Chapter 2.4, Problem 61E

Components of a certain type are shipped to a supplier in batches of ten. Suppose that 50% of all such batches contain no defective components, 30% contain one defective component, and 20% contain two defective components. Two components from a batch are randomly selected and tested. What are the probabilities associated with 0, 1, and 2 defective components being in the batch under each of the following conditions?

1. a. Neither tested component is defective.
2. b. One of the two tested components is defective. [Hint: Draw a tree diagram with three first-generation branches for the three different types of batches.]

To determine

Find the probability that none of the tested component is defective.

Answer to Problem 61E

The probability that the ‘0’ defective in batch giventhat ‘0’ defectives in the sample is 0.578.

The probability that the ‘1’ defective in batch given that ‘0’ defectives in the sample is 0.278.

The probability that the ‘2’ defective in batch given that ‘0 defectives in the sample is 0.144.

Explanation of Solution

Given info:

The information is based on shipping the batches of ten components. A sample of two components in the batch is taken randomly to test the defectives in the batch. The probability that there is no defective components, 1 defective component, and 2 defective components are 0.50, 0.30 and 0.20, respectively.

Calculation:

Here, ‘0’ denotes the batch with no defective,‘1’ denotes the batch with one defective, and ‘2’ denotes the batch with two defective.

The probability of the component with no defective is 0.5, the probability of the component with one defective is 0.30 and the probability of the component with two defectives is 0.20.

Bayes’ rule:

If $A_1,A_2\text{,}\mathrm{...}\text{,}{A}_k$ are k mutually exclusive and exhaustive events with prior probabilities, such that the sum of the probability values of the events is 1 and there is an observed event B, then,

$\begin{array}{c}P\left(A_j|B\right)=\frac{P\left(A_j\cap B\right)}{P\left(B\right)}\\ =\frac{P\left(A_j\right)P\left(B|A_j\right)}{P\left(A_1\right)P\left(B|A_1\right)+P\left(A_2\right)P\left(B|A_2\right)+\mathrm{...}+P\left(A_k\right)P\left(B|A_k\right)}\end{array}$

Condition 1: When there is zero defective in the batch

If there are no defectives in the batch, then there will be no defective in the sample.

The probability of finding no defective in the sample given that no defective in the batch is,

$P\left(\begin{array}{l}\text{'}\text{0'defective in the sample}|\\ \text{'}\text{0'defective in batch}\end{array}\right)=1$.

Condition 2: When there is one defective in the batch

The probability of one defective found inthe sample given that one defective found in the batch is,

$\begin{array}{c}P\left(\begin{array}{l}\text{'}1\text{'defective in the sample}|\\ \text{'}1\text{'defective in batch}\end{array}\right)=\frac{2}{10}\\ =0.2\end{array}$

The chance that no defective found in the samplegiven that one defective found in the batch is,

$\begin{array}{c}P\left(\begin{array}{l}\text{'}0\text{' defective in the sample}|\\ \text{'}1\text{' defective in batch}\end{array}\right)=\frac{8}{10}\\ =0.8\end{array}$

Condition 3: When there are two defectives in the batch

Theprobability of two detectivesfound in the samplegiven that two defective found in the batch is,

$\begin{array}{c}P\left(\begin{array}{l}\text{'}2\text{' defectives in the sample}|\\ \text{'}2\text{' defectives in batch}\end{array}\right)=\frac{2}{10}\times \frac{1}{9}\\ =0.022\end{array}$

Theprobability of no defectivefound in the samplegiven that two defective found in the batch is,

$\begin{array}{c}P\left(\begin{array}{l}\text{'}0\text{' defectives in the sample}|\\ \text{'}2\text{' defectives in batch}\end{array}\right)=\frac{8}{10}\times \frac{7}{9}\\ =0.622\end{array}$

The probability of finding the one defective in the sample given that two defective found in the batch is,

