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What is the probability that a positive integer not exceeding 100 selected at random is divisible by 5 or 7\(?\)

Short Answer

Expert verified
The probability is 0.32.

Step by step solution

01

Define the Set of Positive Integers

Identify the range of numbers: 1 to 100. This is the set of positive integers not exceeding 100.
02

Count Numbers Divisible by 5

Find the numbers in the range that are divisible by 5. These numbers form the sequence: 5, 10, 15, ..., 100. Therefore, there are \(\frac{100}{5} = 20\) numbers divisible by 5 in this range.
03

Count Numbers Divisible by 7

Find the numbers in the range that are divisible by 7. These numbers form the sequence: 7, 14, 21, ..., 98. Therefore, there are \(\frac{100}{7} \approx 14\) numbers divisible by 7 in this range.
04

Count Numbers Divisible by Both 5 and 7 (i.e., 35)

Numbers that are divisible by both 5 and 7 are those divisible by 35. These numbers form the sequence: 35, 70. Therefore, there are \(\frac{100}{35} \approx 2\) numbers divisible by 35 in this range.
05

Apply the Inclusion-Exclusion Principle

We use the principle of inclusion-exclusion to avoid double-counting. The total number of integers divisible by 5 or 7 is \(20 + 14 - 2 = 32\).
06

Calculate the Probability

The probability is the number of favorable outcomes divided by the total number of possible outcomes: \(\frac{32}{100} = 0.32\)

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Key Concepts

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

positive integers
Positive integers are simply the set of all whole numbers greater than 0. If you think about the counting numbers you learned as a kid, those are positive integers—like 1, 2, 3, and so on. In our exercise, we are focusing on positive integers up to 100. This means we only consider numbers from 1 to 100, inclusive.
How does this help? Well, it gives us a clear boundary for our problem of finding which numbers are divisible by 5 or 7.
divisibility
Divisibility refers to the ability of one number to be divided by another without leaving a remainder. For example, a number is divisible by 5 if, when you divide it by 5, the result is a whole number. In math terms, an integer n is divisible by 5 if there exists an integer k such that: \( n = 5k \)
In our problem, we need to find numbers from 1 to 100 that are divisible by both 5 and 7. We can list these numbers in sequences:
* Numbers divisible by 5 include 5, 10, 15, ..., 100
* Numbers divisible by 7 include 7, 14, 21, ..., 98
* Numbers divisible by both 5 and 7 (i.e., 35) include 35, 70.
inclusion-exclusion principle
The inclusion-exclusion principle helps us count the number of items in overlapping groups. It's crucial to avoid double-counting the items that are common to both groups. The formula for two sets A and B is:\[ |A \, \text{or} \, B| = |A| + |B| - |A \, \text{and} \, B| \]
In our exercise, we let A be the set of numbers divisible by 5 and B be the set of numbers divisible by 7. Using the inclusion-exclusion principle, we compute: \[ |A \, \text{or} \, B| = |A| + |B| - |A \, \text{and} \, B| \] \[ |A| = 20, \, |B| = 14, \, |A \, \text{and} \, B| = 2 \]
Hence, \[ |A \, \text{or} \, B| = 20 + 14 - 2 = 32 \]
That is, there are 32 numbers from 1 to 100 that are divisible by either 5 or 7.
sequence counting
Sequence counting involves listing all numbers in a range that fit a particular criterion. We count these numbers by figuring out the pattern they form. For instance, to count numbers divisible by 5 in the range from 1 to 100, we recognize that these numbers form the sequence: \( 5, 10, 15, ..., 100 \)
The general term of this arithmetic sequence can be given by: \[ a_n = 5n \]
To find how many terms are in this sequence, we solve for n: \[ 100 / 5 = 20 \]
Similarly, we do this for the sequence of numbers divisible by 7: \( 7, 14, 21, ..., 98 \)
The general term here is: \[ b_n = 7n \]
Solving for n gives: \[ 100 / 7 \approx 14 \]
This arithmetic sequence approach makes sequence counting an organized method for identifying all qualifying numbers within a set.

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