Question

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, \({\rm{1979: 99--105)}}{\rm{.}}\)

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)?

b. Give an expression for the probability that a flaw will be detected by the end of the nth 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 \({\rm{10\% }}\) of all items contain a flaw (P(randomly chosen item is flawed) . \({\rm{1}}\)). 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 \({\rm{p = 5}}\).

Short answer

a. The probability that it is detected by the end of the second fixation is \({\rm{P(D) = p*(2 - p)}}\);

b. The probability that a flawed item will pass inspection is \({\rm{P(E) = 1 - (1 - p}}{{\rm{)}}^{\rm{n}}}{\rm{;}}\)

c. The probability that a flawed item will pass inspection is \({\rm{P(}}{{\rm{N}}_{\rm{3}}}{\rm{) = (1 - p}}{{\rm{)}}^{\rm{3}}}\);

d. The probability that a randomly chosen item will pass inspection is \({\rm{P(A) = 0}}{\rm{.9 + 0}}{\rm{.1*(1 - p}}{{\rm{)}}^{\rm{3}}}{\rm{ }}\);

e. The probability that it is actually flawed is \({\rm{P(F|P) = 0}}{\rm{.0137}}{\rm{.}}\)

Step-by-step solution

Definition of Independent probability

Independence If the probability of one event is unaffected by the occurrence or non-occurrence of the other, two events are said to be independent in probability. Consider the following example for a better understanding of this definition.

Finding probability that it is detected by the end of the second fixation

Denote events

\({{\rm{A}}_{\rm{1}}}\){flaw detected during first fixation}

\({{\rm{A}}_{\rm{2}}}\){flaw detected during second fixation}

As given in the exercise, denote

\({\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{ = P}}\left( {{{\rm{A}}_{\rm{2}}}} \right){\rm{ = p}}\)

(a):

By the end of the second fixation, denote event D as "detected." Once a flaw is found, the fixation sequence ends, therefore event \({\rm{D}}\)can be represented as the union of two events: \({{\rm{A}}_{\rm{1}}}\) (if the flaw is detected in the first fixation), and \(A_1^\prime \cap {{\rm{A}}_{\rm{2}}}\) (if the defect is detected in the second fixation) (it is not detected in first fixation but in second). Keep in mind that we thought an item had a fault (so the statement above is true). The following is an example of what you can do with this.

\(\begin{align}\text P(D)=&P \left( {{A}_{1}}\cup \left( A_{1}^{\prime }\cap {{A}_{2}} \right) \right) \\ \text =&P \left( {{\text{A}}_{\text{1}}} \right)\text{+P}\left( A_{1}^{\prime }\cap {{A}_{2}} \right)............(1) \\ \text =&P \left( {{\text{A}}_{\text{1}}} \right)\text{+P}\left( A_{1}^{\prime } \right)\cdot P\left( {{A}_{2}} \right).........(2) \\ \text =& p+(1-p)*p............\text{(3)} \\ \text =& p*(2-p) \\ \end{align}\)

Explain the probability  

(1): the events are disjoint,

(2): events \(A_1^\prime \) and \({{\rm{A}}_{\rm{2}}}\)are independent because events \({{\rm{A}}_{\rm{1}}}\)and \({{\rm{A}}_{\rm{2}}}\)are independent (see proposition below). Since they are

independent, we can use the multiplication rule given below,

(3): for any event A, \({\rm{P(A) + P}}\left( {{A^\prime }} \right){\rm{ = 1}}\), therefore, \({\rm{P}}\left( {A_1^\prime } \right){\rm{ = 1 - p}}\).

Multiplication Property: Two events \({\rm{A}}\)and \({\rm{B}}\)are independent if and only if

\(P(A \cap B){\rm{ = P(A)*P(B)}}\)

Proposition: If \({\rm{A}}\)and \({\rm{B}}\)are independent, then

- \({A^\prime }\)and \({\rm{B}}\)are independent;

- \({A^\prime }\)and \({B^\prime }\)are independent;

- \({\rm{A}}\)and \({B^\prime }\)are independent.

Thus, The probability that it is detected by the end of the second fixation is \({\rm{P(D) = p*(2 - p)}}\);

Give an expression for the probability that a flaw will be detected by the end of the nth fixation. 

(b):

Denote events

\({{\rm{A}}_{\rm{1}}}\){flaw detected during 1^st fixation}

\({{\rm{A}}_{\rm{2}}}\){flaw detected during 2^nd fixation }

\({A_n} = \left\{ {} \right.\)flaw detected during 1^st fixation} .

