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Under the normal operating conditions, the probability of the produced items by the machine being defective is 0.01, and the items are independent of probability.

The machine has a memory of production. It can sense and be independent of anything that happens earlier. After producing a defective item, it produces another defective item with probability\(\frac{2}{5}\). Thus, producing a non-defective item produces a defective item with probability\(\frac{1}{{165}}\).

We have to consider that either the machine is operating normally or has a memory for the whole time of our observation.

And the machine is operating normally with a probability \(\frac{2}{3}\)

Consider the events,

B =The event that the machine is operating normally

\({D_i}\)= The event that the inspected i^th item is defective

So, \(\Pr \left( B \right) = \frac{2}{3}\)

We know that the probability of the item being defective is 0.01 independently. And\({D_i}\)is independent of B. So, we can consider\(\Pr \left( {{D_i}\left| B \right.} \right) = 0.01\).

As\({D_i}\)is independent of B, so, we can also consider\(\Pr \left( {{D_i}\left| {{B^c}} \right.} \right) = 0.01,\;\forall \;i \le n\).

Therefore,

\({\bf{Pr}}\left( {{{\bf{D}}_{{\bf{n + 1}}}}\left| {{{\bf{B}}^{\bf{c}}}} \right.} \right){\bf{ = Pr}}\left( {{{\bf{D}}_{{\bf{n + 1}}}}\left| {{{\bf{D}}_{\bf{n}}} \cap {{\bf{B}}^{\bf{c}}}} \right.} \right){\bf{Pr}}\left( {{{\bf{D}}_{\bf{n}}}\left| {{{\bf{B}}^{\bf{c}}}} \right.} \right){\bf{ + Pr}}\left( {{{\bf{D}}_{{\bf{n + 1}}}}\left| {{\bf{D}}_{\bf{n}}^{\bf{c}} \cap {{\bf{B}}^{\bf{c}}}} \right.} \right){\bf{Pr}}\left( {{\bf{D}}_{\bf{n}}^{\bf{c}}\left| {{{\bf{B}}^{\bf{c}}}} \right.} \right)\)鈥�..(a)

By using induction assumption, we can consider,

\(\Pr \left( {{D_n}\left| {{B^c}} \right.} \right) = 0.01\)and\(\Pr \left( {D_n^c\left| {{B^c}} \right.} \right) = 0.99\)

Now from the given information, we can conclude,

\(\Pr \left( {{D_{n + 1}}\left| {{D_n} \cap {B^c}} \right.} \right) = \frac{2}{5}\)and\(\Pr \left( {{D_{n + 1}}\left| {D_n^c \cap {B^c}} \right.} \right) = \frac{1}{{165}}\)

Now substituting the values in equation (a), we get,

\(\begin{aligned}{}\Pr \left( {{D_{n + 1}}\left| {{B^c}} \right.} \right) = \frac{2}{5} \times 0.01 \times \frac{1}{{165}} \times 0.99\\ = 0.01\end{aligned}\)

Thus, by induction, we can say that\(\Pr \left( {{D_i}\left| B \right.} \right)\)that is\(\Pr \left( {{D_i}} \right) = 0.01\)(proved)

In the given event E, out of six items third and fourth item is defective and the other four are non-defective.

Now consider the probability that the event E willoccur given B.

\(\begin{aligned}{}\Pr \left( {E\left| B \right.} \right) &= \Pr \left( {D_1^c \cap D_2^c \cap {D_3} \cap {D_4} \cap D_5^c \cap D_6^c\left| B \right.} \right)\\& = \left[ \begin{aligned}{l}\Pr \left( {D_1^c\left| B \right.} \right) \times \Pr \left( {D_2^c\left| B \right.} \right) \times \Pr \left( {{D_3}\left| B \right.} \right) \times \Pr \left( {{D_4}\left| B \right.} \right) \times \Pr \left( {D_5^c\left| B \right.} \right)\\ \times \Pr \left( {D_6^c\left| B \right.} \right)\end{aligned} \right]\\ &= 0.99 \times 0.99 \times 0.01 \times 0.01 \times 0.99 \times 0.99\\ &= 0.000096\end{aligned}\)

Now consider the probability of event Eoccurring given\({B^c}\),

\(\begin{aligned}{}\Pr \left( {E{{\left| B \right.}^c}} \right) &= \Pr \left( {D_1^c \cap D_2^c \cap {D_3} \cap {D_4} \cap D_5^c \cap D_6^c{{\left| B \right.}^c}} \right)\\ &= \left[ \begin{aligned}{l}\Pr \left( {D_1^c{{\left| B \right.}^c}} \right) \times \Pr \left( {D_2^c\left| {D_1^c \cap {B^c}} \right.} \right) \times \Pr \left( {{D_3}\left| {D_2^c \cap {B^c}} \right.} \right)\\ \times \Pr \left( {{D_4}\left| {{D_3} \cap B} \right.} \right) \times \Pr \left( {D_5^c\left| {{D_4} \cap B} \right.} \right) \times \Pr \left( {D_6^c\left| {D_5^c \cap B} \right.} \right)\end{aligned} \right]\\ &= 0.99 \times \left( {1 - \frac{1}{{165}}} \right) \times \frac{1}{{165}} \times \frac{2}{5} \times \left( {1 - \frac{2}{5}} \right) \times \left( {1 - \frac{1}{{165}}} \right)\\ &= 0.00142\end{aligned}\)

So, our required probability is,

\(\begin{aligned}{}{\bf{Pr}}\left( {{\bf{B}}\left| {\bf{E}} \right.} \right){\bf{ = }}\frac{{{\bf{Pr}}\left( {\bf{B}} \right){\bf{Pr}}\left( {{\bf{E}}\left| {\bf{B}} \right.} \right)}}{{{\bf{Pr}}\left( {\bf{B}} \right){\bf{Pr}}\left( {{\bf{E}}\left| {\bf{B}} \right.} \right){\bf{ + Pr}}\left( {{{\bf{B}}^{\bf{c}}}} \right){\bf{Pr}}\left( {{\bf{E}}{{\left| {\bf{B}} \right.}^{\bf{c}}}} \right)}}\\ = \frac{{\left( {\frac{2}{3}} \right) \times 0.000096}}{{\left( {\left( {\frac{2}{3}} \right) \times 0.000096} \right) + \left( {\left( {1 - \frac{2}{3}} \right) \times 0.00142} \right)}}\\ = 0.1191\end{aligned}\)

Thus, \(\Pr \left( {B\left| E \right.} \right) = 0.1191\).