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I was working on this problem and I can solve it by applying the same technique as suggested in this answer: Inclusion-Exclusion Problem?

"A widget inspector inspects 12 widgets and finds that exactly 3 are defective. The widgets then get all mixed up and the inspector has to find the 3 defective widgets again by testing one.Find the probability that the inspector will now have to test at least 9 widgets"

However, for my own sake I'm trying to also get the same answer by counting the ways the widget inspector doesn't find the defective widgets amongst the first 8 widgets using inclusion-exlusion, however, my answers don't match up so I was hoping someone could point out what I am doing wrong:

$$ \frac{\binom{12}{1}11!- \binom{12}{2}10!+\binom{12}{3}9!-\binom{12}{4}8!+\binom{12}{5}7!-\binom{12}{6}6!+\binom{12}{7}5!-\binom{12}{8}4!}{12!} $$

Edit (Reasoning for the numerator): I wanted to use inclusion-exclusion to count on the numerator the number of ways that the first 8 widgets were non-defective. I can see now that that's not what I have done at all. My reasoning for the first term in the numerator was that if the first widget was non-defective then the remaining widgets could be chosen in 11! ways and the first widgets could be chosen in 12C1 ways. I feel like this is were I have a gap in my understanding which I am trying to fill and Im struggling to make the proper connections.

Sam Andersson
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1 Answers1

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However, for my own sake I'm trying to also get the same answer by counting the ways the widget inspector doesn't find the defective widgets amongst the first 8 widgets using inclusion-exlusion,

No, you want the probability that the inspector does not find the third defective widget among the first eight to be inspected. It is fine to find one or two defective widgets among them, as long as all three are not there.

You can count the favoured case as selecting none, one, or two places for the three defective widgets among the first eight inspections, and places for the remainder among the last four scheduled to be inspected.

$$\dfrac{\dbinom{8}{0}\dbinom{4}{3}+\dbinom{8}{1}\dbinom{4}{2}+\dbinom{8}{2}\dbinom{4}{1}}{\dbinom{12}3}$$

You can count the favoured case by ways to select places for the first two defective widgets when the third defective widget is the ninth, tenth, eleventh, or twelfth inspected.

$$\dfrac{\dbinom{8}{2}+\dbinom{9}{2}+\dbinom{10}{2}+\dbinom{11}{2}}{\dbinom{12}{3}}$$

Or you may use the suggestion of excluding the unfavoured case.

$$\dfrac{\dbinom{12}{3}-\dbinom{8}{3}}{\dbinom{12}{3}}$$

Graham Kemp
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