Question

Consider a machine that produces items in sequence. Under normal operating conditions, the items are independent with a probability of 0.01 of being defective. However, it is possible for the machine to develop a “memory” in the following sense: After each defective item, and independent of anything that happened earlier, the probability that the next item is defective is\(\frac{{\bf{2}}}{{\bf{5}}}\). After each non-defective item, and independent of anything that happened earlier, the probability that the next item is defective is\(\frac{{\bf{1}}}{{{\bf{165}}}}\).

Please assume that the machine is operating normally for the whole time we observe or has a memory for the whole time we observe. LetBbe the event that the machine is operating normally, and assume that\({\bf{Pr}}\left( {\bf{B}} \right){\bf{ = }}\frac{{\bf{2}}}{{\bf{3}}}\). Let\({{\bf{D}}_{\bf{i}}}\)be the event that theith item inspected is defective. Assume that\({{\bf{D}}_{\bf{1}}}\)is independent ofB.

a. Prove that\({\bf{Pr}}\left( {{{\bf{D}}_{\bf{i}}}} \right){\bf{ = 0}}{\bf{.01}}\) for alli. Hint:Use induction.

b. Assume that we observe the first six items and the event that occurs is\({\bf{E = D}}_{\bf{1}}^{\bf{c}} \cap {\bf{D}}_{\bf{2}}^{\bf{c}} \cap {{\bf{D}}_3} \cap {{\bf{D}}_4} \cap {\bf{D}}_{\bf{5}}^{\bf{c}} \cap {\bf{D}}_{\bf{6}}^{\bf{c}}\). The third and fourth items are defective, but the other four are not. Compute\({\bf{Pr}}\left( {{\bf{B}}\left| {\bf{E}} \right.} \right)\).

Short answer

1. \(\Pr \left( {{D_i}} \right) = 0.01\)is proved.
2. \(\Pr \left( {B\left| E \right.} \right) = 0.1191\)

Step-by-step solution

Given information

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 \(\frac{2}{3}\)

State the events

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

(a) Proof with induction

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,

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

(b) Calculate the probability

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