A factory has a total of three manufacturing units, M1, M2, and M3, which produce bulbs independent of each other. The units M1, M2, and M3 produce bulbs in the proportions of 2:2:1, respectively. It is known that 20\% of the bulbs produced in the factory are defective. It is also known that, of all the bulbs produced by M1, 15\% are defective. Suppose that, if a randomly chosen bulb produced in the factory is found to be defective, the probability that it was produced by M2 is 25. If a bulb is chosen randomly from the bulbs produced by M3, then the probability that it is defective is .
Answer
0.27 to 0.33
Solution
P(M1)=P(M2)= 25, P(M3)= 15; let p2=P(D|M2), p3=P(D|M3).
Bayes: P(M2|D)= 25p20.2= 25 p2=0.2.
Total probability: 0.2= 25(0.15)+ 25(0.2)+ 15p3=0.06+0.08+ p35.
p3=0.3
Answer: 0.3