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Chapter 5: Problem 125

When commercial aircraft are inspected, wing cracks are reported as nonexistent, detectable, or critical. The history of a particular fleet indicates that \(70 \%\) of the planes inspected have no wing cracks, \(25 \%\) have detectable wing cracks, and \(5 \%\) have critical wing cracks. Five planes are randomly selected. Find the probability that a. one has a critical crack, two have detectable cracks, and two have no cracks. b. at least one plane has critical cracks.

Short Answer

Expert verified
a. Probability is 0.046. b. Probability is 0.226.

Step by step solution

01

Define the Problem

We are tasked with finding the probability of two conditions involving wing cracks on a fleet of aircrafts where the probabilities are given for critical, detectable, and no cracks as 0.05, 0.25, and 0.70 respectively. We need to determine the probability for a specific distribution of these crack types among five randomly inspected planes.
02

Calculate Probability for part a

We use the multinomial probability formula to solve part a. The problem outlines: one plane has critical cracks, two planes have detectable cracks, and two planes have no cracks. The formula is:\[P(X_1=k_1, X_2=k_2, X_3=k_3) = \frac{n!}{k_1! k_2! k_3!} p_1^{k_1} p_2^{k_2} p_3^{k_3}\]Where:- \(n = 5\) (total planes)- \(k_1 = 1\), \(k_2 = 2\), \(k_3 = 2\) (planes with critical, detectable and no cracks)- \(p_1 = 0.05\), \(p_2 = 0.25\), \(p_3 = 0.70\) (probabilities of the types of cracks)Substitute these values in to get:\[P = \frac{5!}{1!2!2!} (0.05)^1 (0.25)^2 (0.70)^2\]Calculating:\[= \frac{120}{1 \times 2 \times 2} \times 0.05 \times 0.0625 \times 0.49\]\[= 30 \times 0.05 \times 0.0625 \times 0.49\]\[= 0.0459375\]
03

Solve for at least one critical crack - part b

To calculate the probability of at least one plane having a critical crack, we calculate the complement of no planes having critical cracks. The probability of no planes having critical cracks is:\[P( ext{no critical cracks}) = (p_2 + p_3)^5 = (0.95)^5 \]Calculating:\[= 0.95^5 = 0.7737809375\]The probability of at least one plane having critical cracks is:\[P( ext{at least one critical crack}) = 1 - 0.7737809375 = 0.2262190625\]
04

Conclusion

For part a, the probability of one plane having a critical crack, two with detectable cracks, and two with no cracks is approximately 0.046. For part b, the probability of at least one plane having a critical crack is approximately 0.226.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Theory
Probability theory is a branch of mathematics that deals with the likelihood of events happening. In many real-world scenarios, like inspecting aircraft for cracks, probability theory helps us determine how often certain outcomes occur.

It provides a mathematical framework to understand and predict certain outcomes given known probabilities.
  • Probability is quantified with numbers between 0 and 1, where 0 indicates an impossible event and 1 signifies certainty.
  • Probabilities of all possible outcomes of a random process always add up to 1.
In the context of aircraft inspections, knowing that 70% have no cracks, 25% have detectable cracks, and 5% have critical cracks gives us a basis for further calculations. These individual probabilities become crucial when determining the chances of complex scenarios, such as in the exercise given.
Random Selection
Random selection is a process by which samples are chosen with equal probability of being selected. This concept ensures that each member or part of a population has an equal chance of being chosen in a sample.

In situations like inspecting a fleet of aircraft, random selection is essential because it ensures that the sample of planes being inspected is representative of the entire fleet.
  • This randomness removes bias in selecting which planes get inspected.
  • It allows us to use probability distributions, like the multinomial distribution, to make predictions or calculate probabilities.
By selecting five planes randomly, we can apply probability theory to predict the distribution of crack types as seen in the original exercise. The randomness is what gives credibility to the probability calculations that follow.
Complement Rule
The complement rule is a fundamental concept in probability, allowing us to determine the probability of an event's complement鈥攖hat is, the event not occurring. This is often used when it's easier to calculate the probability of something not happening and subtract that from 1.

