How many people live at an address that has a street number that is not divisible by 2 or 3 or 5?
It was assumed that all street numbers had only one or two digits.
(e) 	(i) Given the integers 1- 99, what is the probability that none of these numbers are 	divisible by 2, 3 or 5.	

To solve this problem, we need to determine how many numbers between 1 and 99 are divisible by 2, 3, or 5, and then calculate the probability that a randomly chosen number is not divisible by any of these.

### Step 1: Total number of integers from 1 to 99
There are 99 integers between 1 and 99.

### Step 2: Use Inclusion-Exclusion Principle

We want to find how many numbers between 1 and 99 are divisible by 2, 3, or 5. We can use the inclusion-exclusion principle to count this.

#### A = Numbers divisible by 2:
The numbers divisible by 2 are:
\[ \frac{99}{2} = 49.5 \]
So, the number of integers divisible by 2 is \( 49 \).

#### B = Numbers divisible by 3:
The numbers divisible by 3 are:
\[ \frac{99}{3} = 33 \]
So, the number of integers divisible by 3 is \( 33 \).

#### C = Numbers divisible by 5:
The numbers divisible by 5 are:
\[ \frac{99}{5} = 19.8 \]
So, the number of integers divisible by 5 is \( 19 \).

#### Now we calculate overlaps (intersection of sets A, B, and C):

- Numbers divisible by both 2 and 3 (i.e., divisible by 6):
  \[ \frac{99}{6} = 16.5 \]
  So, there are \( 16 \) numbers divisible by 6.

- Numbers divisible by both 2 and 5 (i.e., divisible by 10):
  \[ \frac{99}{10} = 9.9 \]
  So, there are \( 9 \) numbers divisible by 10.

- Numbers divisible by both 3 and 5 (i.e., divisible by 15):
  \[ \frac{99}{15} = 6.6 \]
  So, there are \( 6 \) numbers divisible by 15.

- Numbers divisible by 2, 3, and 5 (i.e., divisible by 30):
  \[ \frac{99}{30} = 3.3 \]
  So, there are \( 3 \) numbers divisible by 30.

### Step 3: Apply the Inclusion-Exclusion Principle
Now we can apply the inclusion-exclusion principle to find how many numbers between 1 and 99 are divisible by 2, 3, or 5:

\[
\text{Total divisible by 2, 3, or 5} = (49 + 33 + 19) - (16 + 9 + 6) + 3 = 101 - 31 + 3 = 73
\]

So, 73 numbers are divisible by 2, 3, or 5.

### Step 4: Numbers not divisible by 2, 3, or 5
The number of integers between 1 and 99 that are **not divisible** by 2, 3, or 5 is:
\[
99 - 73 = 26
\]

### Step 5: Probability
The probability that a randomly chosen number between 1 and 99 is **not divisible** by 2, 3, or 5 is the ratio of favorable outcomes (numbers not divisible by 2, 3, or 5) to the total outcomes (all numbers between 1 and 99):
\[
\text{Probability} = \frac{26}{99}
\]

So, the probability is \( \frac{26}{99} \).

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Would you like any additional details or explanations?

### Related Questions:
1. What is the probability that a number is divisible by 2 but not by 3 or 5?
2. How many numbers between 1 and 99 are divisible by both 2 and 3 but not by 5?
3. What is the probability that a randomly chosen number between 1 and 50 is not divisible by 2, 3, or 5?
4. How does the probability change if the range is extended from 1 to 200?
5. Can you calculate the probability for three other prime numbers instead of 2, 3, and 5?

### Tip:
The inclusion-exclusion principle is a very useful tool in probability and combinatorics to account for overlapping sets. It helps avoid double-counting elements.

How many people live at an address that has a street number that is not divisible by 2 or 3 or 5? It was assumed that all street numbers had only one or two digits. (i) Given the integers 1- 99, what is the probability that none of these numbers are divisible by 2, 3 or 5.

Explore how to calculate the probability that a randomly selected street number between 1 and 99 is not divisible by 2, 3, or 5. This problem utilizes the Inclusion-Exclusion Principle and basic probability concepts, suitable for Grades 9-12 students.

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