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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

A family consisting of three persons鈥擜, B, and C鈥攇oes to a medical clinic that always has a doctor at each of stations 1, 2, and 3. During a certain week, each memberof the family visits the clinic once and is assigned at random to a station. The experiment consists of recording the station number for each member. One outcome is (1, 2, 1) for Ato station 1, Bto station 2, and Cto station 1.

a. List the 27 outcomes in the sample space.

b. List all outcomes in the event that all three members go to the same station.

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A mutual fund company offers its customers a varietyof funds: a money-market fund, three different bond funds (short, intermediate, and long-term), two stock funds (moderate and high-risk), and a balanced fund.

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Money-market 20% High-risk stock 18%

Short bond 15% Moderate-risk stock 25%

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Long bond 5%

A customer who owns shares in just one fund is randomlyselected.

a. What is the probability that the selected individualowns shares in the balanced fund?

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Three molecules of type A, three of type B, three of type C, and three of type \({\rm{D}}\) are to be linked together to form a chain molecule. One such chain molecule is ABCDABCDABCD, and another is BCDDAAABDBCC.

a. How many such chain molecules are there? (Hint: If the three were distinguishable from one another鈥� \({\rm{A1,\;A2, A3}}\)鈥攁nd the were also, how many molecules would there be? How is this number reduced when the subscripts are removed from the ?)

b. Suppose a chain molecule of the type described is randomly selected. What is the probability that all three molecules of each type end up next to one another (such as in BBBAAADDDCCC)?

A college library has five copies of a certain text onreserve. Two copies (1 and 2) are first printings, and the other three (3, 4, and 5) are second printings. A student examines these books in random order, stopping only when a second printing has been selected. One possible outcome is 5, and another is 213.

a. List the outcomes in S.

b. Let Adenote the event that exactly one book must be examined. What outcomes are in A?

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a. A lumber company has just taken delivery on a shipment of \({\rm{10,000 2 \times 4}}\)boards. Suppose that 20% of these boards (\({\rm{2000}}\)) are actually too green to be used in first-quality construction. Two boards are selected at random, one after the other. Let A 5 {the first board is green} and B 5 {the second board is green}. Compute \({\rm{P(A),P(B)}}\), and \({\rm{P(A}} \cap {\rm{B)}}\) (a tree diagram might help). Are \({\rm{A}}\) and \({\rm{B}}\) independent?

b. With \({\rm{A}}\) and \({\rm{B}}\) independent and \({\rm{P(A) = P(B) = }}{\rm{.2,}}\) what is \({\rm{P(A}} \cap {\rm{B)}}\)? How much difference is there between this answer and \({\rm{P(A}} \cap {\rm{B)}}\)part (a)? For purposes of calculating \({\rm{P(A}} \cap {\rm{B)}}\), can we assume that \({\rm{A}}\)and \({\rm{B}}\) of part (a) are independent to obtain essentially the correct probability?

c. Suppose the shipment consists of ten boards, of which two are green. Does the assumption of independence now yield approximately the correct answer for \({\rm{P(A}} \cap {\rm{B)}}\)? What is the critical difference between the situation here and that of part (a)? When do you think an independence assumption would be valid in obtaining an approximately correct answer to \({\rm{P(A}} \cap {\rm{B)}}\)?
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