Questions: Among 8902 cases of heart pacemaker malfunctions, 532 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 8902 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

Given:

• Total cases of heart pacemaker malfunctions: \( 8902 \)
• Cases caused by firmware: \( 532 \)

The probability of a firmware failure (\( p \)) is calculated as: \[ p = \frac{532}{8902} \approx 0.0598 \]

The probability of no firmware failure (\( q \)) is: \[ q = 1 - p \approx 0.9402 \]

We need to find the probability that there are no failures in 3 randomly selected pacemakers. This can be modeled using the binomial distribution with \( n = 3 \) trials and \( x = 0 \) successes (failures in this context).

The binomial probability formula is: \[ P(X = x) = \binom{n}{x} \cdot p^x \cdot q^{n-x} \]

Substituting the values: \[ P(X = 0) = \binom{3}{0} \cdot (0.0598)^0 \cdot (0.9402)^3 \approx 0.831 \]

The probability that the firmware in the entire batch will be accepted (i.e., no failures in 3 tested pacemakers) is \( 0.831 \).

Since \( 0.831 \) is greater than \( 0.5 \), it is likely that the entire batch will be accepted.

Final Answer