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. Suppose we randomly select a part produced at the factory.

a. Find the probability that the part is defective.

b. If the part is inspected and found to be defective, what’s the probability that it was

produced by Machine B?

We have been given that Machine A produces $60\%$of the parts, while Machines B and C produce $30\%$and $10\%$of the parts, respectively while $10\%$of the parts produced by Machine A are defective, compared with $30\%$for Machine B and $40\%$for Machine C.

We need to find out the probability that the part is defective.

Let A$=$Machine A, B$=$Machine B, C$=$Machine C, D$=$Defective.

$$\begin{gathered} P\left(A\right)=60\%=0.60 \\ P\left(B\right)=30\%=0.30 \\ P\left(C\right)=10\%=0.10 \\ P\left(D\overline{)A}\right)=10\%=0.10 \\ P\left(D\overline{)B}\right)=30\%=0.30 \\ P\left(D\overline{)C}\right)=40\%=0.40 \\ P\left(AandD\right)=P\left(A\right)\times P\left(D\overline{)A}\right)=0.60\times 0.10=0.06 \\ P\left(BandD\right)=P\left(B\right)\times P\left(D\overline{)B}\right)=0.30\times 0.30=0.09 \\ P\left(CandD\right)=P\left(C\right)\times P\left(D\overline{)C}\right)=0.10\times 0.40=0.04 \\ u\sin gtheadditionruleformutuallyexclusiveevents \\ P\left(D\right)=P\left(AandD\right)+P\left(BandD\right)+P\left(CandD\right) \\ =0.06+0.09+0.04=0.19 \\ =19\% \end{gathered}$$

We have been given that Machine A produces $60\%$of the parts, while Machines B and C produce $30\%$and $10\%$of the parts, respectively while$10\%$of the parts produced by Machine A are defective, compared with $30\%$ for Machine B and $40\%$ for Machine C.

We need to find out the probability that a part is inspected and found to be defective is produced by machine B.

Using the concept of conditional probability

$$\begin{gathered} P\left(BandD\right)=0.09 \\ P\left(D\right)=0.19 \\ P\left(B\overline{)D}\right)=\frac{P\left(BandD\right)}{P\left(D\right)} \\ =\frac{0.09}{0.19} \\ =\frac{9}{19} \\ \approx 0.4737 \\ =47.37\% \end{gathered}$$