Defective product from Manufacturing- Probability 2

Question

In a factory, only two machines, A and B, manufacture washers. Neither machine is perfect: machine A produces defective washers 18% of the time, while machine B produces defectives 14% of the time. Machine B is more efficient than machine A and accounts for 70% of the total output of washers. For purposes of quality control, a sample of washers is taken and examined for defectives. Compute the probability that a randomly chosen washer found to be defective was manufactured by machine A. Round your answer to two decimal places.

P(A)=0.3 P(B)=0.7 D=Defective washers P(A|D)=0.18 P(B|D)=0.14

P(D)= P(D∩A)+P(D∩B)

=P(A)*P(D|A)+P(B)*P(D|B)
=0.3*0.18+0.7*0.14
=0.152

P(A|D)=P(A∩D)/P(D)

  =0.054/0.152
  =0.36 (Correct to 2 decimal place) 

Is the calculation correct?

Answer 1

P(A)=0.3 P(B)=0.7 D=Defective washers P(A|D)=0.18 P(B|D)=0.14 P(D)= P(D∩A)+P(D∩B)

=P(A)*P(D|A)+P(B)*P(D|B) =0.3*0.18+0.7*0.14 =0.152 P(A|D)=P(A∩D)/P(D)

=0.054/0.152 =0.36 (Correct to 2 decimal place)

Answer 2

The calculation is correct. It could be also done using Bayes Theorem:

• $P(D) = 0.152$ (as you've already computed)
• $P(A)=0.3$
• $P(D|A)=0.18$

$$P(A|D)=\frac{P(D|A)P(A)}{P(D)} = \frac{0.18\cdot 0.3}{0.152}\approx 0.36$$