A Quality Assurance Problem

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Chapter: Biostatistics for the Health Sciences: Basic Probability

One of the present authors provided consultation services to a medical device company that was shipping a product into the field.


One of the present authors provided consultation services to a medical device company that was shipping a product into the field. Before shipping, the company rou-tinely subjected the product to a sequence of quality control checks. In the field, it was discovered that one item had been shipped with a mismatched label. After checking the specifics, the company identified a lot of 100 items that included the mislabeled item at the time of shipment. These 100 items were sampled in order to test for label mismatches (failures).

The company tested a random sample of 13 out of 100 and found no failures. Al-though the company believed that this one mismatch was an isolated case, they could not be certain. They were faced with the prospect of recalling the remaining items in the lot in order to inspect them all for mismatches. This operation would be costly and time-consuming. On the other hand, if they could demonstrate with high enough assurance that the chances of having one or more mismatched labels in the field is very small, they would not need to conduct the recall.

The lot went through the following sequence of tests:

1. Thirteen out of 100 items were randomly selected for label mismatch checking.

2. No mismatches were found and the 13 were returned to the lot; two items were pulled and destroyed for other reasons.

3. Of the remaining 98 items, 13 were chosen at random and used for a destructive test (one that causes the item to be no longer usable in the field).

4. The remaining 85 items were then released.

In the field, it was discovered that one of these 85 had a mismatched label. A sta-tistician (Chernick) was asked to determine the probability that at least one more of the remaining 84 items in the field could have a mismatch, assuming:

a) Exactly two are known to have had mismatches.

b) The mismatch inspection works perfectly and would have caught any mis-matches.

c) In the absence of any information to the contrary, the two items pulled at the second stage could equally likely have been any of the 100 items.

The statistician also was asked to determine the probability that at least one more of the remaining 84 items in the field could have a mismatch, assuming that exactly three are known to have had mismatches. This problem entails calculating two probabilities and adding them together: (1) the probability that all three mislabeled items passed the inspection, and (2) the probability that one was destroyed among the two pulled while the other two passed. The first of these two probabilities was of primary interest.

In addition, for baseline comparison purposes, the statistician was to consider what the probability was of the outcome that if only one item out of the 100 in the lot were mismatched, it would be among the 85 that passed the sequence of tests. This probability, being the easiest to calculate, will be derived first.

For the one mismatched label to pass with the 85 that survived the series of in-spections, it must not have been selected from the first 13 for label mismatch check; otherwise, it would not have survived (assuming mismatch checking is perfectly ac-curate). Selecting 13 items at random from 100 is the same as drawing 13 one at a time at random without replacement. The probability that the item is not in these 13 is the product of 13 probabilities.

Each of these 13 probabilities represents the probability that among the 13 draws, the item is not drawn. On the first draw, this probability is 99/100. On the second draw, there are now only 99 items to select, resulting in the probability of 98/99 of the items not being selected. Continuing in this way and multiplying these probabilities together, we see that the probability of the item not being drawn in any one of the 13 draws is



This calculation can be simplified greatly by canceling common numerators and de-nominators to 87/100, which gives us the probability that the item survives the first inspection.

The second and third inspections occur independently of the first. The probabili-ty we calculate for the third inspection is conditional on the result of the second in-spection. So we calculate the probability of surviving those inspections and then multiply the three probabilities together to get our final result.

In the second stage, the 13 items that passed the initial inspection are replaced with others. So we again have 100 items to select from. Now, for the item with the mismatched label to escape destruction, it must not be one of the two items that were originally drawn. As we assumed that each item is equally likely to be drawn, the probability that the item with the mismatched label is not drawn is the probabil-ity that it is not the first one drawn multiplied by the probability that it is not the second one drawn, given that it was not the first one drawn. That probability is (98/100)(97/99).

At the third stage, there are only 98 items left and 13 are chosen at random for destructive testing. Consequently, the method to compute the probability is the same as the method used for the first stage, except that the first term in the product is 97/98 instead of 99/100. After multiplication and cancellation, we obtain 85/98.

The final result is then the product of these three probabilities, namely [(87/100)][(98/100)(97/99)][(85/98)]. This simplifies to (87/100)(97/100)(85/99) after cancellation. The result equals 0.72456 or 72.46%. (Note that a proportion also may be expressed as a percentage.)

Next we calculate the probability that there are two items with mismatched la-bels out of the 100 items in the lot. We want to determine the probability that both are missed during the three stages of inspection. Probability calculations that are similar to the foregoing calculations apply. Accordingly, we multiply the three probabilities obtained in the first three stages together.

To repeat, the probabilities obtained in the first three stages (the probability that both mismatched items are missed during inspection) are as follows:

·        The first stage—(87/100)(86/99)

·        The second stage, given that they survive the first stage—(98/100)(97/99)

·        The third stage, given that they are among the remaining 98—(85/98)(84/97)

The final result is (87/100)(86/99)(98/100)(97/99)(85/98)(84/97). This result simplifies to (87/100)(86/99)(85/100)(84/99) = 0.54506 or 54.51%.

In the case of three items with mismatched labels out of the 100 total items in the lot, we must add the probability that all three pass inspection to the probability that two out of three pass. To determine the latter probability, we must have exactly one of the three thrown out at stage two. This differs from the previous calculation in that we are adding the possibility of two passing and one failing.

The first term follows the same logic as the previous two calculations. We com-pute at each stage the probability that all the items with mismatched labels pass in-spection and multiply these probabilities together. The arguments are similar to those presented in the foregoing paragraphs. We present this problem as Exercise 5.22.

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