Probability of defective after seeing test results We have three identical products A,B,C. A test is designed to test any defective products among the three. Each time, exactly 2 of them are tested together. The test shows positive only if there is at least one of them defective. We have tested (A,B) and (A,C) and both tests show positive. What is the minimum possible probability that A is defective.
Here is my idea:
We have 8 equal likely possibilities: (T,T,F), (T,T,T), (T,F,T), (T,F,F), (F,T,T), (F,T,F), (F,F,T), (F,F,F). T represents defective and F represents not defective. By see the test results, we could narrow down our sample space to (T,T,F), (T,T,T), (T,F,T), (T,F,F), (F,T,T). So the probability of A defective should be $\frac{4}{5}$.
I am not sure why it is possible to have a minimum probability for A to be defective. It seems like there should be a range of probabilities that A could have but I got only one number.
 A: 
We have tested (A,B) and (A,C) and both tests show positive. What is the minimum possible probability that A is defective.

I interpret this as asking for the minimum possible conditional probability that "A is defective" given that "at least one of A or B is defective and at least one of A or C is defective".
Let $a,b,c$ (resp.) denote the event that $A,B,C$ (resp.) is defective. With the OP's $8$-element sample space $\Omega=\{(x,y,z): x,y,z\in \{T,F\} \},$ these events are
$$a = \{(x,y,z)\in\Omega:x=T\}\\
b = \{(x,y,z)\in\Omega:y=T\}\\
c = \{(x,y,z)\in\Omega:z=T\}.
$$
Then we have, by definition of the conditional probability: $$\begin{align}
&P(a\mid(a\cup b)\cap(a\cup c))\\[2ex]
&={P(a\cap( (a\cup b)\cap(a\cup c) ) )\over P((a\cup b)\cap(a\cup c))}\tag{1}\\[2ex]
&={P(a)\over P(a\cup (b\cap c))}\tag{2}\\[2ex]
&={P(a)\over P(a)+P(b\cap c)-P(a\cap b\cap c)}\tag{3}\\[2ex]
&={P(a)\over P(a)+P(b)\cdot P(c)-P(a)\cdot P(b)\cdot P(c)}\quad\text{assuming independence}\tag{4}\\[2ex]
&={P(a)\over P(a)+P(a)\cdot P(a)-P(a)\cdot P(a)\cdot P(a)}\quad\text{bec. A,B,C are "identical products"}\tag{5}\\[2ex]
&={1\over 1+P(a)-P(a)^2}\tag{6}\\
\end{align}$$
Note that we can obtain (2) from (1) without actually inspecting the elements of the sets $a,b,c.$ Rather, the numerator and denominator of (2) are obtained by applying the fact that unions are distributive over intersections, and also that $X\cap Y=X$ if $X\subseteq Y;$ thus, for the denominator,
$$(a\cup b)\cap(a\cup c)=a\cup(b\cap c)$$
and hence for the numerator,
$$\begin{align}a\cap( (a\cup b)\cap(a\cup c) )&=a\cap(a\cup(b\cap c))\\ &=a\quad\text{because $a\subseteq(a\cup(b\cap c))$}.\end{align}$$
Here's a plot of the RHS of (6) versus $P(a)(=P(b)=P(c))$:

By examining the first and second derivatives of the RHS w.r.t. $P(a)$, we can confirm that the required probability $P(a\mid(a\cup b)\cap(a\cup c))$ attains a minimum when $P(a)={1\over 2}$:
$$\min_{0\le P(a)=P(b)=P(c)\le 1} P(a\mid(a\cup b)\cap(a\cup c)) = {1\over 1+{1\over 2}-({1\over 2})^2} = {4\over 5}.
$$

Note: When the OP says ...

We have 8 equal likely possibilities: (T,T,F), (T,T,T), (T,F,T), (T,F,F), (F,T,T), (F,T,F), (F,F,T), (F,F,F).

... this tacitly assumes $P(a)=P(b)=P(c)={1\over 2}$, which is why the OP's solution gives the minimum value.
