Among 6643 heart pacemaker malfunctions, 368 were caused by firmware; if 3 pacemakers are randomly tested and the whole batch is accepted only if there are no firmware failures, what is the probability the batch is accepted, and is acceptance likely?

What the acceptance rule is really checking

You accept the entire batch if your sample of 3 contains zero pacemakers whose malfunction was caused by firmware. So we need the probability that a random sample of 3 contains no “firmware-caused” cases.

Turn the counts into “good” and “bad” items

• Total items: $N=6643$
• Firmware-caused (treat as bad for acceptance): $K=368$
• Not firmware-caused (good for acceptance): $N-K=6643-368=6275$
• Sample size: $n=3$

Compute the probability of 0 firmware failures (hypergeometric)

Sampling is without replacement, so:

$$P(\text{accept})=P(X=0)=\frac{\binom{6275}{3}}{\binom{6643}{3}}$$

A convenient way to evaluate it is sequentially:

$$P(\text{accept})=\frac{6275}{6643}\cdot\frac{6274}{6642}\cdot\frac{6273}{6641}\approx 0.843$$

So there is about an $84.3\%$ chance the batch is accepted.

Is the procedure likely to accept the whole batch?

Yes. Since $0.843$ is much larger than $0.5$, this testing rule will accept the batch most of the time.

To see why, notice the firmware-caused rate is $368/6643\approx 0.055$ (about $5.5\%$). With only 3 tested units, it is still quite common to miss that $5.5\%$ and see no failures in the sample.