Chapter 2.5, Problem 85E

A quality control inspector is examining 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).

1. 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)?
2. b. Give an expression for the probability that a flaw will be detected by the end of the nth fixation.
3. 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?
4. 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)?
5. 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.

To determine

Find the probability of detecting the flaw at the end of the second fixation.

Answer to Problem 85E

The probability of detecting the flaw at the end of the second fixation is $\underset{\_}{p\left(2-p\right)}$.

Explanation of Solution

Given info:

The series of independent items were detected for faults by the quality control inspector. The items were inspected for the faults at each fixation for the each fixed time. Here, p denotes the probability of detecting the flaw at any one fixation.

Calculation:

Define the events as given below:

$\begin{array}{l}{D}_1:\text{ The event of detecting the flaw at first fixation}\\ D_2\text{: The event of detecting the flaw at second fixation }\\ p\text{: the probability of the detecting the flaws at any one fixation }\end{array}$

The probabilities of the events are given below:

$P\left(D_1\right)=p$ and $P\left(D_2\right)=p$.

Independent events

If two events are said to be independent then the occurrence of one does not affect the occurrence of the other.

$P\left(A|B\right)=P\left(A\right)$

Multiplication rule for two independent events:

If the events A and B are independent then,

$P\left(A\cap B\right)=P\left(A\right)\times P\left(B\right)$.

The probability of detecting the flaw at the end of the second fixation is obtained as shown below:

$\begin{array}{c}P\left(\text{detects at most 2 fixation }\right)=P\left(D_1\right)+P\left(D_1{}^{\prime }\cap D_2\right)\\ =P\left(D_1\right)+P\left(D_1{}^{\prime }\right)\times P\left(D_2\right)\end{array}$

The probability of the event is obtained as given below:

$\begin{array}{c}P\left(\text{detects at most 2 fixation }\right)=P\left(D_1\right)+P\left(D_1{}^{\prime }\right)\times P\left(D_2\right)\\ =p+\left(1-p\right)p\\ =p+p-p^2\\ =2p-p^2\\ =p\left(2-p\right)\end{array}$

Thus, the probability of the event is $\underset{\_}{p\left(2-p\right)}$.

To determine

Find the probability of detecting the flaw at the end of the $n^{th}$ fixation.

Answer to Problem 85E

The probability of detecting the flaw at the end of the $n^{th}$ fixation is $\underset{\_}{1-{\left(1-p\right)}^n}$.

Explanation of Solution

Calculation:

Define the event as given below:

$D_i\left(i=1,2,\mathrm{...}n\right)$ denotes event of detecting the flaw at the end of the n^th fixation.

The probability of detecting the flaw at the end of the $n^{th}$ fixation is obtained as shown below:

$\begin{array}{c}P\left(\text{detects at most }n\text{ fixation }\right)=P\left(D_1\right)+P\left(D_1{}^{\prime }\cap D_2\right)+\mathrm{...}+P\left(D_1{}^{\prime }\cap D_2{}^{\prime }\cap \mathrm{...}\cap D_{n-1}{}^{\prime }\cap D_n\right)\\ =P\left(D_1\right)+P\left(D_1{}^{\prime }\right)\times P\left(D_2\right)+\mathrm{...}+P\left(D_1{}^{\prime }\right)\times P\left(D_2{}^{\prime }\right)\times \mathrm{...}\times P\left(D_n\right)\\ =p+p\left(1-p\right)+p{\left(1-p\right)}^2+\mathrm{...}+p{\left(1-p\right)}^{n-1}\\ =p\left[1+\left(1-p\right)+{\left(1-p\right)}^2+\mathrm{...}+{\left(1-p\right)}^{n-1}\right]\end{array}$

Based on geometric series: $\left[1+\left(x\right)+{\left(x\right)}^2+\mathrm{...}+{\left(x\right)}^{n-1}\right]=\frac{1-x^n}{1-x}$

The expression $\left[1+\left(1-p\right)+{\left(1-p\right)}^2+\mathrm{...}+{\left(1-p\right)}^{n-1}\right]$ can also be written as $\left[\frac{1-{\left(1-p\right)}^n}{1-\left(1-p\right)}\right]$.

