An aircraft seam requires 21 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. (Round your answers to four decimal places.)
(a) If 16% of all seams need reworking, what is the probability that a rivet is defective?
  
Incorrect: Your answer is incorrect.


(b) How small should the probability of a defective rivet be to ensure that only 12% of all seams need reworking?
  

Let the probability that a single rivet is defective be denoted as \( p \).

The seam will need to be reworked if at least one of the 21 rivets is defective. Therefore, the probability that a seam needs reworking can be found by calculating the probability that at least one rivet is defective, which is the complement of the probability that no rivets are defective.

### Step-by-step breakdown:

#### (a) Given that 16% of all seams need reworking, what is the probability that a rivet is defective?

Let the probability that no rivets are defective be \( (1 - p)^{21} \). The probability that a seam needs reworking is \( 1 - (1 - p)^{21} \).

We are told that 16% of seams need reworking, so:
\[
1 - (1 - p)^{21} = 0.16
\]
Solving for \( p \):

\[
(1 - p)^{21} = 0.84
\]
Now, take the 21st root of both sides:
\[
1 - p = 0.84^{\frac{1}{21}}
\]

Using a calculator, compute \( 0.84^{\frac{1}{21}} \), and then solve for \( p \).

\[
p = 1 - 0.84^{\frac{1}{21}}
\]

#### (b) How small should the probability of a defective rivet be to ensure that only 12% of all seams need reworking?

Now, we are told that the probability of a seam needing reworking is 12%, or 0.12. We use the same equation:

\[
1 - (1 - p)^{21} = 0.12
\]
This gives:
\[
(1 - p)^{21} = 0.88
\]
Again, taking the 21st root of both sides:

\[
1 - p = 0.88^{\frac{1}{21}}
\]
Solve for \( p \):

\[
p = 1 - 0.88^{\frac{1}{21}}
\]

I'll now compute both values of \( p \).The results are as follows:

(a) The probability that a rivet is defective, given that 16% of seams need reworking, is approximately \( p = 0.0083 \) (or 0.83%).

(b) To ensure that only 12% of seams need reworking, the probability that a rivet is defective should be reduced to approximately \( p = 0.0061 \) (or 0.61%).

Would you like more details on the calculations or have any questions?

Here are five follow-up questions:
1. How does reducing the defective rivet probability affect the manufacturing process?
2. What assumptions are made in the model of independent rivet failures?
3. How would the results change if the number of rivets per seam increased?
4. What statistical techniques can be used to estimate the probability of defective rivets from observed data?
5. How could reliability testing improve the defect rate in rivets?

**Tip:** When dealing with probabilities, always check if events are independent or dependent, as it significantly affects the calculations.

An aircraft seam requires 21 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. (Round your answers to four decimal places.) (a) If 16% 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 12% of all seams need reworking?

Learn how to calculate the probability that a rivet is defective to ensure aircraft seams don't need reworking. Understand key probability concepts and formulas, like (1 - p)^21 and the complement rule, with solutions rounded to four decimal places. Suitable for advanced high school and undergraduate students.

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