Question #239808

A part is randomly selected during a quality assurance test and is found to be defective. Calculate the probability that it was supplied by supplier B


1
Expert's answer
2021-09-22T23:01:29-0400

Total problem

Quality assurance is an important aspect of any business. This case study is based on Lyman Limitad, a company that manufactures and sells office furniture. They obtair parts (hinges, chair wheels, ets.) from three different suppliers, A, B and C. Past experience has shown that 20% of the parts are supplied by A, 50% are supplied by B and the rest are supplied by C. It is also known that 10% of the parts supplied by A are defect, 8% of the parts from B are defect and 6% of the parts from C are defect.

Question 1

A part is mly selected during a quality assurance test and is found to be defective. Calculate the probability that it was supplier B.

A= parts from A

P(A) = 20 % = 0.2

B= parts from B

P(B) = 50% = 0.5

C= Parts from C

P(C) = (1-P(A)-P(B)) = 1 -0.2 -0.5 =0.3

P(came from A and defective) = 0.1

P(came from B and defective) = 0.08

P(came from C and defective) = 0.06

P(part was supplied by supplier B) =P(B)×P(came  from  B  and  defective)P(A)×P(came  from  A  and  defective)+P(B)×P(came  from  B  and  defective)+P(C)×P(came  from  C  and  defective)= \frac{P(B) \times P(came \;from\; B \;and\; defective)}{P(A) \times P(came\; from \;A \;and\; defective) + P(B) \times P(came \;from\; B \;and\; defective) + P(C) \times P(came \;from\; C \;and\; defective)}

=0.5×0.080.2×0.1+0.5×0.08+0.3×0.06=0.5128= \frac{0.5 \times 0.08}{0.2 \times 0.1 + 0.5 \times 0.08 + 0.3 \times 0.06} \\ = 0.5128


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