In mathematical terms, if you want to find out the probability of event A happening, and you know the probability of it not happening ( P(A^c) ), then:
P(A) = 1 - P(A^c) .
  • The complement rule simplifies calculations for multiple complex scenarios, particularly when dealing with 'at least' probabilities.
  • It provides a useful shortcut for understanding and calculating probabilities more efficiently.
In our exercise, to find the probability that at least one plane has critical cracks, it was simpler to calculate the probability that none have critical cracks and then use the complement rule to find the result.

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Most popular questions from this chapter

In Exercise 5.8 , we derived the fact that $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} 4 y_{1} y_{2}, & 0 \leq y_{1} \leq 1,0 \leq y_{2} \leq 1 \\ 0, & \text { elsewhere } \end{array}\right.$$ is a valid joint probability density function. Find a. the marginal density functions for \(Y_{1}\) and \(Y_{2}\) b. \(P\left(Y_{1} \leq 1 / 2 | Y_{2} \geq 3 / 4\right)\) c. the conditional density function of \(Y_{1}\) given \(Y_{2}=y_{2}\) d. the conditional density function of \(Y_{2}\) given \(Y_{1}=y_{1}\) e. \(P\left(Y_{1} \leq 3 / 4 | Y_{2}=1 / 2\right)\)

Suppose that the random variables \(Y_{1}\) and \(Y_{2}\) have joint probability density function \(f\left(y_{1}, y_{2}\right)\) given by $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} 6 y_{1}^{2} y_{2}, & 0 \leq y_{1} \leq y_{2}, y_{1}+y_{2} \leq 2 \\ 0, & \text { elsewhere } \end{array}\right.$$ a. Verify that this is a valid joint density function. b. What is the probability that \(Y_{1}+Y_{2}\) is less than \(1 ?\)

A supermarket has two customers waiting to pay for their purchases at counter I and one customer waiting to pay at counter II. Let \(Y_{1}\) and \(Y_{2}\) denote the numbers of customers who spend more than \(\$ 50\) on groceries at the respective counters. Suppose that \(Y_{1}\) and \(Y_{2}\) are independent binomial random variables, with the probability that a customer at counter I will spend more than \$50 equal to .2 and the probability that a customer at counter II will spend more than \(\$ 50\) equal to .3. Find the a. joint probability distribution for \(Y_{1}\) and \(Y_{2}\) b. probability that not more than one of the three customers will spend more than \(\$ 50 .\)

Of nine executives in a business firm, four are married, three have never married, and two are divorced. Three of the executives are to be selected for promotion. Let \(Y_{1}\) denote the number of married executives and \(Y_{2}\) denote the number of never-married executives among the three selected for promotion. Assuming that the three are randomly selected from the nine available, find the joint probability function of \(Y_{1}\) and \(Y_{2}\).

The random variables \(Y_{1}\) and \(Y_{2}\) are such that \(E\left(Y_{1}\right)=4, E\left(Y_{2}\right)=-1, V\left(Y_{1}\right)=2\) and \(V\left(Y_{2}\right)=8\). a. What is \(\operatorname{Cov}\left(Y_{1}, Y_{2}\right) ?\) b. Assuming that the means and variances are correct, as given, is it possible that \(\operatorname{Cov}\left(Y_{1}, Y_{2}\right)=7 ?\left[\text { Hint: } \text { If } \operatorname{Cov}\left(Y_{1}, Y_{2}\right)=7, \text { what is the value of } \rho, \text { the coefficient of correlation? }\right]\) c. Assuming that the means and variances are correct, what is the largest possible value for \(\operatorname{Cov}\left(Y_{1}, Y_{2}\right) ?\) If \(\operatorname{Cov}\left(Y_{1}, Y_{2}\right)\) achieves this largest value, what does that imply about the relationship between \(Y_{1}\) and \(Y_{2} ?\) d. Assuming that the means and variances are correct, what is the smallest possible value for \(\operatorname{Cov}\left(Y_{1}, Y_{2}\right) ?\) If \(\operatorname{Cov}\left(Y_{1}, Y_{2}\right)\) achieves this smallest value, what does that imply about the relationship between \(Y_{1}\) and \(Y_{2} ?\)
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