The probability of the event is obtained as given below:

$\begin{array}{c}P\left({\text{detects at most n}}^{th}\text{ fixation }\right)=p\left[1+\left(1-p\right)+{\left(1-p\right)}^2+\mathrm{...}+{\left(1-p\right)}^{n-1}\right]\\ =p.\frac{1-{\left(1-p\right)}^n}{1-\left(1-p\right)}\\ =p.\frac{1-{\left(1-p\right)}^n}{p}\\ =1-{\left(1-p\right)}^n\end{array}$

Thus the probability of the event is $\underset{\_}{1-{\left(1-p\right)}^n}$.

To determine

Obtain the probability of not detecting the flaw in three fixations.

Answer to Problem 85E

The probability of not detecting the flaws in three fixations is $\underset{\_}{{\left(1-p\right)}^3}$.

Explanation of Solution

Calculation:

$\left(1-p\right)$ denotes the probability of not detecting the flaws.

The probability of not detecting the flaws in three fixations is obtained as shown below:

$\begin{array}{c}P\left(\text{not detecting the flaws in 3 fixation }\right)=P\left(D_1{}^{\prime }\cap D_2{}^{\prime }\cap D_3{}^{\prime }\right)\\ =P\left(D_1{}^{\prime }\right)\times P\left(D_2{}^{\prime }\right)\times P\left(D_3{}^{\prime }\right)\\ =\left(1-p\right)+\left(1-p\right)+\left(1-p\right)\\ =\left[{\left(1-p\right)}^3\right]\end{array}$

Thus the probability of not detecting the flaws in three fixations is $\underset{\_}{{\left(1-p\right)}^3}$

To determine

Obtain the probability of selecting the item that passes the inspection.

Answer to Problem 85E

The probability of selecting the item that pass the inspection is $\underset{\_}{0.9+0.1{\left(1-p\right)}^3}$.

Explanation of Solution

Calculation:

Here, the probability of items flawed is 0.1, the probability of the item not flawed is 0.9, the probability of not detecting the flaws in 3 fixation is $\left[{\left(1-p\right)}^3\right]$.

The probability of selecting the item that pass the inspection is obtained as shown below:

$\begin{array}{c}P\left(\begin{array}{l}\text{items passing }\\ \text{the inspection}\end{array}\right)=P\left(\text{items not flawed}\cup \left(\begin{array}{l}\text{items flawed and }\\ \text{pass the inspection}\end{array}\right)\right)\\ =P\left(\text{items not flawed}\right)+\left(\begin{array}{l}P\left(\text{items flawed}\right)\times \\ P\left(\text{passes the inspection|items flawed}\right)\end{array}\right)\\ =0.9+0.1\times {\left(1-p\right)}^3\end{array}$

Thus the probability of not detecting the flaws in three fixations is $\underset{\_}{0.9+0.1{\left(1-p\right)}^3}$.

To determine

Obtain the probability that the items flawed given that the item has passed the inspection.

Answer to Problem 85E

The probability that the items flawed given that the item has passed the inspection is 0.0137.

Explanation of Solution

Calculation:

Here, the probability of detecting the flaws at any one fixation is 0.5.

From part (d), the probability of selecting the item that pass the inspection is $0.9+0.1{\left(1-p\right)}^3$.

The probability that that the items flawed given that the item has passed the inspection is obtained as given below:

$\begin{array}{c}P\left(\text{Flawed}|\text{passed the inspection}\right)=\frac{P\left(\text{Flawed}\cap \text{passed the inspection}\right)}{P\left(\text{Passed the inspection}\right)}\\ =\frac{P\left(\text{flawed}\right).P\left(\text{passed|flawed}\right)}{P\left(\text{Passed the inspection}\right)}\\ =\frac{0.1{\left(1-p\right)}^3}{0.9+0.1{\left(1-p\right)}^3}\end{array}$

Substitute p as 0.5,

$\begin{array}{c}P\left(\text{Flawed}|\text{passed the inspection}\right)==\frac{0.1{\left(1-0.5\right)}^3}{0.9+0.1{\left(1-0.5\right)}^3}\\ =\frac{0.1{\left(0.5\right)}^3}{0.9+0.1{\left(0.5\right)}^3}\\ =\frac{0.1\times 0.125}{0.9+0.1\times 0.125}\end{array}$

$\begin{array}{l}=\frac{0.0125}{0.9+0.0125}\\ =\frac{0.0125}{0.9125}\\ =0.0137\end{array}$

Thus, the probability that the items flawed given that the item has passed the inspection is 0.0137.