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Chapter 5: Problem 32

Aircraft inspectors (who specialize in mechanical engineering) report wing cracks in aircraft as nonexistent, detectable (but still functional), or critical (needs immediate repair). For a particular model of commercial jet 10 years old, history indicates \(75 \%\) of the planes had no wing cracks, \(20 \%\) had detectable wing cracks, and \(5 \%\) had critical wing cracks. Five planes that are 10 years old are randomly selected. What is the probability that (a) 4 have no cracks, 1 has detectable cracks, and 0 have no critical cracks? (b) 3 have no cracks, I has detectable cracks, and I has critical cracks?

Short Answer

Expert verified
Part (a): Probability = 0.0633; Part (b): Probability = 0.0042.

Step by step solution

01

Understand the Problem

We have a binomial distribution problem where we want to find the probability of certain outcomes given the probabilities of no cracks, detectable cracks, and critical cracks in a sample of 5 planes. For part (a), out of 5 planes, 4 should have no cracks, 1 should have detectable cracks, and 0 should have critical cracks. For part (b), 3 should have no cracks, 1 should have detectable cracks, and 1 should have critical cracks.
02

Define the Probabilities

Let the probability of no cracks be \( P(N) = 0.75 \), the probability of detectable cracks be \( P(D) = 0.20 \), and the probability of critical cracks be \( P(C) = 0.05 \). These probabilities sum up to 1 as each plane can only be in one state.
03

Calculate Part (a) Probability

To find the probability that 4 out of 5 planes have no cracks, 1 has detectable cracks, and 0 have critical cracks, use the multinomial formula: \[ P = \frac{5!}{4!1!0!} \times (0.75)^4 \times (0.20)^1 \times (0.05)^0 \]Calculating each component, we have:- \( \frac{5!}{4!1!0!} = 5 \)- \( (0.75)^4 = 0.31640625 \)- \( (0.20)^1 = 0.20 \)Plug these into the formula:\[ P = 5 \times 0.31640625 \times 0.20 = 0.31640625 \times 0.20 = 0.06328125 \]
04

Calculate Part (b) Probability

For part (b), the probability that 3 have no cracks, 1 has detectable cracks, and 1 has critical cracks is calculated as:\[ P = \frac{5!}{3!1!1!} \times (0.75)^3 \times (0.20)^1 \times (0.05)^1 \]Calculating each component, we have:- \( \frac{5!}{3!1!1!} = 20 \)- \( (0.75)^3 = 0.421875 \)- \( (0.20)^1 = 0.20 \)- \( (0.05)^1 = 0.05 \)Plug these into the formula:\[ P = 20 \times 0.421875 \times 0.20 \times 0.05 = 0.084375 \times 0.05 = 0.00421875 \]

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Key Concepts

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

Binomial Distribution
In statistics, the binomial distribution is used to model the number of successful outcomes in a fixed number of trials when the outcome is binary. This means each trial can only have two possible outcomes: success or failure.
For example, in the context of aircraft wing cracks, the trials involve checking each plane for cracks. For each plane, the outcome can be either that it has no cracks (success) or some level of cracks (failure).

Key features of binomial distribution include:
  • Fixed number of trials: In our example, this is the five planes being inspected.
  • Same probability of success for each trial: For each plane, the probability of not having cracks is constant, specifically 75%.
  • Independent trials: The outcome for one plane does not affect the outcome of another.
The binomial distribution's simplicity makes it a powerful and widely used tool in statistics, useful for understanding processes where results are binary.
Probability
Probability is the measure of the likelihood that an event will occur. It's expressed as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty.
In our aircraft scenario, each type of crack (none, detectable, critical) has its own probability:
  • Probability of no cracks, \( P(N) = 0.75 \)
  • Probability of detectable cracks, \( P(D) = 0.20 \)
  • Probability of critical cracks, \( P(C) = 0.05 \)
These probabilities sum to 1, consistent with the rule that probabilities must cover all possible outcomes.

Understanding probabilities helps in estimating the likelihood of various events, allowing for effective decision-making and predictions about likely outcomes.
Multinomial Formula
The multinomial formula is an extension of the binomial distribution. It's used when each trial can result in one of more than two outcomes, and we are interested in the probability of each possible combination of outcomes.
In the context of the aircraft problem, there are three possible outcomes for each plane: no cracks, detectable cracks, and critical cracks. Hence, we use the multinomial formula to find the probability of a specific combination of these outcomes:
\[P = \frac{n!}{k_1!k_2!...k_m!} \times (p_1)^{k_1} \times (p_2)^{k_2}...\times (p_m)^{k_m}\]
Where:
  • \( n \) is the total number of trials (planes inspected).
  • \( k_i \) is the number of outcomes of each type (e.g., no cracks, detectable cracks).
  • \( p_i \) is the probability of each outcome.
This powerful formula allows us to calculate the combined probability of complex events happening simultaneously.

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Most popular questions from this chapter

Basic Computation: Geometric Distribution Given a binomial experiment with probability of success on a single trial \(p=0.30,\) find the probability that the first success occurs on trial number \(n=2\).

Consider a binomial distribution of 200 trials with expected value 80 and standard deviation of about \(6.9 .\) Use the criterion that it is unusual to have data values more than 2.5 standard deviations above the mean or 2.5 standard deviations below the mean to answer the following questions. (a) Would it be unusual to have more than 120 successes out of 200 trials? Explain. (b) Would it be unusual to have fewer than 40 successes out of 200 trials? Explain. (c) Would it be unusual to have from 70 to 90 successes out of 200 trials? Explain.

For a binomial experiment, how many outcomes are possible for each trial? What are the possible outcomes?

What does the random variable for a binomial experiment of \(n\) trials measure?

A research team at Cornell University conducted a study showing that approximately \(10 \%\) of all businessmen who wear ties wear them so tightly that they actually reduce blood flow to the brain, diminishing cerebral functions (Source: Chances: Risk and Odds in Everyday Life, by James Burke). At a board meeting of 20 businessmen, all of whom wear ties, what is the probability that (a) at least one tie is too tight? (b) more than two ties are too tight? (c) no tie is too tight? (d) at least 18 ties are not too tight?
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