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

**A QUALITY ASSURANCE PROBLEM***

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

(99/100)(98/99)(97/98)(96/97)(95/96)(94/95)(93/94)

(92/93)(91/92)(90/91)(89/90)(88/89)(87/88)

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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