In a shipment of 20 packages, 7 packages were damaged. The packages are randomly inspected, one at a time, with replacement between each selection, until the fourth damaged package is discovered. Calculate the probability that exactly 12 packages are inspected.

11.9%

Step-by-step explanation:

To find the solution, we will do it through combinations.

Which has the following formula:

nCx = n! / [x! * (n-x) !]

Now, to know the probability in the first 11 packages we have to calculate the probability of favorable results divided by the probability of the number of possible results.

Favorable cases:

First, the number of combinations in which 3 damaged packages can be chosen from 7 damaged packages:

7C3 = 7! / [3! * (7-3) !] = 35

Second, the number of combinations in which you can choose 8 good packages from 13 good ones:

13C8 = 13! / [8! * (13-8) !] = 1287

Possible results: Number of combinations in which 11 packages can be taken from the 20 available packages:

20C11 = 20! / [eleven! * (20-11) !] = 167960

Therefore, the probability for the first 11 packages is:

(35 * 1287) / 167960 = 0.2681

For the last package, for package 12 to be damaged it is also 7 - 3 = 4, from 20 - 11 = 9.

That is 4/9, it would be the probability.

Finally, the probability would be the multiplication of the first 11 by the number 12, like this:

0.2681 * 4/9 = 0.119

That is to say that the answer is 11.9%