So I'm doing homework for my probability class and I am almost finished, however now I've got 1 Inclusion-Exclusion problem, which I have never done before.
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.
A) Find the probability that the inspector will now have to test at least 9 widgets B) Find the probability that the inspector will now have to test at least 10 widgets
I'm almost sure that the sample space is (12 C 3), but that's all I can come up with and I don't even know if I have to use that.
Imagine that he lays the $12$ widgets out on the table in a line and works his way down the line until he finds the third defective widget. There are $12!$ possible orders in which he can line them up, and they’re all equally likely. To find the probability that he needs to test at least $9$ of the widgets, we need to count the arrangements in which he hasn’t found all three defective widgets amongst the first $8$ in the line. The easiest way to do this is to count the arrangements in which he does find them amongst the first $8$.
If the defective widgets are amongst the first $8$, there are $\binom83$ possible sets of positions for them. They can occur in any of $3!$ orders within these $3$ positions, and the $9$ good widgets can appear in any of $9!$ orders in the other $9$ positions in the line, so there are altogether
$$\binom83\cdot3!\cdot9!=\frac{8!\cdot3!\cdot9!}{3!\cdot5!}=\frac{8!\cdot9!}{5!}$$
arrangements in which he finds the defective widgets amongst the first $8$. The probability that he does so is therefore
$$\frac{\frac{8!\cdot9!}{5!}}{12!}=\frac{8!\cdot9!}{5!\cdot12!}=\frac{8\cdot7\cdot6}{12\cdot11\cdot10}=\frac{2\cdot7}{11\cdot5}=\frac{14}{55}\;,$$
and the probability that he needs to test at least $9$ of the widgets must then be
$$1-\frac{14}{55}=\frac{41}{55}\;.$$
You can work (B) in exactly the same way.