Calculate Statistic False Positives with Our Examples

$f$ cts on $[a,b]$ and $f(x)\ne <!--\; \ne \; -->0$ for all $x\in <!--\; \in \; -->[a,b]$ implies that $f(x)$ is either always positive or negative on $[a,b]$.

While it makes some sense, it's not clear to me why those are different. If a test, say medical test, is correct 90% of time then chances of it being wrong is 10%.
There are 4 events
1. Test is +, patient has a disease
2. Test is -, patient doesn't have a disease
3. Test is +, patient doesn't have a disease
4. Test is -, patient has a disease

"The population, for a disease D, has a true rate of T%"
"Some Test ST, has false positive rate of FP% and a false negative rate of FN%."
"T, FP, and FN are elements of the set of real numbers."
Was the number T determined by some 100% accurate and possibly expensive test?

The occurrence of a disease is $\frac{1}{100}=P(D)$
The false negative probability is $\frac{6}{100}=P(-<!--\; -\; -->|D)$, and the false positive is $\frac{3}{100}=P(+|\mathrm{\neg <!--\; \neg \; -->}D)$
Compute $P(D|+)$
By bayes formula,
$P(+)=P(+|D)P(D)+P(+|\mathrm{\neg <!--\; \neg \; -->}D)P(\mathrm{\neg <!--\; \neg \; -->}D)=\frac{97}{10000}+\frac{297}{10000}=\frac{394}{10000}$
Similarly $P(D|+)=\frac{P(+|D)P(D)}{P(+)}=\frac{97}{394}=0.246$
Is this correct?

We should also observe that it is important to know precisely where in the sentence a given connecting word is introduced. For example, compare the following two sentences:
1. For every positive number x there exists a positive number y such that y2. There exists a positive number y such that for every positive number z, we have yAlthough these statements may look similar, they do not say the same thing. As a matter of fact, (1) is true and (2) is false.

A large company gives a new employee a drug test. The False-Positive rate is 3% and the False-Negative rate is 2%. In addition, 2% of the population use the drug. The employee tests positive for the drug. What is the probability the employee uses the drug?
What I try:
X={employee uses drugs}
Y={employee tests positive in the drug test}
P(X|Y) = 0.97
P(notX|Y) = 0.03
P(X|notY) = 0.02
P(notX|notY) = 0.98
P(X) = 0.02

$f$ cts on $[a,b]$ and $f(x)>0$ for all $x\in <!--\; \in \; -->[a,b]$ implies that there exists an $m>0$ such that $f(x)\ge <!--\; \ge \; -->m$ for all $x\in <!--\; \in \; -->[a,b]$.
Proposition $1$ seems to obviously be true. If $f(x)$ is not equal to $0$, it has to be positive or negative? Prop $2$ seems to be false because what if the function approaches $0$ as $x$ approaches $\mathrm{\infty <!--\; \infty \; -->}$?

Prove it if it is true or give counterexample if it is false:
let $$a,b,k\in Z$$ and $$n$$ is positive integer, if $$a\equiv b(\mathrm{mod}n)$$ then $$k^a\equiv k^b(\mathrm{mod}n)$$
My try: it is false statement like if $$-<!--\; -\; -->3\equiv 2(\mathrm{mod}5)$$ then $$3^{-<!--\; -\; -->3}\not\equiv 3^2(\mathrm{mod}5)$$

True or false? If true, justify. If false, give counter-example. If $f,g:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ are functions such that $f$ is bounded and positive and $\underset{x\to <!--\; \to \; -->+\mathrm{\infty <!--\; \infty \; -->}}{lim}g(x)=+\mathrm{\infty <!--\; \infty \; -->}$, so $\underset{x\to <!--\; \to \; -->+\mathrm{\infty <!--\; \infty \; -->}}{lim}f(x)g(x)=+\mathrm{\infty <!--\; \infty \; -->}$

True and False?
Let $A$ and $B$ be nonempty sets and $1-<!--\; -\; -->1$ be a $1-<!--\; -\; -->1$ function. Then $f(X\cap <!--\; \cap \; -->Y)=f(X)\cap <!--\; \cap \; -->f(Y)$ for all nonempty subsets $X$ and $Y$ of $A$

Let f and g be functions. Then the domain of f composed with g is the intersection of the domain of f and the domain of g. True or False?

Let $X$ be a locally comapct and Hausdorff space. We say a positive Radon Measure on $X$ is faithful if
$$0\le <!--\; \le \; -->f,\int <!--\; \int \; -->fd\mathrm{\mu <!--\; \mu \; -->}=0\to <!--\; \to \; -->f(x)=0\mathrm{\forall <!--\; \forall \; -->}x\in <!--\; \in \; -->X$$
True or false: If there is a faithful positive Radon measure on $X$ then $X$ has a countable dense subset ?

