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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

The population of a particular country consists of three ethnic groups. Each individual belongs to one of the four major blood groups. The accompanying joint probability table gives the proportions of individuals in the various ethnic group-blood group combinations.

Suppose that an individual is randomly selected from the population, and define events by \({\rm{A = }}\)\{type A selected), \({\rm{B = }}\) (type B selected), and \({\rm{C = }}\) (ethnic group \({\rm{3}}\) selected).

a. Calculate\({\rm{P(A),P(C)}}\), and \({\rm{P(A{C}C)}}\).

b. Calculate both \({\rm{P(A}}\mid {\rm{C)}}\)and \({\rm{P(C}}\mid {\rm{A)}}\), and explain in context what each of these probabilities represents.

c. If the selected individual does not have type B blood, what is the probability that he or she is from ethnic group\({\rm{1}}\)?

Let Adenote the event that the next request for assistance from a statistical software consultant relates to the SPSS package, and let Bbe the event that the next request is for help with SAS. Suppose that P(A)=.30and P(B)=.50.

a. Why is it not the case that P(A)+P(B)=1?

b. Calculate P(A鈥�).

c. Calculate P(A\( \cup \)B).

d. Calculate P(A鈥橽( \cap \)B鈥�).

One of the assumptions underlying the theory of control charting is that successive plotted points are independent of one another. Each plotted point can signal either that a manufacturing process is operating correctly or that there is some sort of malfunction.

A certain system can experience three different types of defects. Let \({{\rm{A}}_{\rm{1}}}\)\({\rm{(i = 1,2,3)}}\)denote the event that the system has a defect of type Suppose that

\(\begin{aligned}P\left( {{A_1}} \right) &= .12\;\;\;P\left( {{A_2}} \right) = .07\;\;\;P\left( {{A_3}}\right) = .05 \hfill \\P\left( {{A_1}E {A_2}} \right) = .13\;\;\;P\left( {{A_1}E {A_3}}\right) &= .14 \hfill \\P\left( {{A_2}E {A_2}} \right) = .10\;\;\;P\left( {{A_1}{C}{A_3}{C}{A_7}} \right) &= .01 \hfill \\\end{aligned} \)

a. What is the probability that the system does not have a type 1 defect?

b. What is the probability that the system has both type 1 and type 2 defects?

c. What is the probability that the system has both type 1 and type 2 defects but not a type 3 defect?

d. What is the probability that the system has at most two of these defects?

The three most popular options on a certain type of newcar are a built-in GPS (A), a sunroof (B), and an automatictransmission (C). If 40% of all purchasers request A, 55% request B, 70% request C, 63% request Aor B,77% request Aor C, 80% request Bor C, and 85% request Aor Bor C, determine the probabilities of the following events. (Hint:鈥淎or B鈥� is the event that at leastone of the two options is requested; try drawing a Venn

diagram and labeling all regions.)

a. The next purchaser will request at least one of thethree options.

b. The next purchaser will select none of the three options.

c. The next purchaser will request only an automatictransmission and not either of the other two options.

d. The next purchaser will select exactly one of thesethree options.
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