$\begin{array}{c}P\left(\begin{array}{l}\text{'}1\text{' defective in the sample}|\\ \text{'}1\text{' defective in batch}\end{array}\right)=1-\left[P\left(\begin{array}{l}\text{'}2\text{'defective in the }\\ \text{sample}|\\ \text{'}2\text{'defective in batch}\end{array}\right)+P\left(\begin{array}{l}\text{'}0\text{' defective in the }\\ \text{sample}|\\ \text{'}2\text{' defective in batch}\end{array}\right)\right]\\ =1-\left[0.022+0.622\right]\\ =0.356\end{array}$

The probability of ‘0’ defective in the batch and ‘0’ defective in the sample is obtained as given below:

$\begin{array}{c}P\left(\begin{array}{l}\text{'0' defective in the batch}\\ \text{and '0' defective in the sample}\end{array}\right)=P\left(\text{'}0\text{'defective in batch}\right)\times P\left(\begin{array}{l}\text{'}0\text{'defective in the}\\ \text{ sample}|\\ \text{'}0\text{'defective in batch}\end{array}\right)\\ =0.5\times 1\\ =0.50\end{array}$

The probability of ‘1’ defective in the batch and ‘0’ defective in the sample is obtained as given below:

$\begin{array}{c}P\left(\begin{array}{l}\text{'1' defective in the batch}\\ \text{and '0' defective in the sample}\end{array}\right)=P\left(\text{'}1\text{'defective in batch}\right)\times P\left(\begin{array}{l}\text{'}0\text{'defective in the }\\ \text{sample}|\\ \text{'}1\text{'defective in batch}\end{array}\right)\\ =0.3\times 0.8\\ =0.24\end{array}$

The probability of the ‘1’ defective in the batch and ‘1’ defective in the sample is obtained as given below:

$\begin{array}{c}P\left(\begin{array}{l}\text{'1' defective in the batch}\\ \text{and '1' defective}\\ \text{in the sample}\end{array}\right)=P\left(\text{'}1\text{' defective in batch}\right)\times P\left(\begin{array}{l}\text{'}1\text{' defective in the sample}|\\ \text{'}1\text{' defective in batch}\end{array}\right)\\ =0.3\times 0.2\\ =0.06\end{array}$

The probability of the ‘2’ defectives in the batch and ‘0’ defective in the sample is obtained as given below:

$\begin{array}{c}P\left(\begin{array}{l}\text{'2' defectives in the batch}\\ \text{and '0' defective}\\ \text{in the sample}\end{array}\right)=P\left(\text{'}2\text{' defectives in batch}\right)\times P\left(\begin{array}{l}\text{'}0\text{'defective in the sample}|\\ \text{'}2\text{' defectives in batch}\end{array}\right)\\ =0.2\times 0.622\\ =0.1244\end{array}$

The probability of the ‘2’ defective in the batch and ‘1’ defective in the sample is obtained as given below:

$\begin{array}{c}P\left(\begin{array}{l}\text{'2' defectives in the batch}\\ \text{and '1' defective}\\ \text{ in the sample}\end{array}\right)=P\left(\text{'}2\text{' defectives in batch}\right)\times P\left(\begin{array}{l}\text{'}1\text{' defective in the sample}|\\ \text{'}2\text{' defectives in batch}\end{array}\right)\\ =0.2\times 0.356\\ =0.0712\end{array}$

The probability of ‘2’defectives in the batch and‘2’ defective in the sample is obtained as given below:

$\begin{array}{c}P\left(\begin{array}{l}\text{'2' defectives in the batch}\\ \text{and '2' defectives}\\ \text{in the sample}\end{array}\right)=P\left(\text{'}2\text{'defective in batch}\right)\times P\left(\begin{array}{l}\text{'}2\text{'defective in the sample}|\\ \text{'}2\text{'defective in batch}\end{array}\right)\\ =0.2\times 0.022\\ =0.0044\end{array}$

The probability that the ‘0’ defective in batch given that the ‘0 defective found in the sample is obtained as given below:

