binary P(Not B) from
$1=P(B)+P(NotB)$
and getting P(A|B) from Bayes
$P(A|B)==P(A)P(B|A)/P(B)$
and then substituting that all back into the first equation to get the result

A population includes a fraction m of individuals carrying a disease exists and has the following characteristics:
P(positive test | individual with the disease) = p
P(positive test | individual without the disease) = r
What is the probability of a false negative: P(individual with the disease | negative test)?
My try
Let W = with the disease, ~W = without the disease;
Let + = positive test, - = negative test
From the given, we have: P(W) = m, P(~W) = 1-m
We want P(W | -) = (P(- |W) P(W))/(P(-)) = (P(- | W)P(W)/(P(- | W)P(W) + P(- ~W)P(~W))
We have: P (+) = P(+|W)P(W) + P(+|~W)P(~W) = p * P(W) + r * P(~W)

Prove that the $F$-measure of any binary classifier is $\le <!--\; \le \; -->{\displaystyle \frac{precision+recall}{2}}$
Let $P=precision$ and $R=recall$
$F$ measure $={\displaystyle \frac{2PR}{P+R}}$
Note that $precision={\displaystyle \frac{tp}{tp+fp}}$ and $recall={\displaystyle \frac{tp}{tp+fn}}$
where $tp=$ true positive, $fp=$ false positive, $tn=$ true negative and $fn=$ false negative

The smallest positive integer in the set $\{6u+9v:u,v=\text{integers}\}$ is $1$.
My thoughts: false. there is no value of $u$ and $v$ (positive or negative) that could result in a remainder of $1$. Smallest positive I get is $3$. Solid foundation to prove it?

"Every quadratic form $x^{T}Ax$ with $A$ an invertible matrix is either positive definite, negative definite, or indefinite."
Is this true or false?

One in two hundred people in a population have a particular disease. A diagnosis test gives a false positive $3$% of the time, and a false negative $2$% of the time. Ross takes the test and the report comes positive. Find the probability that Ross has the disease.

Test for HIV. There's a false positive rate of 0.025 and a false negative rate of 0.08. Let's say a particular patient has a probability of testing positive for HIV of 0.005. The patient gets tested and it's positive. What are the chances that the patient actually has HIV?

True and False?
If the product $A\times <!--\; \times \; -->B$ of two sets $A$ and $B$ is the empty set , then both $A$ and $B$ have to be empty set.

True and False?
There is no one-to-one correspondence between the set of all positive integers and the set of all odd positive integers because the second set is a proper subset of the first.

True and False?
If $(A\cup <!--\; \cup \; -->B\subset <!--\; \subset \; -->A\cup <!--\; \cup \; -->C)$ then $B\subset <!--\; \subset \; -->C$

Simplification step by step (exponential and logarithm) - regarding BLOOM FILTER - FALSE POSITIVE probability
$(1-<!--\; -\; -->\frac{1}{m}{)}^{kn}\approx <!--\; \approx \; -->e^{-<!--\; -\; -->kn/m}$
I don't get how to reach $e^{-<!--\; -\; -->kn/m}$ from $(1-<!--\; -\; -->\frac{1}{m}{)}^{kn}$
I have reviewed logarithm and exponent rules but I always get stuck, here is what I have tried; Starting from $(1-<!--\; -\; -->\frac{1}{m}{)}^{kn}$, I can write:
- $e^{(ln(1-<!--\; -\; -->\frac{1}{m}{)}^{kn})}$
- $e^{(kn\times <!--\; \times \; -->ln(1-<!--\; -\; -->\frac{1}{m}))}$
Or the other way around:
$ln(e^{(1-<!--\; -\; -->\frac{1}{m}{)}^{kn}})$
$ln(e^{(1-<!--\; -\; -->\frac{1}{m})\times <!--\; \times \; -->kn})$
$ln(e^{(1-<!--\; -\; -->\frac{1}{m})})+ln(kn)$
$(1-<!--\; -\; -->\frac{1}{m})+ln(kn)$
After few steps I fall in an unsolvable loop hole playing with $ln$ and $e$ trying to reach $e^{-<!--\; -\; -->kn/m}$.

