Unsure about Conditional Probability question I'm really not understanding this question

So for Exercise 1, I got 0.794 which is correct. My reasoning was that there are only 2 ways to test positive, a "true" positive and false positive. So the probability of them having antibodies given they are infected is 0.8 and the probability of testing positive given they have antibodies is 0.99 so the probability of having antibodes and testing positive given they are infected is 0.990.8. Now we also need to find them not having antibodies and being tested as positive given they are infected which is 1 - p(a|v)(0.01) so that is 0.2*0.01. This gave me 0.792 + 0.002 = 0.794.
Is that reasoning accurate and how would this look mathematically using the formulas of conditional probability?
I am also unsure about Exercise 2.
 A: $$\def\P{\operatorname{\sf P}}\require{enclose}\begin{array}{c}\underline{\text{Virus}}&&\underline{\text{Anitbody}}&&\underline{\text{Test Positive}}\\\enclose{circle}{ V}&\longrightarrow&\enclose{circle}{A}&\longrightarrow&\enclose{circle}{T}\\\P(V)=0.001&&\P(A\mid V)=0.08&&\P(T\mid A)=0.99\\&&\P(A\mid\lnot V)=0.10&&\P(T\mid\lnot A)=0.01\end{array}$$



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*Sensitivity is the probability that the test is positive given that the person is infected. Compute the sensitivity of this diagnostic test.


My reasoning was that there are only 2 ways to test positive, a "true" positive and false positive.
... Is that reasoning accurate and how would this look mathematically using the formulas of conditional probability?

Correctly reasoned, if a little imprecisely phrased.  I'd say: A true positive test for the virus may come from true positive test for antibodies that are present due to the virus, or a false positive test for antibodies that are absent despite the virus.  That is what you evaluated.
This simply the Law of Total Probability, then applying the factorisation described by this Directed Acylic Graph (DAG).
$$\begin{align}\P(T\mid V)&=\P(A,T\mid V)~+~\P(\lnot A,T\mid V)\\[1ex]&=\P(A\mid V)\P(T\mid A)~+~\P(\lnot A\mid V)\P(T\mid\lnot A)\end{align}$$
And this is what you calculated:

So the probability of them having antibodies given they are infected is 0.8 and the probability of testing positive given they have antibodies is 0.99 so the probability of having antibodes and testing positive given they are infected is 0.99$\times$ 0.8.

$$\P(A\mid V)\P(T\mid A) = 0.8\times 0.99\qquad\color{green}{\checkmark}$$

Now we also need to find them not having antibodies and being tested as positive given they are infected which is 1 - p(a|v)(0.01)

Close; watch the association brackets. Although, as you calculate the values correctly, this looks like a typo.
$$\begin{align}\P(\lnot A\mid V)\P(T\mid\lnot A)&=[1-\P(A\mid V)]\P(T\mid\lnot A)\\&=[1-0.8]\times 0.01\qquad\color{green}\checkmark\end{align}$$





*Specificity is the probability that the test is negative, given that the person is not infected.


I am also unsure about Exercise 2.

Second verse, same as the first.  Just do again as you did before.
$$\begin{align}\P(\lnot T\mid\lnot V)&=\P(A,\lnot T\mid\lnot V)+\P(\lnot A,\lnot T\mid\lnot V)\\[1ex]&=\P(A\mid\lnot V)\P(\lnot T\mid A)~+~\P(\lnot A\mid\lnot V)\P(\lnot T\mid\lnot A)\\[1ex]&=\P(A\mid\lnot V)\,[1-\P(T\mid A)]~+~[1-\P(A\mid\lnot V)]\,[1-P(T\mid\lnot A)]\end{align}$$
