Chapter 5.3, Problem 42E

Quality control: A population of 600 semiconductor wafers contains wafers from three lots. The wafers are categorized by lot and by whether they conform to a thickness specification: with die results shown in the following table. A wafer is chosen at random from the population.

What is the probability that a wafer is from Lot A?

What is the probability that a wafer is conforming?

What is the probability that a wafer is from Lot A and is conforming?

Given that the wafer is from Lot A, what is the probability that it is conforming?

Given that the wafer is conforming, what is the probability that it is from Lot A?

Let $E_1$ be the event that the wafer comes from Lot A, and let $E_2$ be the event that the wafer is conforming. Are $E_1$ and $E_2$ independent? Explain

To determine

The probability that randomly chosen wafer belongs to Lot A.

Answer to Problem 42E

  $0.1667$

Explanation of Solution

Given:

The following table categorizes 600 wafer.

Lot	Conforming	Non Conforming
A	88	12
B	165	35
C	260	40

Formula Used:

Probability of an event is given by

  $\frac{\text{Number}\text{of}\text{Favorable}\text{Outcome}}{\text{Number}\text{of}\text{total}\text{outcome}}$

Calculation:

Let $A$ be the event that selectedwafer is from Lot A.

  $\begin{array}{c}n\left(A\right)=88+12=100\\ n\left(S\right)=600\\ P\left(A\right)=\frac{100}{600}\\ =0.1667\end{array}$

Hence, the probability of the event that randomly chosen wafer belongs to Lot A,is $0.1667$ .

To determine

The probability that randomly chosen wafer is conforming.

Answer to Problem 42E

  $0.52$

Explanation of Solution

Given:

The following table categorizes 600 wafer.

Lot	Conforming	Non Conforming
A	88	12
B	165	35
C	260	40

Formula Used:

Probability of an event is given by

  $\frac{\text{Number}\text{of}\text{Favorable}\text{Outcome}}{\text{Number}\text{of}\text{total}\text{outcome}}$

Calculation:

Let $X$ be the event that chosen wafer is conforming.

  $\begin{array}{c}n\left(X\right)=88+165+260=513\\ n\left(S\right)=600\\ P\left(X\right)=\frac{513}{600}\\ =0.855\end{array}$

Hence, the probability of the event that randomly chosen wafer is conforming, is $0.855$ .

To determine

The probability that the wafer is from Lot A and conforming.

Answer to Problem 42E

  $0.1467$

Explanation of Solution

Given:

The following table categorizes 600 wafer.

Lot	Conforming	Non Conforming
A	88	12
B	165	35
C	260	40

Formula Used:

For any two t events $A$ and B , probability of occurring both the events is

  $\begin{array}{c}P\left(A\text{and}B\right)=P\left(A\cap \text{}B\right)\\ =\frac{n\left(A\cap \text{}B\right)}{n\left(S\right)}\end{array}$

Calculation:

Let $A$ be the event that selected wafer is from Lot A.

Let $X$ be the event that chosen wafer is conforming.

So, $A\text{and}X$ be the event that selected wafer is from Lot A and conforming.

There are 88wafer from Lot A who are conforming.

  $\begin{array}{c}P\left(A\cap \text{}X\right)=\frac{n\left(A\cap \text{}X\right)}{n\left(S\right)}\\ =\frac{88}{600}\\ =0.1467\end{array}$

The probability that the wafer is from Lot A and conforming, is $0.1467$ .

To determine

The probability of conforming wafer, given wafer belongs toLot A.

Answer to Problem 42E

  $0.88$

Explanation of Solution

Given:

The following table categorizes 600 wafer.

Lot	Conforming	Non Conforming
A	88	12
B	165	35
C	260	40

Formula Used:

For any two dependent events $A$ and B ,

  $P\left(B/A\right)=\frac{P\left(A\text{and}B\right)}{P\left(A\right)}$

Where

  $P\left(B/A\right)=$ The probability of occurring $B$ under the assumption that $A$ has been occurred.

Calculation:

Let $A$ be the event that selected wafer is from Lot A.

  $\begin{array}{c}n\left(A\right)=88+12=100\\ n\left(S\right)=600\\ P\left(A\right)=\frac{100}{600}\\ =0.1667\end{array}$

Let $X$ be the event that chosen wafer is conforming.

So, $A\text{and}X$ be the event that selected wafer is from Lot A and conforming.