If $m$ is the number of bits in the array, $k$ is the number of hash functions and we have $n$ entries, then the false positive probability upper bound can be approximated as:
$(1-<!--\; -\; -->e^{\frac{-<!--\; -\; -->kn}{m}}{)}^k$
It further states in the wiki article that:
"The number of hash functions, $k$, must be a positive integer. Putting this constraint aside, for a given $m$ and $n$, the value of $k$ that minimizes the false positive probability is:"
$k=\frac{m}{n}\ln <!--\; \; -->2$
How can I find the $k$ value that minimizes the false positive function ?

Consider a disease that is on average caught by $1$ in $N$ people. Let's say that a disease test gives a false positive for $p\mathrm{\%<!--\; \%\; -->}$ of healthy people. For diseased people, it always gives a correct, positive answer.
What's the chance that a person who got a positive result is healthy?

Is there some indicator for false positives on the Legendre symbol?
For example,
$\left(\frac{35}{13}\right)=1$
yet $x^2-<!--\; -\; -->35y^2=13$ has no solutions in integers. For a non-UFD like this, is there some way to take the Legendre symbol, add to it or multiply it by something else, like maybe something to the power of the class number to get the $-<!--\; -\; -->1$ meaning that $13$ is irreducible in this particular domain of algebraic integers?

A cancer test is 90 percent positive when cancer is present. It gives a false positive in 10 percent of the tests when the cancer is not present. If 2 percent of the population has this cancer what is the probability that someone has cancer given that the test is positive?
I multiplied the 90 by 10 divided by 90 times 10 plus 2.

6 real numbers, sum of any 3 consecutive is negative, while sum of any 4 consecutive is positive. Prove false.
A computer programmer claims that he generated six real numbers $a_1,a_2...a_6$ so that the sum of any four consecutive $a_i$ is positive, but the sum of any three consecutive $a_i$ is negative. Prove that his claim is false.

Could Miller-Rabin primality test give false negative, for example when test prime number and gives it as composite?

Use Bayes' Theorem to prove that the rate of false positives is accurate (86%) in the following passage.
What I have done so far is list the following
$P(A)=0.99$
$P(B)=0.05$
As Bayes' Theorem is as follows
$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$
I need $P(B|A)$. I'm stuck on how to find this. Help would be greatly appreciated.
Is it possible to prove this using Bayes'? If not, is there some other sort of mathematics that I can use here?
Roughly 1 per cent of the population suffer from mild cognitive impairment, which might, but doesn’t always, lead to dementia. Suppose that the test is quite a good one, in the sense that 95 per cent of the time it gives the right (negative) answer for people who are free of the condition. That means that 5 per cent of the people who don’t have cognitive impairment will test, falsely, as positive. That doesn’t sound bad. It’s directly analogous to tests of significance which will give 5 per cent of false positives when there is no real effect, if we use a p-value of less than 5 per cent to mean ‘statistically significant’.
But in fact the screening test is not good – it’s actually appallingly bad, because 86 per cent, not 5 per cent, of all positive tests are false positives. So only 14 per cent of positive tests are correct. This happens because most people don’t have the condition, and so the false positives from these people (5 per cent of 99 per cent of the people), outweigh the number of true positives that arise from the much smaller number of people who have the condition (80 per cent of 1 per cent of the people, if we assume 80 per cent of people with the disease are detected successfully).

A store-bought pregnancy test produces a correct result 99% of the time when a woman is actually pregnant but is only 97% accurate when a woman is not pregnant. What are the chances of a false positive and a false negative respectively?
So I have tried to solve it and is the false positive 3%? SO, false positive is when the result shows positive despite her not being pregnant. And for her not being pregnant , the test is 97% accurate. Then, the false positive should be 100-97%?

How to calculate the false negative, the true negative, the false positive and the true positive to form a confusion matrix. A numeric example would clear all doubts I think.