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Chapter 2: Problem 52

A closet contains 10 pairs of shoes. If 8 shoes are randomly selected, what is the probability that there will be (a) no complete pair? (b) exactly 1 complete pair?

Short Answer

Expert verified
The probability of selecting 8 shoes with (a) no complete pair is \(0.045\) and (b) exactly 1 complete pair is approximately \(0.454\).

Step by step solution

01

Calculate the number of ways to select 8 shoes without a complete pair

To achieve this, we need to ensure that we select 1 shoe from each of 8 different pairs and none from the remaining 2 pairs. There are 10!/(2!8!) ways to select 8 pairs out of 10 pairs. From each of the 8 pairs, there are 2 choices to pick 1 shoe (as there are 2 shoes per pair), so there are 2^8 possible shoe selections.
02

Calculate the total possible ways of selecting 8 shoes out of 20

Use the binomial coefficient formula (also known as 'n choose k' or combinations) which is: \({n \choose k}= \frac{n!}{k!(n-k)!}\), where n=20 shoes and k = 8 shoes to be selected. It gives us the total number of ways to select 8 shoes from 20 shoes: \({20 \choose 8} = \frac{20!}{8!(20-8)!}\)
03

Compute the probability of having no complete pairs

Divide the number of ways to select 8 shoes without a complete pair by the total possible ways of selecting 8 shoes: \(\frac{ \frac{10!}{2!8!} \times 2^8}{\frac{20!}{8!(20-8)!}}\) After simplifying the math, we get the probability to be \(\frac{45}{990}\) or \(0.045\). (b) Exactly 1 complete pair
04

Calculate the number of ways to select 8 shoes having exactly 1 complete pair

From the 10 pairs, we choose one pair to have a complete pair which gives us 10 possibilities. Then, we select 1 shoe from each of 6 different pairs out of the remaining 9 pairs which gives us 9!/(3!6!) ways. From each chosen pair, we have 2 possibilities of picking a shoe. So there are \(10 \times \frac{9!}{3!6!} \times 2^6\) ways to select the shoes.
05

Compute the probability of having exactly 1 complete pair

Divide the number of ways to select 8 shoes having exactly 1 complete pair by the total possible ways of selecting 8 shoes (from step 2 of part a): \(\frac{10 \times \frac{9!}{3!6!} \times 2^6}{\frac{20!}{8!(20-8)!}}\) Simplify the math to find the probability, which is \(\frac{135}{297}\) or approximately \(0.454\). So, the probability of selecting 8 shoes with (a) no complete pair is 0.045 and (b) exactly 1 complete pair is approximately 0.454.

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Key Concepts

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

Combinatorics
Combinatorics is a fascinating area of mathematics that deals with counting, arranging, and understanding the structure of sets. In this context, combinatorics helps us determine the number of ways to choose a subset of items from a larger set. This can be useful for solving problems related to probability as it helps in understanding how many possible combinations exist for a given situation.
  • A fundamental concept in combinatorics is the arrangement of items, where the order of selection may or may not matter.
  • Combination is often used when the order does not matter, while permutation is used when order is important.
By mastering these concepts, students can learn how to approach complex problems by breaking them down into simpler counting procedures. In the shoe problem, combinatorics is used to calculate the number of possible ways to choose shoes without forming complete pairs.
Binomial Coefficient
A binomial coefficient, often referred to as "n choose k", is a key part of combinatorics. It helps us calculate the number of ways to choose a subset of items from a larger set without considering the order in which they are selected. The formula for binomial coefficients is: \[ {n \choose k} = \frac{n!}{k!(n-k)!} \]
  • Here, \(n!\) denotes the factorial of \(n\), which is the product of all positive integers up to \(n\).
  • The binomial coefficient is leveraged for calculating combinations.
In the original exercise, the binomial coefficient formula is utilized to find the total number of ways to select 8 shoes from a collection of 20 shoes. This calculation helps in determining the probability of different scenarios occurring in the shoe selection problem.
Probability Calculation
Probability is the quantitative description of how likely an event is to occur. In the context of this problem, it involves calculating the likelihood of selecting shoes that meet certain criteria.
To calculate probability, we use the following formula:\[ P(") = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \]
  • A favorable outcome is one that meets the specific condition we are examining.
  • The total number of possible outcomes is all the ways the event can happen without restrictions.
In our exercise, we use probability to figure out the chances of selecting shoes in such a manner that we avoid complete pairs or achieve exactly one complete pair. Detailed calculations are done using
  • binomial coefficients for total scenarios
  • specific combinatoric strategies for favorable scenarios
to ensure accurate outcomes.
Discrete Mathematics
Discrete mathematics is a branch of mathematics dealing with distinct and separate values or objects. It plays an essential role in understanding and solving problems involving probability and combinatorics. Key areas of discrete mathematics include:
  • Combinatorics: the counting and arrangement of distinct items.
  • Graph Theory: the study of graphs and networks.
  • Logic: the principles of valid reasoning.
  • Set Theory: the study of collections of objects.
In probability problems like our shoe selection exercise, discrete mathematics helps us construct a clear method of counting and organizing our choices in a structured way. Many real-world applications rely on discrete mathematics, from computer science algorithms to network security protocols, underscoring its vital importance in modern technology and decision-making processes.

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Most popular questions from this chapter

A system is composed of 5 components, each of which is either working or failed. Consider an experiment that consists of observing the status of each component, and let the outcome of the experiment be given by the vector \(\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right),\) where \(x_{i}\) is equal to 1 if component \(i\) is working and is equal to 0 if component \(i\) is failed. (a) How many outcomes are in the sample space of this experiment? (b) Suppose that the system will work if components 1 and 2 are both working, or if components 3 and 4 are both working, or if components \(1,3,\) and 5 are all working. Let \(W\) be the event that the system will work. Specify all the outcomes in \(W\). (c) Let \(A\) be the event that components 4 and 5 are both failed. How many outcomes are contained in the event \(A ?\) (d) Write out all the outcomes in the event \(A W\).

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If a die is rolled 4 times, what is the probability that 6 comes up at least once?

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Two dice are thrown. Let \(E\) be the event that the sum of the dice is odd, let \(F\) be the event that at least one of the dice lands on \(1,\) and let \(G\) be the event that the sum is 5. Describe the events \(E F, E \cup F, F G, E F^{c},\) and \(E F G\).
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