From 27 pieces of luggage, an airline luggage handler damages a random sample of four. Suppose 8 of the 27 are insured. Calculate the probability that exactly two of the four damaged pieces are insured.
Let's hold off with the solution to that problem and consider the following problem.
Suppose there are 27 pieces of luggage of which 8 are insured. Randomly pick 4 luggage and find P(exactly 2 of the 4 are insured). My answer, using hypergeometric, is 8C2*19C2/27C4
Now back to the original question. The solution manual states that p(that exactly two of the four damaged pieces are insured) = 8C2*19C2/27C4. But it is not (in my opinion) taking into account that 2 insured is coming from 4 damaged.
Can someone please enlighten me on this?