In a shipment of 20 packages, 7 packages are damaged. The packages are randomly inspected, one at a time, without replacement, until the fourth damaged package is discovered.
Calculate the probability that exactly 12 packages are inspected.
So you have this diagram
_ _ _ _ _ _ _ _ _ _ _D
Where D represents a damaged package and it is fixed at the 12th spot. Let X represent a non damaged package. One possible ordering of the above scenario is DDDXXXXXXXXD. Since the last D is fixed you have 11 choose 3 ways of moving around the Xs and Ds. I'm getting the correct answer $\approx.119$ doing
$${11\choose 3}\frac{7\times6\times5\times4\times13\times12\times11\times10\times9\times8\times7\times6}{20\times19\times18\times17\times16\times15\times14\times13\times12\times11\times10\times9}$$
This is supposed to represent for example:
(chance of choosing the 1st damaged package)AND(chance of choosing 2nd damaged package)AND(chance of choosing 3nd damaged package)AND(chance of choosing 4th damaged package)AND(chance of choosing 1st non-damaged package)AND(chance of choosing 2nd non-damaged package)AND…AND(chance of choosing 8th non-damaged package).
My question: selection "without replacement" is a very typical case where the events are not independent and $P(A\cap B)=P(A)P(B)$ if A and B are independent, but it works in this case, why? Or is this a coincidence?
Best Answer
This is not an instance of the probability multiplication rule because later factors in the product depend on earlier ones. If the events were truly independent, you would be writing down the same probabilities for each damaged package, which is not the case here.
Here the multiplication represents going down the probability tree rather than introducing a new independent random variable.