Question

Confidence of feedback information for improving quality. In the semiconductor manufacturing industry, a key to improved quality is having confidence in the feedback generated by production equipment. A study of the confidence level of feedback information was published in Engineering Applications of Artificial Intelligence(Vol. 26, 2013). At any point in time during the production process, a report can be generated. The report is classified as either “OK” or “not OK.” Let Arepresent the event that an “OK” report is generated in any time period (t).Let Brepresent the event that an “OK” report is generated in the next time period. Consider the following probabilities:

$\mathbf{P}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{8}$,$\mathbf{P}\left(\frac{B}{A}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{9}$, and$\mathbf{P}\left(\frac{B}{{\mathbf{A}}^C}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}$.

a. Express the event B|Ain the words of the problem.

b. Express the event B|${\mathbf{A}}^C$in the words of the problem.

c. Find$\mathbf{P}\mathbf{(}{\mathbf{A}}^C\mathbf{)}$.

d. Find$\mathbf{P}\mathbf{(}\mathbf{A}\cap \mathbf{B}\mathbf{)}$.

e. Find$\mathbf{P}\mathbf{(}{\mathbf{A}}^C\cap \mathbf{B}\mathbf{)}$.

f. Use the probabilities, parts d and e, to find P(B).

g. Use Bayes’ Rule to find P(A|B), i.e., the probability that an “OK” report was generated in one time period(t), given that an “OK” report is generated in the next time period$\left(t+1\right)$.

Short answer

The results are as follows:

1. The event B|A means that OK report id generated in time period$\left(t+1\right)$granted that an OK report is generated time period (t).
2. The event B|${\mathrm{A}}^{\mathrm{C}}$ means that OK report id generated in time period$\left(t+1\right)$granted that not OK report is generated time period (t).
3. The value of compliment is 0.2.
4. The value of$\mathrm{P}\text{(}\mathrm{A}\cap \mathrm{B}\text{)}$ is 0.72.
5. The value of$\mathrm{P}\text{(}{\mathrm{A}}^{\mathrm{C}}\cap \mathrm{B}\text{)}$ is 0.1.
6. The value of (B) is 0.82.
7. The value of P|B is 0.82.

Step-by-step solution

Given information

$\begin{array}{c}\mathrm{P}\left(\mathrm{A}\right)=0.8\\ \mathrm{P}\left(\frac{\mathrm{B}}{\mathrm{A}}\right)=0.9\\ \mathrm{P}\left(\frac{\mathrm{B}}{{\mathrm{A}}^{\mathrm{C}}}\right)=0.5\end{array}$

Let A represent the event that an “OK” report is generated in any time period (t).Let B represent the event that an “OK” report is generated in the next time period .

Express the event B|A in the words of the problem.

A=[in the time period (t) an OK report id generated]

B=[ in the time period$\left(t+1\right)$ an OK report id generated]

So,If an OK report is generated during time period $\left(t+1\right)$, then the event B|A indicates that it was done so (t).

Express the event B|AC in the words of the problem

Thereafter, unless a not-OK report is made during the time period $\left(t+1\right)$, the event B| denotes that an OK report was generated during that time period (t).

Find P(AC)

Apply the complementary rule

$\begin{array}{c}\mathrm{P}\text{(}{\mathrm{A}}^{\mathrm{C}}\text{)}=1-\mathrm{P}\text{(}\mathrm{A}\text{)}\\ =1-0.8\\ =0.2\end{array}$

Accordingly, the value of a compliment is 0.2

Evaluate P(A∩B)

$\begin{array}{c}\mathrm{P}\text{(}\mathrm{A}\cap \mathrm{B}\text{)}=\mathrm{P}\left(\frac{\mathrm{B}}{\mathrm{A}}\right)\times \mathrm{P}\text{(}\mathrm{A}\text{)}\\ =\text{(}0.9\text{)(}0.8\text{)}\\ =0.72\end{array}$

Henceforth, the value of$\mathrm{P}\text{(}\mathrm{A}\cap \mathrm{B}\text{)}$ is 0.72.

Determine P(AC∩B)

$\begin{array}{c}\mathrm{P}\text{(}{\mathrm{A}}^{\mathrm{C}}\cap \mathrm{B}\text{)}=\mathrm{P}\left(\frac{\mathrm{B}}{{\mathrm{A}}^{\mathrm{C}}}\right)\times \mathrm{P}\text{(}{\mathrm{A}}^{\mathrm{c}}\text{)}\\ =\text{(}0.5\text{)(}0.2\text{)}\\ =0.1\end{array}$

So, the value of$\mathrm{P}\text{(}{\mathrm{A}}^{\mathrm{C}}\cap \mathrm{B}\text{)}$ is 0.1.

Find P(B)

$\begin{array}{c}\mathrm{P}\text{(}\mathrm{B}\text{)}=\mathrm{P}\text{(}\mathrm{A}\cap \mathrm{B}\text{)}+\mathrm{P}\text{(}{\mathrm{A}}^{\mathrm{C}}\cap \mathrm{B}\text{)}\\ =0.72+0.1\\ =0.82\end{array}$

Hence, the value of (B) is 0.82.

Find P(A|B)

$\begin{array}{c}\mathrm{P}\left(\mathrm{A}|\mathrm{B}\right)=\frac{\mathrm{P}\text{(}\mathrm{A}\cap \mathrm{B}\text{)}}{\mathrm{P}\text{(}\mathrm{A}\text{)}}\\ =\frac{\mathrm{P}\text{(}\mathrm{A}\text{)}.\mathrm{P}\text{(}\mathrm{B}\text{)}}{\mathrm{P}\text{(}\mathrm{A}\text{)}}\\ =0.82\end{array}$

Therefore, the value of P|B is 0.82.