I have come across a question on joint and conditional probability. 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.

Part of Solution: The requested probability can be determined as P(3 of first 11 damaged)P(12th is damaged | 3 of first 11 damaged).

I don't know how they came up with this part of the solution shown above, can someone explain how they got that?


You need to inspect exactly 12 packages out of 20. The only way this is possible is if you hit 4 damaged packages, with the 4'th one occuring on the 12'th inspection (which would cause you to stop).


$$P(\mbox{stop on package 12}) = P(\mbox{3 of first 11 damaged and 12'th is damaged})=P(\mbox{12'th is damaged|3 of first 11 damaged})P(\mbox{3 of first 11 damaged}).$$


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