I think I find this line of reasoning rather too speculative, I'm afraid. :-)
First, you assert that $P(I) = 1/2$. I don't see any symmetry that justifies this kind of application of the principle of indifference; it's not as though the only difference between being selected for an interview and not being selected is the opportunity to go to the interview. From the job poster's perspective, they have limited time to interview candidates, whether five people apply for the posting, or five hundred.
But suppose we put that aside for the moment. You then assert that $P(J \mid I) = 1/4$, on the assumption that once half of the possibilities have been eliminated (presumably, the $\neg I$ portion), only half of the half remain, or $1/4$. But $P(J \mid I)$ is already a conditional probability—it expresses the probability of the more specific compound event $I$ and $J$ as a fraction of the probability of the condition $I$! If you are to apply the principle of indifference again, you should treat the two possibilities equally; that is, $P(J \mid I) = P(\neg J \mid I)$. And since, by excluded middle, $P(J \mid I)+P(\neg J \mid I) = 1$ necessarily, it should be the case that
$$
P(J \mid I) = P(\neg J \mid I) = \frac{1}{2}
$$
Instead, you have
$$
P(J \mid I) = \frac{1}{4}
$$
which implies
$$
P(\neg J \mid I) = \frac{3}{4}
$$
This has no more basis than the assertion that $P(I) = 1/2$, and violates the principle of indifference to boot. (Of course, I said that the principle of indifference shouldn't be applied here, but since you appear to want to do it...)
In other words, when you arrive at $1/4$, you are determining the probability that you get the interview and you get the job (and that's $P(I, J)$), not the probability that you get the job given that you got the interview (that's $P(J \mid I)$). If you want to analyze the problem that way, there's nothing wrong with that (modulo indifference), but then you should just leave that value alone, and not multiply, again, the probability that you get the interview. That's already been accounted for in the joint probability.
I think the most we can say about it is as follows: Suppose the probability of being selected for a given job posting is $\sigma$. Then, by independence, the probability of being selected for at least one of two job postings is $1-(1-\sigma)^2$; the probability of being selected for at least one of three job postings is $1-(1-\sigma)^3$; and the probability of being selected for at least one of $k$ job postings is $1-(1-\sigma)^k$. Even independence seems rather daring, but I can see a case for it better than I can for indifference.
The probability that there are no defects, i.e. $\Pr[X = 0]$, is equal to the chance that there are no defects of type $A$ and there are no defects of type $B$. Since a defect of type $A$ occurs with probability $0.1$, i.e. $\Pr[A] = 0.1$, then the probability of no defect of type $A$ is $$\Pr[\bar A] = 1 - \Pr[A] = 1 - 0.1 = 0.9.$$ Similarly, the probability of no defect of type $B$ is $$\Pr[\bar B] = 1 - \Pr[B] = 1 - 0.2 = 0.8.$$ Therefore, $$\Pr[X = 0] = \Pr[\bar A \cap \bar B] \overset{\text{ind}}{=} \Pr[\bar A]\Pr[\bar B] = (0.9)(0.8) = 0.72.$$
The probability that there is exactly one defect is $\Pr[X = 1]$; this can happen if and only if there is a defect of type $A$ and no defect of type $B$; or no defect of type $A$ and a defect of type $B$. That is to say, $$\Pr[X = 1] = \Pr[A \cap \bar B] + \Pr[\bar A \cap B].$$
Another way to reason about this problem is to write out all of the joint probabilities $$\Pr[A \cap B], \quad \Pr[\bar A \cap B], \quad \Pr[A \cap \bar B], \quad \Pr[\bar A \cap \bar B].$$ You can do this with a simple contingency table: $$\begin{array}{c|c|c} & A & \bar A \\ \hline B & (0.1)(0.2) & (0.9)(0.2) \\ \hline \bar B & (0.1)(0.8) & (0.9)(0.8) \\ \end{array}$$
Best Answer
Let $a = 0.1$ the probability to get an interview and $b = 0.2 = 2a$ the probability to be hired after an interview.
The probability to be hired at a given submission is $ab$.
The probability you are looking for is:
$$p = \underbrace{ab}_{\text{1st sub. hired}} + \underbrace{(1-ab)}_{\text{1st sub. fail}}\cdot\left(\underbrace{ab}_{\text{2nd sub. hired}} + \underbrace{(1-ab)}_{\text{2nd sub. fail}}\cdot\underbrace{ab}_{\text{3rd sub. hired}}\right) =\\ = a^3b^3 - 3a^2b^2 + 3ab = \\ = a^3 (2a)^3 - 3a^2 (2a)^2 + 3a(2a) = \\ = 8 a^6 - 12 a^4 + 6 a^2 = \\ = 8 \cdot 10^{-6} - 12 \cdot 10^{-4} + 6 \cdot 10^{2} = 0.058808.$$