Acceptance Sampling. With one method of a procedure called acceptance sampling, a sample of items is randomly selected without replacement and the entire batch is accepted if every item in the sample is found to be okay. Exercises 27 and 28 involve acceptance sampling.

Defective Pacemakers Among 8834 cases of heart pacemaker malfunctions, 504 were found to be caused by firmware, which is software programmed into the device (based on data from “Pacemaker and ICD Generator Malfunctions,” by Maisel et al., Journal of the American Medical Association, Vol. 295, No. 16). If the firmware is tested in three different pacemakers randomly selected from this batch of 8834 and the entire batch is accepted if there are no failures, what is the probability that the firmware in the entire batch will be accepted? Is this procedure likely to result in the entire batch being accepted?

The number of heart pacemakers in a batch is 8834.

Out of these, 504 malfunctioned were found to be caused by firmware.

Three selections are made from a batch of 8834, without replacement.

The probability of a number of events to occur together,when each of the events occurs independently, is represented and computed as follows.

$P\left(AandBandCand...\right)=P\left(A\right)\times P\left(B\right)\times P\left(C\right)\times ...$

The entire batch of 8834 is accepted if the all three randomly selected pacemakers contain firmware that do not malfunction.

Let A be the event of selecting the first pacemaker that do not malfunction.

Let B be the event of selecting the second pacemaker that do not malfunction.

Let C be the event of selecting the third pacemaker that do not malfunction.

The total number of pacemakers is equal to 8834.

The number of defective cases of firmware is equal to 504.

The number of non-defective cases is:

$8834-504=8330$

As the selection is done without replacement, the probability of successive non-defective firmwares changes as follows:

$$\begin{gathered} P\left(A\right)=\frac{8330}{8834} \\ P\left(B\right)=\frac{8329}{8833} \\ P\left(C\right)=\frac{8329}{8832} \end{gathered}$$

The probability of selecting three non-defective pacemakers, without replacement, is given by:

$$\begin{gathered} P\left(AandBandC\right)=P\left(A\right)\times P\left(B\right)\times P\left(C\right) \\ =\frac{8330}{8834}\times \frac{8329}{8833}\times \frac{8328}{8832} \\ \approx 0.838 \end{gathered}$$

Therefore, the probability of selecting three non-defective cases is equal to 0.838.

As the probability value for getting all non-defective firmware is high for the sample of size three, the entire batch can be accepted even though there are a substantial number of defective cases.