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Chapter 7: Problem 26

A standard piano keyboard has 88 different keys. Find the probability that a cat, jumping on 4 keys in sequence and at random (possibly with repetition), will strike the first four notes of Beethoven's Fifth Symphony. (Leave your answer as a formula.)

Short Answer

Expert verified
The probability of a cat randomly striking the first four notes of Beethoven's Fifth Symphony on a standard piano with 88 keys is \(P(\text{Beethoven's Fifth}) = \frac{1}{88^4}\).

Step by step solution

01

Identify the relevant notes of Beethoven's Fifth Symphony

The first four notes of Beethoven's Fifth Symphony are \(G_3, G_3, G_3\), and \(D_3\), which are all different keys on a standard piano. To simplify this problem, we can use placeholder non-specific names for these keys: Note A, Note A, Note A, and Note B.
02

Calculate the probability of each event in the sequence

Since there are 88 different keys on the piano and the cat can hit any of them, the probability of hitting the first note (or Note A) on one random jump at any given time is given by: \[P(\text{Note A}) = \frac{1}{88}\] The cat needs to hit Note A three times in a row, then hit Note B. So, we can calculate the probability of each event in the sequence as follows: - Probability of hitting Note A first time: \(P(\text{Note A}) = \frac{1}{88}\) - Probability of hitting Note A second time: \(P(\text{Note A}) = \frac{1}{88}\) - Probability of hitting Note A third time: \(P(\text{Note A}) = \frac{1}{88}\) - Probability of hitting Note B on the fourth jump: \(P(\text{Note B}) = \frac{1}{88}\)
03

Determine the probability of the entire sequence

Since these events are independent, we can calculate the probability of the entire sequence happening by multiplying the probabilities of each individual event. The probability of playing the first four notes of Beethoven's Fifth Symphony is given by: \[P(\text{Beethoven's Fifth}) = P(\text{Note A}) \times P(\text{Note A}) \times P(\text{Note A}) \times P(\text{Note B})\] \[P(\text{Beethoven's Fifth}) = \frac{1}{88} \times \frac{1}{88} \times \frac{1}{88} \times \frac{1}{88}\]
04

Simplify the probability formula

Now, we can simplify the probability formula by combining the fractions: \[P(\text{Beethoven's Fifth}) = \frac{1^4}{88^4}\] So, the probability of the cat playing the first four notes of Beethoven's Fifth Symphony on a standard piano is: \[P(\text{Beethoven's Fifth}) = \frac{1}{88^4}\]

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Key Concepts

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

Independent Events Probability
When we talk about independent events in probability, we refer to situations in which the outcome of one event does not affect the outcome of another. In the provided exercise, a cat jumping on piano keys is a classic example. Each jump can result in any one of the 88 keys being pressed, and one jump does not influence the next.

For instance, if the cat presses one key, the chance of it pressing the same or a different key on the next jump remains the same, assuming the cat's jumps are random. This is what makes the events independent. The probability of multiple independent events occurring in a specific sequence is calculated by multiplying the probabilities of each individual event. This is because the outcome of one does not change the probability of the others occurring.
Probability Calculations
The process of probability calculations involves determining the likelihood of an event occurring. For an event E, the probability is generally expressed as P(E), which can range from 0 (impossible event) to 1 (certain event).

In our piano keys scenario, the probability of the cat hitting a specific key (Note A or B) is expressed as a simple probability \(P(\text{Note A}) = \frac{1}{88}\) because there is an equal chance to land on any of the 88 keys. Moreover, to find the probability that several independent events will occur in sequence (like hitting several specific notes in a row), we multiply their individual probabilities together. This approach assumes that the occurrence of each note is independent of the others, leading to our formula for the first four notes of Beethoven's Fifth Symphony.
Applied Combinatorics
The field of applied combinatorics is vital in solving problems involving arrangements or groupings, where the order of things is sometimes of the essence. Combinatorics becomes applicable to our piano exercise as it involves determining the number of ways a particular sequence of events can occur.

Combining combinatorics with probability, we can calculate the likelihood of complex events. Since the cat's selections are random and with repetition allowed, each keypress is considered a separate combinatorial choice from 88 options. Thus, when a cat makes a series of jumps, we can calculate the total probability of a specific sequence, like the iconic symphony notes, by using our knowledge of combinatorics and the probabilities of independent events.

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Most popular questions from this chapter

Based on the following table, which shows the performance of a selection of 100 stocks after one year. (Take S to be the set of all stocks represented in the table.) $$ \begin{array}{|r|c|c|c|c|} \hline & \multicolumn{3}{|c|} {\text { Companies }} & \\ \cline { 2 - 4 } & \begin{array}{c} \text { Pharmaceutical } \\ \boldsymbol{P} \end{array} & \begin{array}{c} \text { Electronic } \\ \boldsymbol{E} \end{array} & \begin{array}{c} \text { Internet } \\ \boldsymbol{I} \end{array} & \text { Total } \\ \hline \begin{array}{r} \text { Increased } \\ \boldsymbol{V} \end{array} & 10 & 5 & 15 & 30 \\ \hline \begin{array}{r} \text { Unchanged }^{*} \\ \boldsymbol{N} \end{array} & 30 & 0 & 10 & 40 \\ \hline \begin{array}{r} \text { Decreased } \\ \boldsymbol{D} \end{array} & 10 & 5 & 15 & 30 \\ \hline \text { Total } & 50 & 10 & 40 & 100 \\ \hline \end{array} $$ If a stock stayed within \(20 \%\) of its original value, it is classified as "unchanged." \(\nabla\) Calculate \(\frac{n(D \cap I)}{n(D)}\). What does the answer represent?

\(\nabla\) Two distinguishable dice are rolled. Could there be two mutually exclusive events that both contain outcomes in which the numbers facing up add to 7 ?

Employment You have worked for the Department of Administrative Affairs (DAA) for 27 years, and you still have little or no idea exactly what your job entails. To make your life a little more interesting, you have decided on the following course of action. Every Friday afternoon, you will use your desktop computer to generate a random digit from 0 to 9 (inclusive). If the digit is a zero, you will immediately quit your job, never to return. Otherwise, you will return to work the following Monday. a. Use the states (1) employed by the DAA and (2) not employed by the DAA to set up a transition probability matrix \(P\) with decimal entries, and calculate \(P^{2}\) and \(P^{3}\). b. What is the probability that you will still be employed by the DAA after each of the next three weeks? c. What are your long-term prospects for employment at the DAA? HIIIT [See Example 5.]

According to a study conducted by the Harvard School of Public Health, a child seated in the front seat who was wearing a seatbelt was \(31 \%\) more likely to be killed in an accident if the car had an air bag that deployed than if it did not. \({ }^{50}\) Let the sample space \(S\) be the set of all accidents involving a child seated in the front seat wearing a seatbelt. Let \(K\) be the event that the child was killed and let \(D\) be the event that the airbag deployed. Fill in the missing terms and quantities: \(P(\longrightarrow \mid \longrightarrow)=\longrightarrow P(\longrightarrow \mid \longrightarrow)\). HINT [When we say "A is \(31 \%\) more likely than \(\mathrm{B}\) " we mean that the probability of \(\mathrm{A}\) is \(1.31\) times the probability of B.]

Describe an interesting situation that can be modeled by the transition matrix $$ P=\left[\begin{array}{ccc} .8 & .1 & .1 \\ 1 & 0 & 0 \\ .3 & .3 & .4 \end{array}\right] $$
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