Question

A product test is designed in such a way that for a defective product to be undiscovered, al four inspections would have to fail to catch the defect. The probability of catching the defect in inspection 1 is 90 in inspection 2, BON in inspection 3, 12% and in inspection 4,95%. What is the probably of catching a defect? 

 

Answer 1

Answer:

• Option d: 99.9%

Explanation:

• Probability of not catching a defect in inspection 1 = A =100-90=10%
• Probability of not catching a defect in inspection 2 = B=100-80=20%
• Probability of not catching a defect in inspection 3 = C=100-12=88%
• Probability of not catching a defect in inspection 4 = D =100-95=5%

Now since if we catch the defect in any of the one inspection defects will be known, hence all four inspections will act as in parallel.:

• Probability of not catching a defect = E =A*B*C*D= 0.1*0.2*0.88*0.05 = 0.00088=0.088%= 0.01%
• Hence the probability of catching a defect = 100-E= 100-0.01= 99.9 % (Answer)