A number x is chosen at random from first n natural numbers. What is the probability that the number chosen satisfies x + \(\frac{1}{\text{x}}\) > 2 ?
Understanding the Probability Problem
The question asks for the probability that a number x, chosen randomly from the first n natural numbers, satisfies the inequality \(x + \frac{1}{\text{x}} > 2\). To solve this probability problem, we first need to understand the set of possible outcomes and then identify the outcomes that meet the specified condition.
Identifying the Sample Space and Total Outcomes
The number x is chosen from the first n natural numbers. The set of the first n natural numbers is \(\{1, 2, 3, \dots, \text{n}\}\). This set represents our sample space.
The total number of possible outcomes when choosing one number from this set is the count of numbers in the set, which is n.
Total number of outcomes = n
Analyzing the Inequality \(x + \frac{1}{\text{x}} > 2\)
We need to find which numbers x from the sample space \(\{1, 2, 3, \dots, \text{n}\}\) satisfy the inequality \(x + \frac{1}{\text{x}} > 2\).
Let's work with the inequality:
\(x + \frac{1}{\text{x}} > 2\)
Since x is a natural number, x is always positive. We can multiply both sides of the inequality by x without changing the direction of the inequality sign:
\(x \left(x + \frac{1}{\text{x}}\right) > 2x\)
\(x^2 + 1 > 2x\)
Now, let's rearrange the inequality to get all terms on one side:
\(x^2 - 2x + 1 > 0\)
The expression on the left side is a perfect square trinomial. It can be factored as \((x - 1)^2\).
\((x - 1)^2 > 0\)
This inequality \((x - 1)^2 > 0\) is true for any real number x as long as \((x - 1)^2\) is not equal to 0. The square of any non-zero real number is positive.
The expression \((x - 1)^2\) is equal to 0 only when \(x - 1 = 0\), which means \(x = 1\).
So, the inequality \((x - 1)^2 > 0\) is satisfied by all real numbers x EXCEPT x = 1.
Identifying Favorable Outcomes in the Sample Space
Our sample space is the set of first n natural numbers: \(\{1, 2, 3, \dots, \text{n}\}\).
The condition \(x + \frac{1}{\text{x}} > 2\) is equivalent to \((x - 1)^2 > 0\). This is true for all natural numbers x except for x = 1.
Therefore, the numbers from the sample space \(\{1, 2, 3, \dots, \text{n}\}\) that satisfy the condition are all numbers in the set except 1.
- If \(n=1\), the sample space is \(\{1\}\). The number 1 does NOT satisfy the condition \((1-1)^2 > 0\). There are 0 favorable outcomes.
- If \(n > 1\), the sample space is \(\{1, 2, 3, \dots, \text{n}\}\). The number 1 does NOT satisfy the condition. The numbers \(2, 3, \dots, \text{n}\) DO satisfy the condition.
The set of favorable outcomes for \(n > 1\) is \(\{2, 3, \dots, \text{n}\}\).
To count the favorable outcomes: The count of numbers from 2 to n is \(n - 2 + 1 = n - 1\).
If \(n=1\), favorable outcomes = 0.
If \(n > 1\), favorable outcomes = \(n - 1\).
Let's consider the edge case \(n=1\). Total outcomes = 1. Favorable outcomes = 0. Probability = \(0/1 = 0\).
Let's consider \(n > 1\). Total outcomes = \(n\). Favorable outcomes = \(n - 1\). Probability = \(\frac{(\text{n}-1)}{\text{n}}\).
Does the formula \(\frac{(\text{n}-1)}{\text{n}}\) work for \(n=1\)? \(\frac{(1-1)}{1} = \frac{0}{1} = 0\). Yes, it works for \(n=1\) as well.
So, the number of favorable outcomes is always \(n - 1\) for \(n \ge 1\).
Calculating the Probability
Probability is defined as:
Probability = \(\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\)
Number of favorable outcomes = \(n - 1\)
Total number of outcomes = \(n\)
Probability = \(\frac{(n - 1)}{n}\)
| Case | Sample Space | Condition \(x + \frac{1}{\text{x}} > 2\) (i.e., \(x \ne 1\)) | Favorable Outcomes | Number of Favorable Outcomes | Total Outcomes | Probability |
|---|---|---|---|---|---|---|
| \(n=1\) | \(\{1\}\) | \(x \ne 1\) | \(\emptyset\) | 0 | 1 | \(\frac{0}{1} = 0\) |
| \(n > 1\) | \(\{1, 2, \dots, \text{n}\}\) | \(x \ne 1\) | \(\{2, 3, \dots, \text{n}\}\) | \(n - 1\) | \(n\) | \(\frac{(\text{n}-1)}{\text{n}}\) |
| Overall \((n \ge 1)\) | \(\{1, 2, \dots, \text{n}\}\) | \(x \ne 1\) | \(\{x \in \{1, \dots, \text{n}\} \mid x \ne 1\}\) | \(n - 1\) | \(n\) | \(\frac{(\text{n}-1)}{\text{n}}\) |
The probability that the number chosen satisfies \(x + \frac{1}{\text{x}} > 2\) is \(\frac{(\text{n}-1)}{\text{n}}\).
Probability Revision Table
| Concept | Definition/Explanation | How it Applies Here |
|---|---|---|
| Natural Numbers | The positive integers starting from 1: \(\{1, 2, 3, \dots\}\) | Our sample space is the first \(n\) natural numbers: \(\{1, 2, \dots, \text{n}\}\) |
| Sample Space | The set of all possible outcomes of an experiment. | The set of numbers from which \(x\) is chosen: \(\{1, 2, \dots, \text{n}\}\). Total outcomes = \(n\). |
| Favorable Outcomes | The outcomes that satisfy the specific condition or event. | Numbers \(x\) in the sample space satisfying \(x + \frac{1}{\text{x}} > 2\). Found to be \(\{2, 3, \dots, \text{n}\}\). Number of favorable outcomes = \(n - 1\). |
| Probability | Ratio of favorable outcomes to total outcomes. \(P(\text{Event}) = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}}\) | Probability = \(\frac{(\text{n}-1)}{\text{n}}\) |
Additional Information on Inequalities and Probability
The inequality \(x + \frac{1}{\text{x}} > 2\) for positive values of x is a classic result often related to the AM-GM inequality. The Arithmetic Mean (AM) of two positive numbers \(a\) and \(b\) is \(\frac{(a+b)}{2}\), and their Geometric Mean (GM) is \(\sqrt{\text{ab}}\). The AM-GM inequality states that for non-negative numbers, \(\frac{(a+b)}{2} \ge \sqrt{\text{ab}}\), with equality if and only if \(a=b\).
For positive x, we can consider the numbers x and \(\frac{1}{\text{x}}\). Their AM is \(\frac{(x + \frac{1}{\text{x}})}{2}\) and their GM is \(\sqrt{x \cdot \frac{1}{\text{x}}} = \sqrt{1} = 1\).
By AM-GM:
\(\frac{(x + \frac{1}{\text{x}})}{2} \ge 1\)
\(x + \frac{1}{\text{x}} \ge 2\)
The equality \(x + \frac{1}{\text{x}} = 2\) holds if and only if \(x = \frac{1}{\text{x}}\), which for positive x means \(x^2 = 1\), so \(x = 1\).
Thus, for positive x, \(x + \frac{1}{\text{x}} > 2\) is true for all \(x \ne 1\).
Since natural numbers are positive, this confirms our algebraic solution: the condition \(x + \frac{1}{\text{x}} > 2\) is satisfied by all natural numbers except 1.
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