There are 88 wafer from Lot A who are conforming.

  $\begin{array}{c}P\left(A\cap \text{}X\right)=\frac{n\left(A\cap \text{}X\right)}{n\left(S\right)}\\ =\frac{88}{600}\\ =0.1467\end{array}$

Therefore, $X/A$ be the event that a senator of being republican, given that she is wo conforming wafer, given wafer belongs to Lot A.

So, the probability of the event “conforming wafer, given wafer belongs to Lot A” is calculated as follows:

  $\begin{array}{c}P\left(X/A\right)=\frac{P\left(X\text{and}A\right)}{P\left(A\right)}\\ =\frac{88}{100}\\ =0.88\end{array}$

The probability of conforming wafer, given wafer belongs to Lot A, is $0.88$ .

To determine

The probability to a waferbelongs to Lot A, given that wafer is conforming.

Answer to Problem 42E

  $0.1715$

Explanation of Solution

Given:

The following table categorizes 600 wafer.

Lot	Conforming	Non Conforming
A	88	12
B	165	35
C	260	40

Formula Used:

For any two dependent events $A$ and B ,

  $P\left(B/A\right)=\frac{P\left(A\text{and}B\right)}{P\left(A\right)}$

Where

  $P\left(B/A\right)=$ The probability of occurring $B$ under the assumption that $A$ has been occurred.

Calculation:

Let $X$ be the event that chosen wafer is conforming.

  $\begin{array}{c}n\left(X\right)=88+165+260=513\\ n\left(S\right)=600\\ P\left(X\right)=\frac{513}{600}\\ =0.855\end{array}$

Let $A$ be the event that selected wafer is from Lot A.

So, $A\text{and}X$ be the event that selected wafer is from Lot A and conforming.

There are 88 wafer from Lot A who are conforming.

  $\begin{array}{c}P\left(A\cap \text{}X\right)=\frac{n\left(A\cap \text{}X\right)}{n\left(S\right)}\\ =\frac{88}{600}\\ =0.1467\end{array}$

Therefore, $A/X$ be the event that a wafer belongs to Lot A, given that wafer is conforming.

So, the probability of the event “a wafer belongs to Lot A, given that wafer is conforming” is calculated as follows:

  $\begin{array}{c}P\left(A/X\right)=\frac{P\left(A\text{and}X\right)}{P\left(X\right)}\\ =\frac{88}{513}\\ =0.1715\end{array}$

The probability to a wafer belongs to Lot A, given that wafer is conforming, is $0.1715$ .

To determine

To explain: Whether the events $E_1$ and $E_2$ are independent.

Answer to Problem 42E

No, the events $E_1$ and $E_2$ are not independent.

Explanation of Solution

Given:

The following table categorizes 600 wafer.

Lot	Conforming	Non Conforming
A	88	12
B	165	35
C	260	40

Formula Used:

For any two independent events $A$ and B ,

  $P\left(A\text{and}\text{}B\right)=P\left(A\right)\cdot P\left(B\right)$

Calculation:

Let $E_2$ be the event that chosen wafer is conforming.

  $\begin{array}{c}n\left(E_2\right)=88+165+260=513\\ n\left(S\right)=600\\ P\left(E_2\right)=\frac{513}{600}\\ =0.855\end{array}$

Let $E_1$ be the event that selected wafer is from Lot A.

  $\begin{array}{c}n\left(E_1\right)=88+12=100\\ n\left(S\right)=600\\ P\left(E_1\right)=\frac{100}{600}\\ =0.1667\end{array}$

So, $E_1\text{and}{E}_2$ be the event that selected wafer is from Lot A and conforming.

There are 88 wafer from Lot A who are conforming.

  $\begin{array}{c}P\left(E_1\text{and}{E}_2\right)=\frac{n\left(E_1\text{and}{E}_2\right)}{n\left(S\right)}\\ =\frac{88}{600}\\ =0.1467\end{array}$

Now $P\left(E_1\right)\cdot P\left(E_2\right)$ can be calculated as

  $\begin{array}{c}P\left(E_1\right)\cdot P\left(E_2\right)=0.1667\times 0.855\\ =0.1425\\ \ne P\left(E_1\text{and}{E}_2\right)\end{array}$

Therefore, the two events $E_1\text{and}{E}_2$ not satisfying the condition of being independent events.

Hence, the events $E_1$ and $E_2$ are not independent.