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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.

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

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:

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

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.