Questions: Among 6872 cases of pacemaker malfunctions, 318 were found to be caused by firmware, which is software programmed into the device. If the firmware is tested in 3 different pacemakers from a batch of 6872 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?

Solution Steps

Given the total number of cases \( n = 6872 \) and the number of firmware failures \( x = 318 \), the probability of a firmware failure is calculated as:

\[ p = \frac{x}{n} = \frac{318}{6872} \approx 0.0463 \]

Consequently, the probability of success (no firmware failure) is:

\[ q = 1 - p \approx 1 - 0.0463 = 0.9537 \]

In the testing procedure, \( n = 3 \) pacemakers are tested. We want to find the probability of observing \( x = 0 \) failures (i.e., all tested pacemakers function correctly). The probability is given by the binomial distribution formula:

\[ P(X = x) = \binom{n}{x} \cdot p^x \cdot q^{n-x} \]

Substituting the values:

\[ P(X = 0) = \binom{3}{0} \cdot (0.0463)^0 \cdot (0.9537)^3 = 1 \cdot 1 \cdot (0.9537)^3 \approx 0.8675 \]

The probability that the firmware in the entire batch will be accepted is approximately \( 0.8675 \). Since this value is greater than \( 0.5 \), we conclude that the procedure is likely to be accepted.

Final Answer