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Chapter 2: Q77E (page 89)

An aircraft seam requires \({\rm{25}}\) rivets. The seam will have to be reworked if any of these rivets is defective. Suppose rivets are defective independently of one another, each with the same probability.

a. If \({\rm{15\% }}\)of all seams need reworking, what is the probability that a rivet is defective?

b. How small should the probability of a defective rivet be to ensure that only \({\rm{10\% }}\) of all seams need reworking?

Short Answer

Expert verified
  1. The probability that \({\rm{P( defective ) = 0}}{\rm{.64797\% }}{\rm{.}}\)
  2. The probability \({\rm{P(defective) = 0}}{\rm{.0042056 = 0}}{\rm{.42056\% }}\)

Step by step solution

01

Definition of Independence

Independence If the probability of one event is unaffected by the occurrence or non-occurrence of the other, two events are said to be independent in probability. Consider the following example for a better understanding of this definition.

02

Finding the probability that a rivet is defective?

Given: If any of the \({\rm{25}}\) rivets are defective, the seam must be redone. The rivets are on their own.

\(\begin{array}{c}P({\rm{ At least one of the rivits is defective }}) = 15\% \\ = 0.15\end{array}\)

Complement rule:

\({\rm{P(notA) = 1 - P(A)}}\)

Multiplication rule for independent events:

\({\rm{P(A and B) = P(A)*P(B)}}\)

Use the complement rule:

\({\rm{P}}\left( {{\rm{ No rivits are defective }}} \right){\rm{ = 1 - P}}\left( {{\rm{ At least one of the rivits is defective }}} \right)\)

\({\rm{ = 1 - 0}}{\rm{.15 = 0}}{\rm{.85}}\)

We can use the multiplication rule for independent events because the rivets are independent. The likelihood of a single rivet not being defective is represented by \({\rm{P}}\) (not defective).

\(\begin{array}{*{20}{l}}{{\rm{P}}\left( {{\rm{No rivets are defective}}} \right)}\\{{\rm{ = P}}\left( {{\rm{not defective}}} \right){\rm{*P}}\left( {{\rm{ not defective }}} \right){\rm{ *}}....{\rm{*P}}\left( {{\rm{ not defective }}} \right)}\end{array}\)

\({\rm{ = (P( not defective )}}{{\rm{)}}^{{\rm{25}}}}\)

We also know that the likelihood of this happening is \(0.85{\rm{ }}:\)

\({{\rm{(P( not defective ))}}^{{\rm{25}}}}{\rm{ = 0}}{\rm{.85}}\)

Take the 25th root of each side of the equation:

\({\rm{P( not defective ) = }}\sqrt({{\rm{25}}}){{{\rm{0}}{\rm{.85}}}} \approx 0.{\rm{9935203}}\)

Use the complement rule again:

\(\begin{array}{c}\rm P( defective ) &=& 1 - P( not defective ) \\ \rm &=& 1 - 0{\rm{.9935203}}\\ \rm &=& 0{\rm{.0064797}}\\ \rm &=& 0{\rm{.64797\% }}{\rm{.}}\end{array}\)

Thus, the probability that \({\rm{P( defective ) = 0}}{\rm{.64797\% }}{\rm{.}}\)

03

The probability of a defective rivet be to ensure that only \({\rm{10\% }}\) 

Given: If any of the 25 rivets are defective, the seam must be redone. The rivets are self-contained.

\(\begin{array}{c}{\rm{P}}\left( {{\rm{ At least one of the rivits is defective }}} \right) \rm &=& 10\% \\ \rm &=& 0{\rm{.10}}\end{array}\)

Complement rule:

\({\rm{P(notA) = 1 - P(A)}}\)

For independent events, use the following multiplication rule:

\({\rm{P(A and B) = P(A)*P(B)}}\)

Use the complement rule to help you:

\({\rm{P}}\left( {{\rm{No rivits are defective }}} \right){\rm{ = 1 - P}}\left( {{\rm{ At least one of the rivits is defective }}} \right)\)

\({\rm{ = 1 - 0}}{\rm{.10 = 0}}{\rm{.90}}\)

We can use the multiplication rule for independent events because the rivets are independent. The likelihood of a single rivet not being defective is represented by \({\rm{P}}\) (not defective).

04

Finding probability

\({\rm{P(No rivets are defective)}}\)

\(\begin{array}{c} \rm &=& P(not defective)*P(not defective)*...{\rm{*P(not defective)}}\\ \rm &=& (P(not defective){{\rm{)}}^{{\rm{25}}}}\end{array}\)

We also know that this probability is equal to \(0.90\):

\({\rm{P(not defective)}}{{\rm{)}}^{{\rm{25}}}}{\rm{ = 0}}{\rm{.90}}\)

Take the\({\rm{\;25th}}\)root of each side of the equation:

\({\rm{P(not defective) = }}\sqrt({{\rm{25}}}){{{\rm{0}}{\rm{.90}}}} \approx {\rm{0}}{\rm{.9957944}}\)

Use the complement rule again:

\(\begin{array}{c} \rm P(defective) &=& 1 - P(not defective) &=& 1 - 0{\rm{.9957944}}\\ \rm &=& 0{\rm{.0042056}}\\ \rm P(defective) &=& 1 - P(notdefective)\\ \rm &=& 1 - 0 {\rm{.9957944}}\\ \rm &=& 0{\rm{.0042056}}\\ \rm &=& 0{\rm{.42056\% }}\end{array}\)

Therefore, the probability \({\rm{P(defective) = 0}}{\rm{.0042056 = 0}}{\rm{.42056\% }}\)

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Most popular questions from this chapter

A quality control inspector is examining newly produced items for faults. The inspector searches an item for faults in a series of independent fixations, each of a fixed duration. Given that a flaw is actually present, let p denote the probability that the flaw is detected during any one fixation (this model is discussed in 鈥淗uman Performance in Sampling Inspection,鈥� Human Factors, \({\rm{1979: 99--105)}}{\rm{.}}\)

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