As given in the exercise, denote

\({\rm{P}}\left( {{{\rm{A}}_{\rm{i}}}} \right){\rm{ = p}}\)

for \({\rm{i = 1,2, \ldots ,n}}\).

Remember that we know an object has a problem and that the series of fixations ends when a flaw is found, just as we did in (a). We can represent event E= By the end of the nth fixation , will be identified as a union of \({\rm{n}}\) disjoint events.

\(\begin{align}P(E)=& P\left( {{A}_{1}}\cup \left( A_{1}^{\prime }\cap {{A}_{2}} \right)\cup \left( A_{1}^{\prime }\cap A_{2}^{\prime }\cap A_{3}^{\prime } \right)\cup \ldots \cup \left( A_{1}^{\prime }\cap A_{2}^{\prime }\cap \ldots \cap A_{n-1}^{\prime }\cap {{A}_{n}} \right) \right)..............(1) \\ =& P\left( {{A}_{1}} \right)+P\left( A_{1}^{\prime }\cap {{A}_{2}} \right)+\ldots +P\left( A_{1}^{\prime }\cap A_{2}^{\prime }\cap \ldots \cap A_{n-1}^{\prime }\cap {{A}_{n}} \right)..............(2) \\ =& P\left( {{A}_{1}} \right)+P\left( A_{1}^{\prime } \right)\cdot P\left( {{A}_{2}} \right)+P\left( A_{1}^{\prime } \right)\cdot P\left( A_{2}^{\prime } \right)\cdot P\left( {{A}_{3}} \right)+\ldots +P\left( A_{1}^{\prime } \right)\cdot P\left( A_{2}^{\prime } \right)\cdot \ldots \cdot P\left( A_{n-1}^{\prime } \right)\cdot P\left( {{A}_{n}} \right)..............(3) \\ =& \text{p+(1-p)*p+(1-p}{{\text{)}}^{\text{2}}}\text{*p++(1-p}{{\text{)}}^{\text{n-1}}}\text{*p} \\ \text =& p*\left( \text{1+(1-p)+(1-p}{{\text{)}}^{\text{2}}}\text{++(1-p}{{\text{)}}^{\text{n-1}}} \right) \\ \text =&p* \frac{\text{1-(1-p}{{\text{)}}^{\text{n}}}}{\text{1-(1-p)}}..................(4) \\ \end{align}\)

\(=1-{{(1-p)}^{n}}\)

Finding the expression

(1): note the explanations in (a) and (b) at the start of (b),

(2): the sequence of events is disjointed,

(3): the following multiplication rule and statement,

(4): the following geometric series.

Proposition: If \({\rm{A}}\)and \({\rm{B}}\)are independent, then

- \({A^\prime }\)and \({\rm{B}}\)are independent;

- \({A^\prime }\)and \({B^\prime }\)are independent;

- \({\rm{A}}\)and \({B^\prime }\)are independent.

This is also true for \({\rm{n}}\) events (any combination of events \({{\rm{A}}_{\rm{i}}}\)and their complements is treated as a separate event).

Multiplication Property:

For events \({{\rm{A}}_{\rm{1}}}{\rm{,}}{{\rm{A}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{A}}_{\rm{n}}}{\rm{,n}} \in {\rm{N}}\)we say that they are mutually independent if

\(P\left( {{A_{{i_1}}} \cap {A_{{i_2}}} \ldots {A_{{i_k}}}} \right) = P\left( {{A_{{i_1}}}} \right) \cdot P\left( {{A_{{i_2}}}} \right) \cdot \ldots \cdot P\left( {{A_{{i_k}}}} \right)\)

for every \({\rm{k}} \in {\rm{\{ 2,3, \ldots ,n\} }}\), and every subset of indices \({{\rm{i}}_{\rm{1}}}{\rm{,}}{{\rm{i}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{i}}_{\rm{k}}}\).

Formula can be used to determine the sum of finite geometric series.

where \({\rm{n}}\)is number of terms, \({a_1}\) is the first term and \({\rm{q}}\) is the common ratio. In our case \({{\rm{a}}_{\rm{1}}}{\rm{ = 1}}\) and \({\rm{q = 1 - p < 1}}\).

Thus, The probability that a flawed item will pass inspection is \({\rm{P(E) = 1 - (1 - p}}{{\rm{)}}^{\rm{n}}}{\rm{;}}\)

Finding  the probability that a flawed item will pass inspection?

