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Consider the set of all permutations of the numbers \(1,2,3 .\) If you select a permutation at random, what is the probability that the number 2 is in the middle position? In the first position? Do your answers suggest a simple way of answering the same questions for the set of all permutations of the numbers 1 to \(7 ?\)

Short Answer

Expert verified
The probability that 2 is in any specific position in the set {1,2,3} is \(\frac{1}{3}\). For any set of 7 numbers, it is \(\frac{1}{7}\).

Step by step solution

01

- Determine the Total Number of Permutations

The total number of permutations of three numbers \(1, 2, 3\) is calculated using the factorial operation. Since there are 3 numbers, there are \(3!\) permutations.\[3! = 3 \times 2 \times 1 = 6\]
02

- Calculate the Probability That Number 2 is in the Middle Position

To find the permutations where 2 is in the middle position, we fix 2 in the middle, leaving 1 and 3 to be arranged in the remaining two positions. The number of such arrangements is \(2!\).\[2! = 2 \times 1 = 2\] The probability is therefore:\[P(2 \text{ in middle}) = \frac{2}{3!} = \frac{2}{6} = \frac{1}{3}\]
03

- Calculate the Probability That Number 2 is in the First Position

Similarly, to find the permutations where 2 is in the first position, we fix 2 in the first position, leaving 1 and 3 to be arranged in the remaining two positions. The number of such arrangements is \(2!\).\[2! = 2 \times 1 = 2\] The probability is therefore:\[P(2 \text{ in first}) = \frac{2}{3!} = \frac{2}{6} = \frac{1}{3}\]
04

- Generalize to a Larger Set of Numbers

Given the symmetry and equal probability for each position in a permutation set, the probability that any specific number (e.g., 2) is in a given position is \(\frac{1}{n}\) for \(n\) numbers. This principle can be applied to any set of permutations, including the numbers from 1 to 7. Therefore, the probability for any designated number to be in any specific position out of 7 is:\[P = \frac{1}{7}\]

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Key Concepts

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

Factorial in Permutations
Understanding permutations begins with understanding a key concept in mathematics called the factorial. The factorial of a number, represented as \(n!\), is the product of all positive integers less than or equal to \(n\). For example, \(3! = 3 \times 2 \times 1 = 6\). In permutations, factorials help us determine the total number of ways to arrange a set number of objects. For a set of 3 distinct numbers, the total number of permutations is \(3!=6.\) Similarly, for a set of 7 distinct numbers, the total number of permutations is \(7!\).
Understanding this concept is crucial for solving problems involving various arrangements or sequences.
Equal Probability
In permutations, each possible arrangement of a set of numbers is equally likely. This concept of equal probability means that any specific position in the permutation has an identical chance of being occupied by any number. For a set of 3 numbers, each of the 6 permutations is equally likely.
Hence, if we seek the probability of a specific event, like a particular number being in a specific position, the probability remains the same for each position. This idea simplifies calculating probabilities for permutations involving larger sets.
Symmetry in Probability
Symmetry plays a key role in making probability calculations easier. When permuting a set of numbers, each position is equally likely to be occupied, creating a symmetrical probability distribution. This symmetry ensures that each number has the same chance of appearing in each possible position.
For example, in the permutation of numbers \(1, 2, 3\), number 2 is just as likely to be in the first position as it is to be in the second or third position. This symmetry across all positions helps to generalize our understanding of probabilities in permutations.
Generalizing Probability
With the insights from factorials and symmetry, we can generalize the concept of probability for any number of permutations. For a set of \(n\) numbers, the probability of a specific number appearing in any given position is always \(\frac{1}{n}\). Regardless of the size of the set, this ratio remains constant.
For numbers ranging from 1 to 7, the probability that any specific number is in a particular position is \(\frac{1}{7}\). This generalization helps in solving more complex permutation problems efficiently.
Permutations of Numbers
A permutation refers to an arrangement of all elements in a particular order. When dealing with numbers, it involves arranging them in various sequences. The total permutations are given by the factorial of the count of numbers. For instance, with 3 numbers, the total permutations are \(3! = 6\).
Understanding permutations allows us to calculate the probability of events like placing a specific number in a specific position. By knowing total permutations, we can easily determine how often an event can occur and convert this into a useful probability measure.

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Most popular questions from this chapter

Suppose it is known that \(1 \%\) of the population have a certain kind of cancer. It is also known that a test for this kind of cancer is positive in \(99 \%\) of the people who have it but is also positive in \(2 \%\) of the people who do not have it. What is the probability that a person who tests positive has cancer of this type?

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Some transistors of two different kinds (call them \(N\) and \(P\) ) are stored in two boxes. You know that there are \(6 N\) 's in one box and that \(2 N\) 's and \(3 P\) 's got mixed in the other box, but you don't know which box is which. You select a box and a transistor from it at random and find that it is an \(N ;\) what is the probability that it came from the box with the \(6 \mathrm{N}\) 's? From the other box? If another transistor is picked from the same box as the first, what is the probability that it is also an \(N ?\)

(a) One box contains one die and another box contains two dice. You select a box at random and take out and toss whatever is in it (that is, toss both dice if you have picked box 2 ). Let \(x=\) number of 3 's showing. Set up the sample space and associated probabilities for \(x.\) (b) What is the probability of at least one \(3 ?\) (c) If at least one 3 turns up, what is the probability that you picked the first box? (d) Find \(\bar{x}\) and \(\sigma\)

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