An industry usaes three methods M1 M2 and M3 to manufacture

An industry usaes three methods, M1, M2, and M3 to manufacture a part. Of All the parts manufactured, 45% are produced by method M1, 32% are produced by method M2, and the rest 23% by method M3. Further it has been noted that 3% of the parts manufactured by method M1 are defective, while 2% manufactured by method M2 and 1.5% by method M3 are defective. A randomly selected part is found to be defective. Find the probability that the part was manufactured by (a) method M1, (b) method M2.

Solution

Let

D= defective

1,2,3 = the method number

Thus,

P(D) = P(1) P(D|1) + P(2) P(D|2) + P(3) P(D|3)

= 0.45*0.03 + 0.32*0.02 + 0.23*0.015

= 0.02335

Therefore:

a)

As

P(1|D) = P(1) P(D|1) / P(D)

= 0.45*0.03/0.02335

= 0.578158458 [ANSWER]

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b)

As

P(2|D) = P(2) P(D|2) / P(D)

= 0.32*0.02/0.02335

= 0.274089936 [ANSWER]