Consider a firm that has two assembly lines, 1 and 2, both producing calculator. Assume that you have purchased a calculator and it turns out to be defective. And the line 1 produces 60% of all calculators produced.
L1: event that the calculator is produced on line 1
L2: event that the calculator is produced on line 2
Suppose that your are given the conditional information:
10% of the calculator produced on line 1 is defective
20% of the calculator produced on line 2 is defective
Q: If we choose one defective, what is the probability that the defective calculator comes from Line 1 and Line2?
L1 = event that the calculator is produced on line 1 = 0.6
L2 = event that the calculator is produced on line 2 = 1 - 0.6 = 0.4
D = Defective
D|L1 Defective from Line 1 = 0.1
D|L2 = Defective from Line 2 = 0.20
Defective from Line 1
P(L1|D) = P(L1)P(D/L1) / [ P(L1)P(D/L1) + P(L2)P(D/L2)]
P(L1|D) = (.60)(.10) /[(.60)(.10)+ (.40)(.20)]
P(L1|D) = 0.4286
Defective from Line 2
P(L2|D) = P(L2)P(D/L2) / [ P(L1)P(D/L1) + P(L2)P(D/L2)]
P(L2|D) = (.40)(.20) /[(.60)(.10)+ (.40)(.20)]
P(L2|D) = 0.5714
L1: event that the calculator is produced on line 1
L2: event that the calculator is produced on line 2
Suppose that your are given the conditional information:
10% of the calculator produced on line 1 is defective
20% of the calculator produced on line 2 is defective
Q: If we choose one defective, what is the probability that the defective calculator comes from Line 1 and Line2?
L1 = event that the calculator is produced on line 1 = 0.6
L2 = event that the calculator is produced on line 2 = 1 - 0.6 = 0.4
D = Defective
D|L1 Defective from Line 1 = 0.1
D|L2 = Defective from Line 2 = 0.20
Defective from Line 1
P(L1|D) = P(L1)P(D/L1) / [ P(L1)P(D/L1) + P(L2)P(D/L2)]
P(L1|D) = (.60)(.10) /[(.60)(.10)+ (.40)(.20)]
P(L1|D) = 0.4286
Defective from Line 2
P(L2|D) = P(L2)P(D/L2) / [ P(L1)P(D/L1) + P(L2)P(D/L2)]
P(L2|D) = (.40)(.20) /[(.60)(.10)+ (.40)(.20)]
P(L2|D) = 0.5714