Comment (and solution of a simple special case.) This has been here for a while, apparently without helpful comments.
This appears to be a generalization of a 'multivariate hypergeometric'
distribution.
You might start with a simplified set of weights. Let an urn contain
balls labeled from 1 through 8. And suppose their respective weights are $w = (2, 2, 1, 1, 1, 1, 1, 1)/10.$ If you withdraw $k = 2$ balls from the urn without replacement, what is the probability
you get the ball labeled '$1$'?
Get 1 on the first draw: $P(\text{1 on 1st}) = (2/10)(8/8) = .2.$
Get 1 on the second draw: Either 21, or something other than 1 or 2
on the first, then 1 on the second.
$P(\text{2 then 1}) = (2/10)(2/8) = .05.$
$P(\text{3 then 1}) = (1/10)(2/9) = 2/90 \approx 0.0222.$
$P(\text{1 on 2nd}) = 0.05 + 6(2/90) \approx 0.05 + 0.1333 = 0.1833.$
Finally, $P(\text{1 in two draws}) \approx 0.2 + 0.1833 = 0.3833.$
Even this simple problem turned out to surprise me by its intricacy and lack of symmetry. But perhaps, you can find patterns to simplify more complicated outcomes.
R statistical software does weighted random sampling in a way that
would allow you to check some of your analytic solutions. As a prototype, here is a
simulation of the simple example just above. Results are mainly
accurate to three places.
m = 10^6; d1 = d2 = numeric(m)
n = 2; pop = 1:8; w = c(2,2,1,1,1,1,1,1)/10
for (i in 1:m) {
draw = sample(pop, n, prob=w)
d1[i] = draw[1]; d2[i] = draw[2] }
mean(d1 ==1 | d2 ==1) # '|' signifies union
## 0.383483
round(table(d1)/m,3)
## d1
## 1 2 3 4 5 6 7 8
## 0.200 0.199 0.100 0.100 0.100 0.100 0.100 0.100
round(table(d2)/m,3)
## d2
## 1 2 3 4 5 6 7 8
## 0.184 0.184 0.105 0.105 0.106 0.106 0.105 0.105
round(table(d1,d2)/m,3)
## d2
## d1 1 2 3 4 5 6 7 8
## 1 0.000 0.050 0.025 0.025 0.025 0.025 0.025 0.025
## 2 0.050 0.000 0.025 0.025 0.025 0.025 0.025 0.025
## 3 0.022 0.022 0.000 0.011 0.011 0.011 0.011 0.011
## 4 0.022 0.022 0.011 0.000 0.011 0.011 0.011 0.011
## 5 0.022 0.022 0.011 0.011 0.000 0.011 0.011 0.011
## 6 0.022 0.022 0.011 0.011 0.011 0.000 0.011 0.011
## 7 0.022 0.022 0.011 0.011 0.011 0.011 0.000 0.011
## 8 0.022 0.022 0.011 0.011 0.011 0.011 0.011 0.000
Denote events $D$, $\bar D$ as the patient having the defect and not having the defect, respectively. Let $P_i$ denote the event that test $i$ is positive, and $\bar P_i$ the event that test $i$ is negative.
Then the desired probability is $$\Pr[P_2 \mid P_1] = \frac{\Pr[P_2 \cap P_1]}{\Pr[P_1]} = \frac{\Pr[P_2 \cap P_1 \mid D]\Pr[D] + \Pr[P_2 \cap P_1 \mid \bar D]\Pr[\bar D]}{\Pr[P_1 \mid D]\Pr[D] + \Pr[P_1 \mid \bar D]\Pr[\bar D]}.$$ Since $P_i$ are conditionally independent given the defect status, we have $$\Pr[P_2 \cap P_1 \mid D] = (\Pr[P_i \mid D])^2.$$ Then, given $$\Pr[D] = 0.01, \quad \Pr[P_i \mid D] = 0.999, \quad \Pr[P_i \mid \bar D] = 0.05,$$ we easily obtain $$\Pr[P_2 \mid P_1] = \frac{(0.999)^2(0.01) + (0.05)^2(1-0.01)}{(0.999)(0.01) + (0.05)(1-0.01)} \approx 0.209363.$$ This number is small because the prevalence of defects is so rare, and the false positive rate is much higher than the prevalence. Therefore, a positive result is more likely to result from a false positive, and a second test is not terribly likely to come back positive.
Best Answer
$$ P(B_1\mid F^c_1) = \frac{P(B_1, F^c_1)}{P(F^c_1)} =\frac{P(F^c_1\mid B_1)P(B_1)}{P(F^c_1)}. $$ If the defect is certainly found in $B_1$, the probability is proportional to the fraction inspected, or: the number of objects found are $Hypergeo(m, 1, n)$, $$ P(F_1\mid B_1) = n/m. $$ Further, $$ P(F_1) = P(F_1\mid B_1)P(B_1) + P(F_1\mid B^c_1)P(B_1) = \frac{n}{m}\frac{1}{k} + 0 = \frac{n}{mk} $$ so, since $P(B_j) = 1/k$, $$ P(B_1\mid F^c_1) = \frac{(1 - n/m)/k}{1 - n/(mk)} = \frac{m-n}{km -n} $$ i.e.
"remaining objects in 1" / "total objects remaining".The second part is also the same: $$ P(B_2\mid F^c_1) = \frac{P(B_2, F^c_1)}{P(F^c_1)} =\frac{P(F^c_1\mid B_2)}{P(F^c_1)}. $$ $P(F^c_1\mid B_2) = 1 - 0$ so $$ P(B_2\mid F^c_1) = \frac{1/k}{1 - n/(mk)} = \frac{m}{km -n} $$ or,
"objects in 2"/"total objects remaining".In conclusion, these answers agree with the original question and with intuition.
Another, perhaps simplifying, formulation is to set two vectors: observed $X$ and unobserved $Y$. There are $m$ copies of each and exactly one takes the value 1 and all others are zero. Define $M=mk$ and set $n_y = m-n$ so that $P(Y_k=1) = n_y/M$ and $P(X_k=1) = n/M.$
This formulation separates the boxes from each other so $P(B_1\mid F^c_1) = P(Y_1 = 1\mid X_1 = 0) = n_y/(M-n)$. This can be visualized as the area occupied by a single $Y$ divided by the area that the 1 could be in. Finally, $$ P(B_2\mid F^c_1) = P(X_2 + Y_2 = 1\mid X_1 = 0) = (n_y+n)/(M-n) = m/(km-n). $$