Questions: Among 6782 cases of heart pacemaker malfunctions, 372 were found to be caused by firmware, which is software programmed into the device. If the firmware is tested in 3 different pacemakers randomly selected from this batch of 6782 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 probability is This procedure is to result in the entire batch being accepted. (Round to three decimal places as needed.)

Solution Steps

The total number of heart pacemaker cases is \( N = 6782 \). Among these, the number of firmware-related failures is \( F = 372 \). The probability of a firmware failure is calculated as:

\[ p = \frac{F}{N} = \frac{372}{6782} \approx 0.0548 \]

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

\[ q = 1 - p \approx 1 - 0.0548 = 0.9452 \]

We need to find the probability of having no firmware failures when testing 3 randomly selected pacemakers. This can be modeled using the binomial distribution, where the probability of exactly \( x = 0 \) successes (failures in this case) in \( n = 3 \) trials is given by:

\[ P(X = 0) = \binom{n}{x} \cdot p^x \cdot q^{n-x} \]

Substituting the values:

\[ P(X = 0) = \binom{3}{0} \cdot (0.0548)^0 \cdot (0.9452)^3 = 1 \cdot 1 \cdot (0.9452)^3 \approx 0.844 \]

The probability of accepting the entire batch is \( P(X = 0) \approx 0.844 \). Since this value is greater than \( 0.5 \), we conclude that the procedure is likely to result in the entire batch being accepted.

Final Answer