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Chapter 2: Problem 31

A boxcar contains six complex electronic systems. Two of the six are to be randomly selected for thorough testing and then classified as defective or not defective. a. If two of the six systems are actually defective, find the probability that at least one of the two systems tested will be defective. Find the probability that both are defective. b. If four of the six systems are actually defective, find the probabilities indicated in part (a).

Short Answer

Expert verified
a. At least one defective: \( \frac{3}{5} \), both defective: \( \frac{1}{15} \). b. At least one defective: \( \frac{14}{15} \), both defective: \( \frac{2}{5} \).

Step by step solution

01

Define the problem setup

We have a boxcar containing 6 electronic systems, out of which 2 are selected randomly for testing. We need to calculate probabilities for two cases: when 2 systems are defective and when 4 systems are defective.
02

Calculate probabilities for 2 defective systems (a)

First, determine the total number of ways to choose 2 systems from 6. This is calculated using combinations: \( \binom{6}{2} = 15 \). Next, calculate the probability that at least one system is defective. There are 2 defective systems, \( \binom{2}{1} = 2 \) ways to pick 1 defective and 1 non-defective system (\( \binom{4}{1} = 4 \)). Total ways to pick at least one defective: \( \binom{2}{1} \times \binom{4}{1} + \binom{2}{2} = 8 + 1 = 9 \). Probability is \( \frac{9}{15} = \frac{3}{5} \).
03

Probability both tested are defective (a)

For both systems to be defective, choose both from the 2 defective ones: \( \binom{2}{2} = 1 \). Probability is \( \frac{1}{15} \).
04

Calculate probabilities for 4 defective systems (b)

If 4 systems are defective, to find the probability of at least one defective, we first find the probability that none is defective (2 non-defective from 2 non-defective systems): \( \binom{2}{2} = 1 \). Probability both are non-defective: \( \frac{1}{15} \). So, probability at least one is defective: \( 1 - \frac{1}{15} = \frac{14}{15} \).
05

Probability both tested are defective (b)

Both defective implies choosing 2 from the 4 defective: \( \binom{4}{2} = 6 \). Probability is \( \frac{6}{15} = \frac{2}{5} \).

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Key Concepts

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

Combinations
In probability theory, combinations are essential when calculating how many ways we can select items from a larger set, without considering the order of the items. This is different from permutations, where order matters.

For example, when dealing with combinations, the notation \( \binom{n}{r} \) is used, which represents the number of ways to choose \( r \) items from \( n \) items.

In our exercise, we used this concept to solve how many ways we could select 2 systems from a total of 6 electronic systems in the boxcar. Specifically, we calculated \( \binom{6}{2} \), which is equal to 15. This means there are 15 different ways to choose 2 systems out of the 6. Using combinations helps to simplify the process of computing probabilities when the order does not matter.
  • This crucial calculation helps to determine the probability of different events.
  • Combinations are used when the focus is on "how many" ways we can form a group, rather than "in what order" those groups are made.
Understanding combinations is key to efficiently and correctly solving problems like those involving defective and non-defective systems.
Defective and Non-defective Systems
In the exercise, we are concerned with identifying defective and non-defective systems during testing. It is crucial to understand that these categories can impact the probabilities of certain outcomes.

A defective system is one that does not meet the operational standards required for proper functioning, while a non-defective system operates as expected. In our exercise, it was determined that:
  • If there are 2 defective systems, the probability of selecting at least one defective system out of 2 chosen systems is calculated by finding all scenarios where at least one is defective.
  • If there are 4 defective systems, finding the probability of at least one defective system increases significantly due to the higher number of defective systems available.
The classification helps identify likely outcomes when conducting quality control checks. Understanding this distinction allows for more accurate calculations in determining the probabilities of testing outcomes, like finding at least one or both systems defective.
Mathematical Statistics
Mathematical statistics is a vital area of mathematics used to describe, analyze, and infer probabilities related to different types of data. Within our exercise, it provides the tools to predict and quantify the likelihood of different outcomes regarding the electronic systems.

In part (a) of our exercise, mathematical statistics was applied to calculate the probability that at least one of the tested systems will be defective or both are defective, based on a known number of defective systems.

Using statistical methods, probabilities were derived as follows:
  • For 2 defective systems, the probability of finding at least one defective was \( \frac{3}{5} \).
  • For 4 defective systems, this probability increased to \( \frac{14}{15} \).
In both instances, the application of probabilities provides insight into quality control processes, informing decisions and predictions in operational settings. Mastering these statistical concepts enhances our ability to make accurate and reliable predictions about real-world situations, particularly in manufacturing and quality testing scenarios.

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Most popular questions from this chapter

The Bureau of the Census reports that the median family income for all families in the United States during the year 2003 was \(\$ 43,318\). That is, half of all American families had incomes exceeding this amount, and half had incomes equal to or below this amount. Suppose that four families are surveyed and that each one reveals whether its income exceeded \(\$ 43,318\) in 2003 . a. List the points in the sample space. b. Identify the simple events in each of the following events: A: At least two had incomes exceeding \(\$ 43,318\). \(B:\) Exactly two had incomes exceeding \(\$ 43,318\) C: Exactly one had income less than or equal to \(\$ 43,318\). c. Make use of the given interpretation for the median to assign probabilities to the simple events and find \(P(A), P(B),\) and \(P(C)\).

Suppose that \(A\) and \(B\) are mutually exclusive events, with \(P(A)>0\) and \(P(B)<1 .\) Are \(A\) and \(B\) independent? Prove your answer.

If \(A\) and \(B\) are two events, prove that \(P(A \cap B) \geq 1-P(\bar{A})-P(\bar{B})\). \([\) Note: This is a simplified version of the Bonferroni inequality.]

Suppose two balanced coins are tossed and the upper faces are observed. a. List the sample points for this experiment. b. Assign a reasonable probability to each sample point. (Are the sample points equally likely?) c. Let \(A\) denote the event that exactly one head is observed and \(B\) the event that at least one head is observed. List the sample points in both \(A\) and \(B\). d. From your answer to part \((c),\) find \(P(A), P(B), P(A \cap B), P(A \cup B),\) and \(P(\bar{A} \cup B)\)

Show that, for any integer \(n \geq 1\), $$\begin{aligned} &\text { a. }\left(\begin{array}{l} n \\ n \end{array}\right)=1 . \text { Interpret this result. }\\\ &\text { b. }\left(\begin{array}{l} n \\ 0 \end{array}\right)=1 . \text { Interpret this result. }\\\ &\text { c. }\left(\begin{array}{l} n \\ r \end{array}\right)=\left(\begin{array}{c} n \\ n-r \end{array}\right) . \text { Interpret this result. }\\\ &\mathrm{d} \cdot \sum_{i=0}^{n}\left(\begin{array}{l} n \\ i \end{array}\right)=2^{n}\left[\text { Hint: Consider the binomial expansion of }(x+y)^{n} \text { with } x=y=1 .\right] \end{aligned}$$
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