$\begin{array}{c}P\left(\begin{array}{l}\text{'0' defective in the batch |}\\ \text{'0' defective in the sample }\end{array}\right)=\frac{P\left(\begin{array}{l}\text{'0' defective in the batch }\cap \\ \text{'0' defective in the sample}\end{array}\right)}{P\left(\text{'0'defective in the sample}\right)}\\ =\frac{0.5}{0.5+0.24+0.1244}\\ =\frac{0.5}{0.8644}\\ =0.578\end{array}$

The probability that the ‘1’ defective in batch given that ‘0’ defective in the sample is obtained as:

$\begin{array}{c}P\left(\begin{array}{l}\text{'1' defective in the batch |}\\ \text{'0' defective in the sample }\end{array}\right)=\frac{P\left(\begin{array}{l}\text{'1' defective in the batch }\cap \\ \text{'0' defective in the sample}\end{array}\right)}{P\left(\text{'0'defective in the sample}\right)}\\ =\frac{0.24}{0.5+0.24+0.1244}\\ =\frac{0.24}{0.8644}\\ =0.278\end{array}$

The probability that the ‘2’ defective in batch given that there is ‘0’ defective in the sample is obtained as:

$\begin{array}{c}P\left(\begin{array}{l}\text{'2' defectives in the batch |}\\ \text{'0' defective in the sample }\end{array}\right)=\frac{P\left(\begin{array}{l}\text{'2' defectives in the batch }\cap \\ \text{ '0' defective in the sample}\end{array}\right)}{P\left(\text{'0'defective in the sample}\right)}\\ =\frac{0.1244}{0.5+0.24+0.1244}\\ =\frac{0.1244}{0.8644}\\ =0.144\end{array}$

Thus, the probabilities of the components tested with no defectives are 0.578, 0.278 and 0.144.

To determine

Find the probability that the tested components have one defective.

Construct a tree diagram with three first-generation branches.

Answer to Problem 61E

The probability that ‘0’ defective in batch given that ‘1’ defective in the sample is 0.

The probability that ‘1’ defective in batch given that ‘1’ defective in the sample is0.457.

The probability that ‘2’ defectives in batch given that ‘1’ defective in the sample is 0.543.

Explanation of Solution

Calculation:

Using the probabilities obtained in part (a) the tree diagram is obtained as given below:

The probability that the ‘0’ defective in batch given that‘1’ defective in the sample is obtained as:

$\begin{array}{c}P\left(\begin{array}{l}\text{'0' defective in the batch |}\\ \text{'1' defective in the sample }\end{array}\right)=\frac{P\left(\text{'0' defective in the batch }\cap \text{ '1' defective in the sample}\right)}{P\left(\text{'1'defective in the sample}\right)}\\ =0\end{array}$

The probability that the ‘1’ defective in batch given that ‘1’ defective in the sample is obtained as:

$\begin{array}{c}P\left(\begin{array}{l}\text{'1' defective in the batch |}\\ \text{'1' defective in the sample }\end{array}\right)=\frac{P\left(\text{'1' defective in the batch }\cap \text{ '1' defective in the sample}\right)}{P\left(\text{'1'defective in the sample}\right)}\\ =\frac{0.06}{0.06+0.0712}\\ =\frac{0.06}{0.1312}\\ =0.457\end{array}$

The probability that the ‘2’ defective in batch given that‘1’ defective in the sample is obtained as:

$\begin{array}{c}P\left(\begin{array}{l}\text{'2' defective in the batch |}\\ \text{'1' defective in the sample }\end{array}\right)=\frac{P\left(\text{'2' defective in the batch }\cap \text{ '1' defective in the sample}\right)}{P\left(\text{'1'defective in the sample}\right)}\\ =\frac{0.0712}{0.06+0.0712}\\ =\frac{0.0712}{0.1312}\\ =0.543\end{array}$

Thus, the probabilities of the components tested with one defective are0, 0.457 and 0.543.