Question

A piece of string is cut into two pieces. The point at which the string is cut was chosen at random. What is the probability that the longer piece is at least three times as long as the shorter piece? (1) \(1 / 4\) (2) \(1 / 3\) (3) \(2 / 5\) (4) \(1 / 2\)

Short answer

1/2

Step-by-step solution

- Understanding the Problem

We need to find the probability that when a string is cut into two pieces at a random point, the longer piece is at least three times as long as the shorter piece.

- Define Variables

Let the total length of the string be 1 unit. If the string is cut at a point x units from one end, the two pieces will have lengths x and (1-x) units respectively.

- Formulate Inequality

For the longer piece to be at least three times the length of the shorter piece we need to set up the inequalities: 1) x ≥ 3(1-x) or 2) (1-x) ≥ 3x.

- Solve First Inequality

Let's solve the first inequality: x ≥ 3(1-x). Distributing the 3, we get: x ≥ 3 - 3x. Adding 3x to both sides: 4x ≥ 3. Dividing by 4: x ≥ 3/4.

- Solve Second Inequality

Now let's solve the second inequality: 1-x ≥ 3x. Adding x to both sides: 1 ≥ 4x. Dividing by 4: x ≤ 1/4.

- Combine Solutions and Interval Calculation

From the solutions of inequalities, x can be in the interval [0, 1/4] or [3/4, 1]. Each of these intervals represents a possible cut point, giving two favorable intervals: [0, 1/4] and [3/4, 1].

- Calculate Total Favorable Interval Length

The total length of the favorable intervals is (1/4 - 0) + (1 - 3/4) = 1/4 + 1/4 = 1/2.

- Compute Probability

The probability that the longer piece is at least three times as long as the shorter piece is the total favorable length divided by the entire length of the unit interval, which is 1. So, the probability is (1/2) / (1) = 1/2.

Key concepts

Probability

Probability measures how likely an event is to occur. It is calculated as the number of favorable outcomes divided by the total number of possible outcomes. This is expressed mathematically as:
\[ P(Event) = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}} \]
In this exercise, our event is cutting the string in a way that the longer piece is at least three times the length of the shorter piece.

• We identified the two intervals [0, 1/4] and [3/4, 1] where the length condition is satisfied.
• The favorable lengths of these intervals are 1/4 each, making the total favorable outcome length 1/2.

To find the probability, we take this total favorable length (1/2) and divide it by the total length of the string (1). Therefore, the probability is
\[ \frac{1/2}{1} = \frac{1}{2} \]

Geometric Probability

Geometric probability involves the chance of an event happening in a geometric space. This could be points on a line, areas in a plane, or volumes in a space. In this problem, our geometric space is the length of the string, and we are interested in specific points along the string.
By choosing any random point \( x \) on a unit-length string, we can evaluate the geometric probability by identifying the ranges (intervals) that meet our condition.

• The condition is that either \( x \geq 3(1-x) \text { or} (1-x) \geq 3x \) must hold.
• Through solving, we found the intervals [0, 1/4] and [3/4, 1].
• The intervals represent the favorable geometrical regions.

The geometric probability is the total favorable interval length divided by the total length of the string. Hence,
\[ \frac{1/2}{1} = \frac{1}{2} \].

Inequality

Inequalities help us to express the condition that one quantity is larger or smaller than another. In this exercise, we used inequalities to find where one piece of the string is at least three times longer than the other. We set up the following inequalities:
\(x \geq 3(1-x) \) or \( (1-x) \geq 3x \)

• The first inequality simplifies to \( x \geq 3/4 \)
• The second inequality simplifies to \( x \leq 1/4 \)

These inequalities help us find the intervals where the condition is met: [0, 1/4] and [3/4, 1]. By combining the solutions, we covered the complete range from 0 to 1 for a string of unit length.