Computing a probability of finding defects by random sampling This is a problem from my semester end exams (which have got over). I suspect that the problem below is vague or open for misinterpretation. I would really like to know the actual answer to the problem if they are correctly asked, if not, I would like how can you also interpret it & the solution to that misinterpretation.
Note - The questions are as-is in the paper (no changes at all)
Question 1 -

An assembly line produces continues product that is 3% defective. If 100 pieces are randomly checked, find the probability that (i) exactly 3 are defective & (ii) none are defective.

My solution (as I recall)
(i) $\left(\binom{100}{3} \times 0.03^{3}\right) \times 0.97^{97}  = 0.2275$
(ii) $\left(\binom{100}{0} \times 0.03^{0}\right) \times 0.97^{100} = 0.0476$
Question 2 -

A stack of construction material contains 8 good sample & 2 defective samples. A part is chosen at random and inspected by 4 different inspectors. Find the probability that two of the inspectors found defective sample?

My solution
$\left(\binom{8}{2} \times \binom{2}{2}\right) \left/ \binom{10}{4}\right. = 0.1333$
Does permutation matter in any of the above?!
Thanks!
 A: The probability $p$ of selecting one defective part is:
$$p=\frac{2}{10}=\frac{1}{5}=0.2$$
This means that two inspectors must pick a defective part and two inspectors must pick a non-defective part. Therefore,
$$P=\binom{2}{2}p^2\binom{8}{2}(1-p)^2=\binom{8}{2}p^2(1-p)^2$$
A: just some thoughts:
i think your first solution is ok. for the second it is trickier to identify the appropriate sample space. it is possible that, say two, of the inspectors just happen to inspect the same sample. 
the fact that a part happens to be inspected by all four inspectors is so unlikely makes our "common sense" wish to exclude it, but nothing in the stated situation rules this out.
if 3 or 4 inspectors choose the same part, it is impossible for exactly two inspectors to find a defective. 
so a positive outcome can occur in these cases:
A) 2 inspectors choose (the same) defective part and 2 choose (the same) part which is not defective
B1) 2 choose one part and it is defective, the others choosing different non-defectives
B2) 2 choose (the same) non-defective part, but the other 2 choose different parts which are both defective
C) all 4 inspectors choose different parts, exactly two of which are defective.
