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!
Best Answer
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, $$P=\binom{2}{2}p^2\binom{8}{2}(1-p)^2=\binom{8}{2}p^2(1-p)^2$$