Question

• Suppose the probability of a part being manufactured by Machine A is 0.6

• Suppose the probability that a part was manufactured by Machine A and the part is defective is 0.09

• Suppose the probability that a part was NOT manufactured by Machine A and the part IS defective is 0.13

Find the probability that Machine A produced a specific part, given that the part was defective. Round your final answer to 2 decimals, if needed.

Answer 1

Solution

Back-up Theory

If A and B are two events such that probability of B is influenced by occurrence or otherwise of A, then Conditional Probability of B given A, denoted by P(B/A) = P(B ∩ A)/P(A)……....….(1)

Now to work out the solution,

Let A represent the event that a part is manufactured by Machine A and D represent the event that a part is defective. Then, trivially, A^C represents the event that a part is not manufactured by Machine A and D represent the event that a part is not defective.

With this definitions, the given probabilities are:

• The probability that a part was manufactured by Machine A and the part is defective is 0.09
• => P(A ∩ D) = 0.09
• The probability that a part was NOT manufactured by Machine A and the part is defective is 0.13
• => P(A^C ∩ D) = 0.13

The above two probabilities => P(D) = 0.09 + 0.13 = 0.22.

Now, probability that Machine A produced a specific part, given that the part was defective

= P(A/D)

= P(A ∩ D)/P(D) [vide (1)]

= 0.09/0.22

= 0.4091

= 0.41   Answer

DONE