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Chapter 2: Q19E (page 65)

Human visual inspection of solder joints on printed circuit boards can be very subjective. Part of the problem stems from the numerous types of solder defects (e.g., pad non wetting, knee visibility, voids) and even the degree to which a joint possesses one or more of these defects. Consequently, even highly trained inspectors can disagree on the disposition of a particular joint. In one batch of 10,000 joints, inspector A found 724 that were judged defective, inspector B found 751 such joints, and 1159 of the joints were judged defective by at least one of the inspectors. Suppose that one of the 10,000 joints is randomly selected.

a. What is the probability that the selected joint was judged to be defective by neither of the two inspectors?

b. What is the probability that the selected joint was judged to be defective by inspector B but not by inspector A?

Short Answer

Expert verified

a. The probability that the selected joint was judged to be defective by neither of the two inspectors is 0.8841.

b. The probability that the selected joint was judged to be defective by inspector B but not by inspector A is 0.0435.

Step by step solution

01

Given information

The number of joints randomly selected is 10000.

The number of defective joints found by inspector A is 724.

The number of defective joints found by inspector B is 751.

The number of defective joints that were found by at least one of the inspectors is 1159.

02

Compute the probability

Let A be the event that the number of defective joints is found by inspector A.

Let B be the event that the number of defective joints is found by inspector B.

The event \(A \cup B\) represents the event of number of defective joints that were found by at least one of the inspectors.

The probability of event A is computed as,

\(\begin{aligned}P\left( A \right) &= \frac{{724}}{{10000}}\\ &= 0.0724\end{aligned}\)

Therefore, the probability of event A is 0.0724.

The probability of event B is computed as,

\(\begin{aligned}P\left( B \right) &= \frac{{751}}{{10000}}\\ &= 0.0751\end{aligned}\)

Therefore, the probability of event B is 0.0751.

The probability of event \(A \cup B\) is computed as,

\(\begin{aligned}P\left( {A \cup B} \right) &= \frac{{1159}}{{10000}}\\ &= 0.1159\end{aligned}\)

Therefore, the probability of event \(A \cup B\) is 0.1159.

Using the formula for the probability of union of the two events,

\(P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)\)

The probability that the number of defective joints that were found by inspector A and inspector B is computed as,

\(\begin{aligned}P\left( {A \cap B} \right) &= P\left( A \right) + P\left( B \right) - P\left( {A \cup B} \right)\\ &= 0.0724 + 0.0751 - 0.1159\\ &= 0.0316\end{aligned}\)

Therefore, the probability that the number of defective joints that were found by inspector A and inspector B is 0.0316

a.

The probability that the selected joint was judged to be defective by neither of the two inspectors is computed as,

\(\begin{aligned}P\left( {A \cup B} \right)' &= 1 - P\left( {A \cup B} \right)\\ &= 1 - 0.1159\\ &= 0.8841\end{aligned}\)

Therefore, the probability that the selected joint was judged to be defective by neither of the two inspectors is 0.8841.

b.

The probability that the selected joint was judged to be defective by inspector B but not by inspector A is computed as,

\(\begin{aligned}P\left( {B \cup A'} \right) &= P\left( B \right) - P\left( {A \cap B} \right)\\ &= 0.0751 - 0.0316\\ &= 0.0435\end{aligned}\)

Therefore, the probability that the selected joint was judged to be defective by inspector B but not by inspector A is 0.0435

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Most popular questions from this chapter

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