Suppose a new Internet company Mumble.com was to require all employees to take a drug test. Mumble.com can aord only the inexpensive drug test { the one with a 5% false positive and a 10% false negative rate. (That means that ve percent of those who are not using drugs will incorrectly test positive, and ten percent of those who are actually using drugs will test negative.) Suppose that 10% of those who work for Mumble.com are using the drugs for which Mumble.com is checking. An employee is chosen at random.
(a) (3 marks). What is the probability the employee both uses drugs and tests positive?
(b) (2 marks). What is the probability the employee does not use drugs but tests positive anyway?
(c) (2 marks). What is the probability the employee tests positive?
(d) (2 marks). If the employee has tested positive, what is the probability he or she uses drugs?

If 𝐴 is indefinite and 𝐵 is positive definite, then 𝐴+𝐵 is indefinite.

True or False: If $x\notin <!--\; \notin \; -->\mathbb{Q}$ then $\underset{m\ge <!--\; \ge \; -->0}{\sum <!--\; \sum \; -->}m{x}^{m-<!--\; -\; -->1}\notin <!--\; \notin \; -->\mathbb{Q},$, where $|x|<1.$
Considered the contra-positive of the above statement:
If $\underset{m\ge <!--\; \ge \; -->0}{\sum <!--\; \sum \; -->}m{x}^{m-<!--\; -\; -->1}\in <!--\; \in \; -->\mathbb{Q}$ then $x\in <!--\; \in \; -->\mathbb{Q}.$
Now if the contra-positive is true/false then the statement is true/false.
Thus, if
$$\underset{m\ge <!--\; \ge \; -->0}{\sum <!--\; \sum \; -->}m{x}^{m-<!--\; -\; -->1}\in <!--\; \in \; -->\mathbb{Q}$$
$$⟹<!--\; ⟹\; -->1+2x+3x^2+...\in <!--\; \in \; -->\mathbb{Q}$$
$$⟹<!--\; ⟹\; -->1^{\prime }+x^{\prime }+(x^2{)}^{\prime }+(x^3{)}^{\prime }+...\in <!--\; \in \; -->\mathbb{Q}$$
$$⟹<!--\; ⟹\; -->(1+x+x^2+x^3+...{)}^{\prime }\in <!--\; \in \; -->\mathbb{Q}$$
$$⟹<!--\; ⟹\; -->(\frac{1}{1-<!--\; -\; -->x}{)}^{\prime }\in <!--\; \in \; -->\mathbb{Q}$$
$$⟹<!--\; ⟹\; -->\frac{-<!--\; -\; -->1}{(1-<!--\; -\; -->x{)}^2}\in <!--\; \in \; -->\mathbb{Q}$$
Now from the last step we can conclude that $x\in <!--\; \in \; -->\mathbb{Q}$, since $|x|<1.$ Therefore the contrapositive is true, so the given statement is also true.
Is my analysis correct?

A virus has been spread around a population. The prevalence of this virus is 84%. A diagnostic test, with a specificity of 94% and sensitivity of 15%, has been introduced. If a patient is drawn randomly from the population, what is the probability that:
a) a person has the virus, given that they tested positive?
b) a person has the virus, given that they tested negative?

True and False?
Let $A$ and $B$ be nonempty sets and $f:A\to <!--\; \to \; -->B$ be a function. Then if $f(X\cap <!--\; \cap \; -->Y)=f(X)\cap <!--\; \cap \; -->f(Y)$ for all nonempty subsets $X$ and $Y$ of $A$, then $f$ must be $1-<!--\; -\; -->1$.

Consider the language $L=\{+,\cdot <!--\; \cdot \; -->,0,1\}$ of rings. It is easy to show using compactness that if $\mathrm{\sigma <!--\; \sigma \; -->}$ is a sentence that holds in all fields of characteristic $0$, there is some $N\in <!--\; \in \; -->\mathbb{N}$ such that $\mathrm{\sigma <!--\; \sigma \; -->}$ holds for all fields of characteristic $p\ge <!--\; \ge \; -->N$. A sort of converse would be, if $\mathrm{\sigma <!--\; \sigma \; -->}$ is a sentence that holds in all fields of positive characteristic, $\mathrm{\sigma <!--\; \sigma \; -->}$ is true in all fields of characteristic $0$.

True and False?
If $A$, $B$, and $C$ are three sets, then the only way that $A\cup <!--\; \cup \; -->C$ can equal $B\cup <!--\; \cup \; -->C$ is $A=B$.

For any positive real number $r,r^n$ converges.
False: Take any positive $r$, then as $n\to \infty r$ diverges.
If $x_n\ast <!--\; \ast \; -->y_n$ converges, then $x_n$ and $y_n$ both converge.
False: Suppose $x_n$ or $y_n$ dont converge. Then take $x_n=n^2$ and $y_n=\frac{1}{n}$. Then $x_n\ast <!--\; \ast \; -->y_n=n$, which diverges. Thus it does not converge.
True or False?