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Chapter 3: Problem 103

A warehouse contains ten printing machines, four of which are defective. A company selects five of the machines at random, thinking all are in working condition. What is the probability that all five of the machines are non- defective?

Short Answer

Expert verified
The probability that all five machines are non-defective is \( \frac{1}{42} \).

Step by step solution

01

Understand the Problem

We have 10 printing machines in total. Out of these, 4 are defective and 6 are non-defective. The company selects 5 machines. We need to find the probability that all 5 machines selected are non-defective.
02

Total Ways to Select Machines

First, calculate the total number of ways to select 5 machines out of 10. This is a combination problem which can be represented as \( \binom{10}{5} \).
03

Calculate Total Combinations

Use the formula for combinations: \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \). So, \( \binom{10}{5} = \frac{10!}{5!5!} = 252 \).
04

Select Non-Defective Machines

Next, calculate the number of ways to select all 5 machines from the 6 non-defective machines. This is given by \( \binom{6}{5} \).
05

Calculate Non-Defective Combinations

Apply the combination formula: \( \binom{6}{5} = \frac{6!}{5!1!} = 6 \).
06

Calculate Probability

The probability that all selected machines are non-defective is the number of favorable outcomes over the total outcomes. Therefore, \( P(\text{all non-defective}) = \frac{\binom{6}{5}}{\binom{10}{5}} = \frac{6}{252} \).
07

Simplify Probability

Simplify \( \frac{6}{252} \) by dividing both the numerator and the denominator by their greatest common divisor, which is 6: \( \frac{1}{42} \).

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Key Concepts

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

Combinations
Combinations are a fundamental concept in probability and statistics, used to determine how many ways a group of items can be selected from a larger set. When order does not matter, we use combinations.
  • In our exercise, we need to select 5 machines out of a total of 10.
  • This is described as a combination problem, and we solve it using the combination formula.
The formula for combinations is given by:\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]where:
  • \( n \) is the total number of items.
  • \( r \) is the number of items to choose.
  • \(!\) denotes factorial, e.g., \( 5! = 5 \times 4 \times 3 \times 2 \times 1 \).
In our case:
  • \( n = 10 \) and \( r = 5 \) for the total number of ways to select machines.
  • The calculation is \( \binom{10}{5} = 252 \) ways.
Non-defective machines
Non-defective machines are those that are in working condition and do not have any malfunctions. In our scenario:
  • Out of 10 machines, 6 are non-defective.
  • To find the specific number of combinations where all selected machines are non-defective, we choose all 5 from these 6 non-defective machines.
The calculation for this is a matter of selecting 5 from 6:
  • We calculate \( \binom{6}{5} = 6 \).
  • This gives us 6 favorable combinations where all selected machines are non-defective.
This step focuses on isolating the event of selecting only non-defective machines, forming the basis for our probability calculation.
Probability calculation
Probability calculation involves determining how likely an event is to occur. It is expressed as a fraction of the number of successful outcomes over the total possible outcomes. In this exercise, that means calculating the probability all selected machines are non-defective:
  • Favorable outcomes: Selecting 5 non-defective machines from the 6 available \( \binom{6}{5} = 6 \).
  • Total outcomes: All possible ways to select 5 machines out of 10 \( \binom{10}{5} = 252 \).
The probability \( P \) is:\[ P(\text{all non-defective}) = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{6}{252} \]Finally, simplifying the fraction:
  • Using the greatest common divisor (GCD), we find \( \frac{1}{42} \).
This gives us the probability that all five machines selected are non-defective.
Defective machines
Defective machines are those that do not meet operational standards or have functional issues. In our problem:
  • We have 4 defective machines out of 10.
  • When selecting from the entire pool, the presence of these machines affects the probability calculations.
Though our primary goal is to find the cases where no defective machines are selected, understanding their count helps manage expectations and calculate probabilities accurately.
By account of the number of defective machines, it's clear why non-defective probabilities are much smaller. Only a fraction of combinations exclude defective machines.

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Most popular questions from this chapter

A fire-detection device utilizes three temperature-sensitive cells acting independently of each other in such a manner that any one or more may activate the alarm. Each cell possesses a probability of \(p=.8\) of activating the alarm when the temperature reaches \(100^{\circ}\) Celsius or more. Let Y equal the number of cells activating the alarm when the temperature reaches \(100^{\circ}\). a. Find the probability distribution for \(Y\). b. Find the probability that the alarm will function when the temperature reaches \(100^{\circ}\).

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A multiple-choice examination has 15 questions, each with five possible answers, only one of which is correct. Suppose that one of the students who takes the examination answers each of the questions with an independent random guess. What is the probability that he answers at least ten questions correctly?

If \(Y\) is a discrete random variable that assigns positive probabilities to only the positive integers, show that $$E(Y)=\sum_{i=1}^{\infty} P(Y \geq k)$$.
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