Chapter 4.3, Problem 42E

a.

To determine

Compute the probability that a wafer is from lot A.

Answer to Problem 42E

The probability that a wafer is from lot A is 0.1667.

Explanation of Solution

Calculation:

There were a total of 600 semiconductor wafers. There are three lots and each of them is either conforming or nonconforming. The table provides number corresponding to each category.

The probability of the event A can be obtained by the formula:

$P\left(A\right)=\frac{\text{Number of outcomes in }A}{\text{Number of outcomes in}\text{sample space}}$

Here, event A denotes that wafer is from lot A. Event A contains ‘lot A and conforming’ and ‘lot A and nonconforming’.

$\begin{array}{c}\text{Number of wafers from lot A}=88+12\\ =100\end{array}$

Among the 600 wafers, 100 of them are from lot A.

Substitute 100 for “number of outcomes in A” and 600 for “Number of outcomes in sample space” in the probability formula.

Therefore,

$\begin{array}{c}P\left(A\right)=\frac{100}{600}\\ =0.1667\end{array}$

Thus, the probability that a wafer is from lot A is 0.1667.

b.

To determine

Compute the probability that a wafer is conforming.

Answer to Problem 42E

The probability that a wafer is conforming is 0.8550.

Explanation of Solution

Calculation:

Here, event B denotes that selected wafer is conforming. Event B contains ‘lot A and conforming’, ‘lot B and conforming’ and ‘lot C and conforming’.

$\begin{array}{c}\text{Number of conforming wafers}=88+165+260\\ =513\end{array}$

Among the 600 wafers, 513 of them are conforming.

Substitute 513 for “number of outcomes in B” and 600 for “Number of outcomes in sample space” in the probability formula.

Therefore,

$\begin{array}{c}P\left(B\right)=\frac{513}{600}\\ =0.855\end{array}$

Thus, the probability that a wafer is conforming is 0.8550.

c.

To determine

Compute the probability that a wafer is from lot A and is conforming.

Answer to Problem 42E

The probability that a wafer is from lot A and is conforming is 0.1467.

Explanation of Solution

Calculation:

Here, event ‘A and B’ denotes that selected wafer is from lot A and is conforming.

Among the 600 wafers, 88 of them are conforming.

Substitute 88 for “number of outcomes in (A and B) ” and 600 for “Number of outcomes in sample space” in the probability formula.

Therefore,

$\begin{array}{c}P\left(A\text{and }B\right)=\frac{88}{600}\\ =0.1467\end{array}$

Thus, the probability that a wafer is from lot A and is conforming is 0.1467.

d.

To determine

Compute the probability that wafer is conforming given that it is from lot A.

Answer to Problem 42E

The probability that wafer is conforming given that it is from lot A is 0.88.

Explanation of Solution

Calculation:

The formula for conditional probability is given by:

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

Event A denotes that wafer is from lot A and event B denotes that wafer is conforming.

From part (a) it is clear that, $P\left(A\right)=0.1667$.

From part (c), it is clear that, $P\left(A\text{and}B\right)=0.1467$.

Substitute these values in the formula for conditional probability.

Therefore,

$\begin{array}{c}P\left(B|A\right)=\frac{0.1667}{0.1467}\\ =0.88\end{array}$

Thus, the probability that wafer is conforming given that it is from lot A is 0.88.

e.

To determine

Compute the probability that wafer is from lot A given that it is conforming.

Answer to Problem 42E

The probability that wafer is from lot A given that it is conforming is 0.1715.

Explanation of Solution

Calculation:

The formula for conditional probability is given by:

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

Event A denotes that wafer is from lot A and event B denotes that wafer is conforming.

From part (b) it is clear that, $P\left(B\right)=0.855$.

From part (c), it is clear that, $P\left(A\text{and}B\right)=0.1467$.

Substitute these values in the formula for conditional probability.

Therefore,

$\begin{array}{c}P\left(A|B\right)=\frac{0.1467}{0.855}\\ =0.1715\end{array}$

Thus, the probability that wafer is from lot A given that it is conforming is 0.1715.

f.

To determine

Check whether the event $E_1\text{and }{E}_2$ are independent or not.

Explain the answer.

Answer to Problem 42E

Event $E_1\text{and }{E}_2$ are not independent.

Explanation of Solution

Calculation:

Event $E_1$ denotes that wafer is from lot A and $E_2$ wafer is conforming.

Independent events:

Two events $E_1$ and $E_2$ are said to be independent if occurrence of one does not affect the other. If the events $E_1$ and $E_2$ are independent then $P\left(E_1\text{and}{E}_2\right)=P\left(E_1\right)\times P\left(E_2\right)$.

Here, event $E_1$ denotes that wafer is from lot A which is same as event A in part (a) and event $E_2$ denotes that wafer is conforming which is same as event B in part (b), and event ‘$E_1\text{and}{E}_2$’ denotes that wafer is from lot A and is conforming which is same as event ‘A and B’ in part (c).

From part (a) it is clear that,

$\begin{array}{c}P\left(E_1\right)=P\left(A\right)\\ =0.1667\end{array}$.

From part (b) it is clear that,

$\begin{array}{c}P\left(E_2\right)=P\left(B\right)\\ =0.855\end{array}$.

From part (c) it is clear that,

$\begin{array}{c}P\left(E_1\text{and }{E}_2\right)=P\left(A\text{and}B\right)\\ =0.1467\end{array}$.

Substitute these values in the formula for conditional probability.

Therefore,

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

That is, $P\left(E_1\text{and}{E}_2\right)\ne P\left(E_1\right)\times P\left(E_2\right)$

Thus, the events $E_1\text{and }{E}_2$ are not independent.