91影视

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.

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

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

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