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 $\frac{26}{99}$.

---

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.