问题及解答

A quality control inspector is inspecting newly produced items for faults. The inspector searches an item for faults in a series of independent fixations, each of a fixed duration. Given that a flaw is actually present, let $p$ denote the probability that the flaw is detected during any one fixation (this model is discussed in "Human Performance in Sampling Inspection", Human Factors, 1979: 99--105).

(a) Assuming that an item has a flaw, what is the probability that it is detected by the end of the second fixation (once a flaw has been detected, the sequence of fixations terminates)? (Hint: $\{\text{flaw detected in at most two fixations}\}$=$\{\text{flaw detected on the first fixation}\}$ $\cup$ $\{\text{flaw undetected on the first and detected on the second}\}$.)

(b) Give an expression for the probability that a flaw will be detected by the end of the $n$th fixation.

(c) If when a flaw has not been detected in three fixations, the item is passed, what is the probability that a flawed item will pass inspection?

(d) Suppose $10\%$ of all items contain a flaw [$P$(randomly chosen item is flawed)$=.1$]. With the assumption of part (c), what is the probability that a randomly chosen item will pass inspection (it will automatically pass if it is not flawed, but could also pass if it is flawed)?

(e) Given that an item has passed inspection (no flaws in three fixations), what is the probability that it is actually flawed? Calculate for $p=.5$.

Remark: The exercise is copied from the reference book Devore2017B.

(a)

Since the sequence of fixations terminates when a flaw is detected, we have $P(\{\text{flaw detected on the first fixation}\})=p$.

If flaw is undetected on the first fixation and detected on the second, then the probability is $(1-p)p$.

Thus, $P(\text{flaw detected in at most two fixations})=p+(1-p)p=p(2-p)$.

Another method.

The probability of event that the flaw was not detected in two fixations is equal to $(1-p)(1-p)$. Then, 

$P(\text{flaw detected in at most two fixations})=1-(1-p)^2=p(2-p)$.

(b) 

Same as (a), the probability of event that the flaw was not detected in $n$ fixations is equal to $(1-p)^n$. Thus, the probability that a flaw will be detected by the end of the $n$th fixation is $1-(1-p)^n$.

(c)

A flawed item will pass inspection if and only if it will never be detected during the first three detection. Thus the probability is $(1-p)^3$.

(d)

Randomly choose an item, if it is not flawed, it will automatically pass the inspection. We denote this event is $A$. And let $B$ denote the event that a flawed item passes the inspection. Thus, the probability of a randomly chosen item will pass inspection is 

\[
P(A\cup B)=P(A)+P(B)=0.9+0.1\times(1-p)^3
\]

(e)

\[
P(\{\text{flawed}\}|\{\text{passed the inspection}\})=\frac{P(\{\text{flawed}\}\cap\{\text{passed the inspection}\})}{P(\{\text{passed the inspection}\})}
\]

Thus, when $p=0.5$, the probability of an item that it is actually flawed while it has passed the inspection is

\[
\frac{0.1\times(1-0.5)^3}{0.9+0.1\times(1-0.5)^3}\approx 0.0137
\]