Question

Components arriving at a distributor are checked for defects by two different inspectors (each component is checked by both inspectors). The first inspector detects \({\rm{90\% }}\)of all defectives that are present, and the second inspector does likewise. At least one inspector does not detect a defect on \({\rm{20\% }}\)of all defective components. What is the probability that the following occur?

a. A defective component will be detected only by the first inspector? By exactly one of the two inspectors?

b. All three defective components in a batch escape detection by both inspectors (assuming inspections of different components are independent of one another)?

Short answer

\(\begin{array}{l}{\rm{a}}{\rm{. }}P\left( {{A_1} \cap A_2^\prime } \right) &=& 0.1;\\P\left( {{E_1}} \right) &=& 0.2;{\rm{ }}\\{\rm{b}}{\rm{. }}P\left( {{A_3}} \right) &=& 0.\end{array}\)

Step-by-step solution

Definition of Independent probability

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 the probability that the following occur

Denote events

\(\begin{array}{l}{{\rm{A}}_{\rm{1}}}{\rm{ = \{ first inspector detects \} }}\\{{\rm{A}}_{\rm{2}}}{\rm{ = \{ second inspector detects \} }}\end{array}\)

We are given

\(\begin{array}{c}{\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{ = 0}}{\rm{.9}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}} \right){\rm{ = 0}}{\rm{.9}}\end{array}\)

Also stated is the likelihood that at least one inspector would miss a flaw, which is the union of events \(A_1^\prime \)and \({{\rm{B}}_{\rm{2}}}^{\rm{'}}\).

\({\rm{P}}\left( {A_1^\prime \cup A_2^\prime } \right){\rm{ = 0}}{\rm{.2}},\)

from which we can obtain

\(\begin{align} P\left( {{A}_{1}}\cap {{A}_{2}} \right) =& 1-P\left[ {{\left( {{A}_{1}}\cap {{A}_{2}} \right)}^{\prime }} \right].........(1) \\ =& 1-P\left( A_{1}^{\prime }\cup A_{2}^{\prime } \right).........(2) \\ =& 1-0.2 \\ =& 0.8 \end{align}\)

(1): for any event A, \({\rm{P}}\left( {{A^\prime }} \right){\rm{ + P(A) = 1}}\),

(2): De Morgan's law, this stands for every two events.

Finding probability

(a):

We are asked to find probability of intersection of event \({{\rm{A}}_{\rm{1}}}\) (detected by first inspector) and event \(A_2^\prime \) (not detected by second inspector

\(\begin{align} P\left( {{A}_{1}}\cap A_{2}^{\prime } \right)\text =& P\left( {{\text{A}}_{\text{1}}} \right)\text{-P}\left( {{A}_{1}}\cap {{\text{A}}_{\text{2}}} \right).........(1) \\ \text=&0 \text{.9-0}\text{.8=0}\text{.1} \end{align}\)

(1): This can represent any two events; for a better understanding, make a Venn Diagram.

Similarly, the likelihood that one of the two inspectors will identify a defective component is the sum of the probability we just calculated and the probability that only the second inspector will detect a bad component (those are two disjoint events, this is why we have sum of those two events)

\(\begin{align} P\left( {{A}_{1}}\cap A_{2}^{\prime } \right)\text =& P\left( {{\text{A}}_{\text{1}}} \right)\text{-P}\left( {{A}_{1}}\cap {{\text{A}}_{\text{2}}} \right) \\ \text=&0 \text{.9-0}\text{.8=0}\text{.1} \end{align}\)

Denote event \({{\rm{E}}_{\rm{1}}}\)= {exactly one inspector detects }. Now we have

Thus , we have

\(\begin{array}{c}P\left( {{E_1}} \right) &=& P\left( {{A_1} \cap A_2^\prime } \right) + P\left( {A_1^\prime \cap {A_2}} \right)\\ &=& {\rm{0}}{\rm{.1 + 0}}\rm.1 \\ &=& 0{\rm{.2}}\end{array}\)

Step 4: All three defective components in a batch escape detection by both inspectors

(b):

We are given independence of events now. The probability we need in order to find is probability of complement of union of events \({{\rm{A}}_{\rm{1}}}\)and \({{\rm{A}}_2}\) (probability that neither of inspectors detects a defect)

\(\begin{align}P\left[ {{\left( {{A}_{1}}\cup {{A}_{2}} \right)}^{\prime }} \right]=& 1-P\left( {{A}_{1}}\cup {{A}_{2}} \right).........(1) \\ =& 1-\left[ P\left( {{A}_{1}}\cap {{A}_{2}} \right)\text{+P}\left( {{\text{E}}_{\text{1}}} \right) \right].........(2) \\\text =& 1-0\text{.8-0}\text.2&=&0 \\ \end{align}\)

(1): for any event \({\rm{A,P(A) + P}}\left( {{A^\prime }} \right) = 1\);

(2): The aggregate of these three events can always be written as the union of two events (note: \({{\rm{E}}_{\rm{1}}}\)is made up of two events, see (a).

As a result, the chances of neither inspector spotting a flaw are nil. The likelihood that all three defective components escape discovery by both inspectors is the product of three probabilities that neither inspector identifies a defect using independence (see multiplication property below). Indicate the occurrence.

\({{\rm{A}}_{\rm{3}}}{\rm{ = \{ }}\)all three defective components escape detection },

we have

\(\begin{align}P\left( {{A}_{3}} \right)=& P\left[ {{\left( {{A}_{1}}\cup {{A}_{2}} \right)}^{\prime }} \right]\cdot P\left[ {{\left( {{A}_{1}}\cup {{A}_{2}} \right)}^{\prime }} \right]\cdot P\left[ {{\left( {{A}_{1}}\cup {{A}_{2}} \right)}^{\prime }} \right].........(1) \\ =& 0\cdot 0\cdot 0\\=&0

\end{align}\)

(1): see explanation above. We use independence and multiplication property!

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

\({\rm{P}}\left( {{A_{{i_1}}} \cap {A_{{i_2}}} \ldots {A_{{i_k}}}} \right){\rm{ = P}}\left( {{{\rm{A}}_{{{\rm{i}}_{\rm{1}}}}}} \right){\rm{*P}}\left( {{{\rm{A}}_{{{\rm{i}}_{\rm{2}}}}}} \right){\rm{* \ldots *P}}\left( {{{\rm{A}}_{{{\rm{i}}_{\rm{k}}}}}} \right)\)

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