(c):

Denote event \({{\rm{N}}_{\rm{3}}}{\rm{ = \{ }}\)the flas has no been detected in 3 fixations \(\} \). This event is intersection of events \(A_1^\prime \) (no detected in first), \(A_2^\prime \) (not detected in second), and \(A_3^\prime \) (not detected in third).

Therefore

\(\begin{aligned}{\rm{P(}}{{\rm{N}}_{\rm{3}}} \rm) =& P({{A'}_1} \cap {{A'}_2} \cap {{A'}_3}) \rm &=& P({{A'}_1})*{\rm{P}}({{A'}_2})*{\rm{P}}({{A'}_3})\\ \rm =& (1 - p{{\rm{)}}^{\rm{3}}}\end{aligned}\)

(1): we employ the multiplication rule, and events are independent of one another \({A'_1},{A'_2}\) and \({A'_3}\)and the propositions mentioned in (b).

Thus, The probability that a flawed item will pass inspection is \({\rm{P(}}{{\rm{N}}_{\rm{3}}}{\rm{) = (1 - p}}{{\rm{)}}^{\rm{3}}}\);

Finding  the probability that a randomly chosen item will pass inspection ?

(d): Denote event A = { randomly chosen item passes inspection } It is possible for an item to be defective and pass or not be flawed and pass, which is a combination of occurrences NF = { not flawed and passes } and FP={ flawed and passes },

Therefore

\(\begin{align}\text P(A)=&P(NF \cup \text{FP})=\text{P(NF)+P(FP)}........\text{(1)} \\ \text =&P(NF)+P(F)*P(P\!\!|\!\!\text{ F)}...........\text{(2)} \\ \text =&0 \text{.9+0}\text{.1*(1-p}{{\text{)}}^{\text{3}}}.................(3) \\ \end{align}\)

(1): events NF and FP are disjoint;

(2): here we use the multiplication rule given below;

(3): Because the likelihood of a randomly chosen object being flawed is 0.1, the probability of it not being flawed is \(1 - 0.1 = 0.9\). We also utilise what we computed in (c) - P(P|F) (recall that in (c), we assumed we had a flawed item, thus we calculated the conditional probability of an event-flawed product passing provided that the event-flawed product has occurred).

The Multiplication Rule

\({\rm{P(A}} \cap {\rm{B) = P(A|B)*P(B) }}\)

Thus, The probability that a randomly chosen item will pass inspection is \({\rm{P(A) = 0}}{\rm{.9 + 0}}{\rm{.1*(1 - p}}{{\rm{)}}^{\rm{3}}}{\rm{ }}\);

Finding the probability that it is actually flawed? Calculate for p =5

(e): Given that the event P (passes) has occurred, we must calculate the conditional probability of F (flawed). This can be calculated using the definition of conditional probability.

\({\rm{P}}\left( {\rm{B}} \right){\rm{ > 0}}\)is the conditional probability of A given the occurrence of event B.

\({\rm{P(A|B) = }}\frac{{{\rm{P(A}} \cap {\rm{B)}}}}{{{\rm{P}}({\rm{B}})}}{\rm{ }}\)

for any two event A and B.

The following is true

\(\begin{align}\text{P(F }\!\!|\!\!\text P)=& \frac{\text{P(F}\cap \text{P)}}{\text{P(P})} \\ =&\frac{\text{0}\text{.1*(1-p}{{\text{)}}^{\text{3}}}}{\text{0}\text{.9+0}\text{.1*(1-p}{{\text{)}}^{\text{3}}}}\text{,}.........\text{(1)} \end{align}\)

(1): we have calculated the probabilities in (d).

For\({\rm{p = 0}}{\rm{.5}}\), we have

\(\begin{aligned} \rm P(F|P) =&\frac{{{\rm{0}}{\rm{.1*(1 - p}}{{\rm{)}}^{\rm{3}}}}}{{{\rm{0}}{\rm{.9 + 0}}{\rm{.1*(1 - p}}{{\rm{)}}^{\rm{3}}}}}\\ \rm =& \frac{{{\rm{0}}{\rm{.1*(1 - 0}}{\rm{.5}}{{\rm{)}}^{\rm{3}}}}}{{{\rm{0}}{\rm{.9 + 0}}{\rm{.1*(1 - 0}}{\rm{.5}}{{\rm{)}}^{\rm{3}}}}}\\ \rm =& 0{\rm{.0137}}{\rm{.}}\end{aligned}\)

Thus, The probability that it is actually flawed is \({\rm{P(F|P) = 0}}{\rm{.0137}}{\rm{.}}\)