I am reading through Finan's Notes for Probability and was stuck for a bit on question $4.1.14$ of his revised $2020$ book (which you can find here https://faculty.atu.edu/mfinan/actuarieshall/PV2020.pdf) - available to the public at no cost. The content of the question is the following:
A store has $80$ modems. $30$ come from source A and the remainder from source B. Of the modems from A $20$% are defective, $8$% from B. Calculate the probability that exactly $2$ out of a random sample of $5$ modems from the stores inventory are defective.
My approach to this is the following:
I want to calculate the probability that any $2$ out of $5$ randomly selected would be defective from each source. Then form a linear combination of the two source probabilities with their respective ratio of inventory as the multiplicative constant. I.e., if the probability of finding $2$ out of $5$ modems from A are defective is $x$ and for B it's $y$ then the total probability of finding $2$ out of $5$ modems defective from the inventory would be $$0.375x + 0.625y$$
I started by attempting the probability calculation for source $A$. I did this by simply multiplying the probabilities of selecting 2 defectives from the total number of modems, then 3 working modems which looked like: $$\frac{6}{30}\frac{5}{29}\frac{24}{28}\frac{23}{27}\frac{22}{88} = \frac{55}{3393}$$ I was concerned that the order of selection may have mattered so I tried calculating again by doing the opposite (selecting the working modems first then the defective) and found $$\frac{24}{30}\frac{23}{29}\frac{22}{28}\frac{6}{27}\frac{5}{26} = \frac{506}{23751}$$ So, clearly that's not the correct approach. I have a suspicion that this involves the binomial coefficient $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ but am not exactly sure how to implement it. I thought I might try the following: $$P = \frac{\binom{30}{2} \cdot \binom {28}{3}}{\binom{30}{5}} = 10$$ However as you can see, $10$ is not within the allowable range for probabilities.
My $2$ questions are as follows:
Is my overall approach correct? I.e., finding the probability of selecting $2$ out of $5$ defective from each source then establishing a linear combination with their ratio of inventory stock as the weight.
If so, how can I approach the calculation of each individual probability?
We can calculate directly that the total number of defective modems is
$$30\cdot 0.2 + 50\cdot0.08 = 10$$
From here this is just combinatorics. The probability of drawing exactly $2$ defective modems out of a sample of $5$ is the number of ways to draw $2$ defective and $3$ healthy modems, divided by the total number of ways to select $5$ modems from the whole group.
$$\frac{{10 \choose 2}{70 \choose 3}}{{80\choose 5}} \approx 10.25\%$$