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Chapter 2: Problem 35

A particular iPod playlist contains 100 songs, of which 10 are by the Beatles. Suppose the shuffle feature is used to play the songs in random order (the randomness of the shuffling process is investigated in "Does Your iPod Really Play Favorites?" (The Amer. Statistician, 2009: 263 - 268)). What is the probability that the first Beatles song heard is the fifth song played?

Short Answer

Expert verified
The probability is approximately 0.0362.

Step by step solution

01

Understanding the Problem

We need to find the probability that among the first five songs played, exactly one Beatles song appears and that it specifically appears as the fifth song in the sequence. We have 100 total songs, with 10 being by the Beatles.
02

Setting up the Scenario

Consider the first five songs played: among these, the fifth song must be a Beatles song. Therefore, the first four songs must all be non-Beatles songs. This means selecting 4 non-Beatles songs from the 90 non-Beatles songs available.
03

Calculating Combinations for Non-Beatles Songs

We first choose 4 non-Beatles songs from the 90 non-Beatles songs. This is given by the combination formula: \[ \binom{90}{4} \]
04

Calculating Combinations for the Beatles Song

Since the fifth song must be a Beatles song, we choose 1 Beatles song from the 10 available. This can be done in: \[ \binom{10}{1} \]
05

Total Combinations for Desired Outcome

The total ways to choose 4 non-Beatles songs and 1 Beatles song for these first 5 positions is given by the product: \[ \binom{90}{4} \times \binom{10}{1} \]
06

Total Combinations for Any Five Songs

The total ways to choose any 5 songs from 100 is given by: \[ \binom{100}{5} \]
07

Calculating the Probability

The probability that the first Beatles song heard is the fifth song played is the ratio of the specific case to the total possible outcomes:\[ \frac{\binom{90}{4} \times \binom{10}{1}}{\binom{100}{5}} \]
08

Simplification and Final Result

Calculate the numeric values for the combinations and the probability. Use a calculator to determine each binomial coefficient and carry out the division to get the probability.

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Key Concepts

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

Combinatorics
When dealing with probability, the study of combinatorics is essential. It enables us to count the number of possible outcomes efficiently.
In our problem, we use combinatorics to count the number of ways to choose songs in a specific order.
Combinatorics can involve different methods such as:
  • Permutations: Arranging objects in a specific order.
  • Combinations: Selecting objects where order does not matter.
In this exercise, combinations are needed.
We want to know how many ways we can select particular songs from a larger pool, not worrying about the specific order among them for the most part.
Evaluating combinatorial problems surfaces repeatedly in probability, making this concept quite important.
Binomial Coefficient
The binomial coefficient is a fundamental tool in combinatorics.
It's used to determine the number of ways to choose a specific number of items from a larger set.
It is denoted by the formula:\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]Here, \( n \) is the total number of items, and \( k \) is the number you wish to choose.
Factorials (denoted by \(!\)) are a product of an integer and all the integers below it.
In the original problem, the binomial coefficient was used in several crucial steps:
  • Choosing 4 non-Beatles songs from 90: \( \binom{90}{4} \)
  • Choosing 1 Beatles song from 10: \( \binom{10}{1} \)
  • Selecting any 5 songs from 100: \( \binom{100}{5} \)
This approach helps quickly compute the number of ways various parts of a problem can be resolved.
Discrete Mathematics
Discrete mathematics involves study areas where solutions or sets of possible answers are discrete. This means they can be counted distinctly, unlike continuous functions.
Key applications include:
  • Graphs and networking
  • Logic statements
  • Counting and probability theory
In our playlist problem, the discrete nature comes from the distinct songs and specific counting of selection patterns.
Probability calculations often rely on discrete mathematics because it deals with finite, countable outcomes.
Understanding this as a foundational pillar means recognizing how sequences and choices lead to generating probabilities.
Random Events
In probability theory, a random event is one where the actual outcome is uncertain, but the possible outcomes are well-defined.
This is key in understanding how probability keeps us aware of the likelihood of various outcomes.
For instance, shuffling a playlist involves randomness since any song could play at any time. Key insights about random events include:
  • The certainty that all outcomes are possible, yet not predictable.
  • Uniform probability distribution if all outcomes are equally likely.
In the playlist issue, randomness dictates the sequence of songs played.
We calculate probability to quantify the chance we get the specific desired sequence of random events.
This comprehension of randomness helps predict likelihood even if specific results can never be fully anticipated.

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Most popular questions from this chapter

A system consists of two components. The probability that the second component functions in a satisfactory manner during its design life is \(.9\), the probability that at least one of the two components does so is \(.96\), and the probability that both components do so is \(.75\). Given that the first component functions in a satisfactory manner throughout its design life, what is the probability that the second one does also?

A box contains the following four slips of paper, each having exactly the same dimensions: (1) win prize \(1 ;(2)\) win prize \(2 ;(3)\) win prize \(3 ;(4)\) win prizes 1,2 , and 3 . One slip will be randomly selected. Let \(A_{1}=\\{\) win prize 1\(\\}, A_{2}=\\{\) win prize 2\(\\}\), and \(A_{3}=\\{\) win prize 3\(\\} .\) Show that \(A_{1}\) and \(A_{2}\) are independent, that \(A_{1}\) and \(A_{3}\) are independent, and that \(A_{2}\) and \(A_{3}\) are also independent (this is pairwise independence). However, show that \(P\left(A_{1} \cap A_{2} \cap A_{3}\right) \neq P\left(A_{1}\right) \cdot P\left(A_{2}\right) \cdot P\left(A_{3}\right)\), so the three events are not mutually independent.

Fifteen telephones have just been received at an authorized service center. Five of these telephones are cellular, five are cordless, and the other five are corded phones. Suppose that these components are randomly allocated the numbers 1 , \(2, \ldots, 15\) to establish the order in which they will be serviced. a. What is the probability that all the cordless phones are among the first ten to be serviced? b. What is the probability that after servicing ten of these phones, phones of only two of the three types remain to be serviced? c. What is the probability that two phones of each type are among the first six serviced?

Consider four independent events \(A_{1}, A_{2}, A_{3}\), and \(A_{4}\) and let \(p_{i}=P\left(A_{i}\right)\) for \(i=1,2,3,4\). Express the probability that at least one of these four events occurs in terms of the \(p_{i}\) 's, and do the same for the probability that at least two of the events occur.

The three major options on a car model are an automatic transmission \((A)\), a sunroof \((B)\), and an upgraded stereo \((C)\). If \(70 \%\) of all purchasers request \(A, 80 \%\) request \(B, 75 \%\) request \(C, 85 \%\) request \(A\) or \(B, 90 \%\) request \(A\) or \(C, 95 \%\) request \(B\) or \(C\), and \(98 \%\) request \(A\) or \(B\) or \(C\), compute the probabilities of the following events. [Hint: "A or \(B^{\prime \prime}\) is the event that at least one of the two options is requested; try drawing a Venn diagram and labeling all regions.] a. The next purchaser will request at least one of the three options. b. The next purchaser will select none of the three options. c. The next purchaser will request only an automatic transmission and neither of the other two options. d. The next purchaser will select exactly one of these three options.
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