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Worked solution

Weight defect rates by production share

A factory has three machines A,B,C producing large numbers of an item. Their shares of daily production are 50%,30%,20%, respectively.

Their respective defect rates are 2%,3%,4%. The occurrence of a defective item is independent of all other items.

One item is chosen at random from the day’s total output.

(i) Show that its probability of being defective is 0.027.

(ii) Given that it is defective, find the probability that it was produced on A.

CJC · Lecture · Self-Practice 7

(i) Add the machine-and-defect contributions

Let D mean defective. The machines partition the output, so P(D)=P(A)P(D∣A)+P(B)P(D∣B)+P(C)P(D∣C)=0.5(0.02)+0.3(0.03)+0.2(0.04)=0.010+0.009+0.008=0.027P(D)=P(A)P(D\mid A)+P(B)P(D\mid B)+P(C)P(D\mid C)=0.5(0.02)+0.3(0.03)+0.2(0.04)=0.010+0.009+0.008=0.027, as required.

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(ii) Reverse the conditioning

P(A∣D)=P(A∩D)/P(D)=[0.5(0.02)]/0.027=0.010/0.027=10/27P(A\mid D)=P(A\cap D)/P(D)=[0.5(0.02)]/0.027=0.010/0.027=10/27. The printed answers 0.027 and 10/27 are retained; these derivations supplement the answer-only source. The stated independence concerns defects of different items, not independence of machine identity and defect status for one item.

How did this part go?

Common mistakes

Do not take the unweighted average of the three defect rates. Choosing uniformly from all output gives machine probabilities equal to production shares. The factory’s different conditional defect rates must remain distinct.

Read the hint