Question

among 6707 cases of heart pacemaker malfunctions, 364 were found to be caused by firmware, which is software programmed into the device. if the firmware is tested in 3 different pac randomly selected from this batch of 6707 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 pr 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 neede likely unlikely certain

Explanation:

Step1: Calculate the number of non - failure cases

The number of non - failure cases is \(6707 - 364=6343\).

Step2: Calculate the probability of no failure in the first selection

The probability of selecting a non - failure pacemaker in the first selection is \(P_1=\frac{6343}{6707}\).

Step3: Calculate the probability of no failure in the second selection

Since one non - failure pacemaker is already selected (without replacement), the probability of selecting a non - failure pacemaker in the second selection is \(P_2=\frac{6342}{6706}\).

Step4: Calculate the probability of no failure in the third selection

Since two non - failure pacemakers are already selected (without replacement), the probability of selecting a non - failure pacemaker in the third selection is \(P_3=\frac{6341}{6705}\).

Step5: Calculate the combined probability

The probability that all three selected pacemakers are non - failures (using the multiplication rule for dependent events) is \(P = \frac{6343}{6707}\times\frac{6342}{6706}\times\frac{6341}{6705}\)

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Using a calculator: \(P\approx0.847\)

A probability of \(0.847\) (where \(0.5\lt0.847\lt1\)) means the event is likely.

Answer:

The probability is \(0.8\). This procedure is likely to result in the entire batch being accepted.