Suppose that a calculator manufacturer buys integrated circuits from three different suppliers: supplier I, supplier II, and supplier III. Based on past experience, it is known that 2% of circuits from supplier I are defective, 1% of circuits from supplier II are defective, and 1.6% of circuits from supplier III are defective. The calculator manufacturer buys 20% of its circuits from supplier I, 35% of its circuits from supplier II, and 45% of its circuits from supplier III. What is the probability that a randomly selected integrated circuit is found to be defective?

Approach Let D represent the event that the integrated circuit is defective. Let S₁ , S₂ , and S₃ represent the events that the circuit came from suppliers I, II, or III, respectively.

We have the following probabilities:

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P(S1) = 0.2 P(S2) = 0.35 P(D S2) = 0.01 P(S3) = 0.45 P(D|S3) = 0.016 P(D| S1) = 0.02 We wish to know the probability that the integrated circuit is defective. Because the sample space is partitioned into three disjoint events corresponding to suppliers I, II, and III, the probability of obtaining a defective circuit is the probability that the circuit is defective and comes from supplier I or defective and comes from supplier II or defective and comes from supplier III. P(D) = P((DS) U (DNS2) U (DNS3)) These three events are disjoint, so we have P(D) = P(DS) + P(DNS) + P(DNS3) Using the General Multiplication Rule, this simplifies to . P(D) = P(S) P(D|S) + P(S2) P(D|S2) + P(S3) P(D | S3) (3)