Question

Three machines—A, B, and C—are used to produce a large quantity of identical parts at a factory. Machine A produces $60\%$ of the parts, while Machines B and C produce $30\%$ and $10\%$ of the parts, respectively. Historical records indicate that $10\%$ of the parts produced by Machine A are defective, compared with $30\%$ for Machine B and $40\%$ for Machine C.

(a) Draw a tree diagram to represent this chance process.

(b) If we choose a part produced by one of these three machines, what’s the probability that it’s defective? Show your work.

(c) If a part is inspected and found to be defective, which machine is most likely to have produced it? Give appropriate evidence to support your answer.

Short answer

Part (b) The probability is $0.19$

Part (c) Machine B.

Part (a) The tree diagram is

Step-by-step solution

Part (a) Step 1. Given

Machine A produces $60\%$ of the products, which equals $0.60$

Machine B produces $30\%$ of the products, which is $0.30$

The percentage of goods generated by machine C is $10\%=0.10$ while the percentage of items produced by machine A is $10\%=0.10$

The percentage of defective goods generated by machine B is $30\%$, or $0.30$

The percentage of defective goods generated by machine C is $40\%$, or $0.40$

 Part (a) Step 2. Concept used 

The Bayes theorem $P\left(A\right|B)=\frac{P(A\cap B)}{P\left(B\right)}$

Part (a) Step 3. Calculation

Let A signify the occurrence of a machine-made object. A. B denotes the occurrence of a machine-made item. The events of a machine-made item are denoted by the letters B. C. Defective items are denoted by the letters C. D. If an item is not faulty, the code ND is used. The following is a tree diagram with the accompanying probabilities:

Part (b) Step 1. Calculation

As stated in the sub part a,

$$\begin{gathered} P\left(A\right)=0.60 \\ P\left(B\right)=0.30 \\ P\left(C\right)=0.10 \end{gathered}$$

The conditional probabilities are

$$\begin{gathered} P\left(D\right|A)=0.10 \\ P(D\left|B\right)=0.30 \\ P\left(D\right|C)=0.40 \end{gathered}$$

Using the formula above,

$$\begin{gathered} P(D\cap A)=0.60\times 0.10=0.06 \\ P(D\cap B)=0.30\times 0.30=0.09 \\ P(D\cap C)=0.10\times 0.40=0.04 \end{gathered}$$

Therefore,

$$\begin{gathered} P\left(D\right)=0.06+0.09+0.04 \\ P\left(D\right)=0.19 \end{gathered}$$

Thus, the required probability is$0.19$

 Part (c) Step 1. Calculation

Part b of the subparts

$P\left(D\right)=0.19$

The likelihood that machine A produced the damaged item selected.

$$\begin{gathered} P\left(A\right|D)=P(A\cap D)/P(D) \\ P(A\left|D\right)=0.06/0.19 \\ =0.3158 \end{gathered}$$$$\begin{gathered} P\left(B\right|D)=\frac{P(B\cap D)}{P\left(D\right)} \\ P(B\left|D\right)=0.09/0.19=0.4737 \end{gathered}$$$$\begin{gathered} P\left(C\right|D)=\frac{P(C\cap D)}{P\left(D\right)} \\ P(AC\left|D\right)=\frac{0.04}{0.19} \\ =0.2105 \end{gathered}$$

The likelihood that machine B produced the damaged item selected.