Question

What is the probability that a number \(n, 1 \leq n \leq 99,\) is divisible by both 6 and \(10 ?\) By either 6 or 10 or both?

Short answer

Probability that a number is divisible by both 6 and 10: \ \frac{1}{99}. Probability that a number is divisible by either 6 or 10 or both: \ \frac{16}{99}.

Step-by-step solution

- Identify numbers divisible by 6

A number is divisible by 6 if it is divisible by both 2 and 3. This means the number must be even and the sum of its digits must be divisible by 3. For numbers from 1 to 99, calculate and identify the numbers that meet these criteria.

- Identify numbers divisible by 10

A number is divisible by 10 if it ends in 0. For numbers from 1 to 99, list all those numbers that end in 0 (e.g., 10, 20, 30, ..., 90).

- Identify numbers divisible by both 6 and 10

For a number to be divisible by both 6 and 10, it must satisfy both divisibility rules simultaneously. Since 10 requires the number to end in 0 and 6 requires the number to also be even and the sum of its digits to be divisible by 3, identify the numbers that meet both conditions.

- Identify union of numbers divisible by 6 or 10 or both

Use the principle of inclusion and exclusion: Add the number of multiples of 6, the number of multiples of 10, and subtract the number of multiples of both 6 and 10.

- Calculate the probabilities

Find the probability by dividing the number of favorable outcomes by the total number of outcomes (which is 99).

- Perform the calculations and get the results

Calculate the exact numbers for the total numbers divisible by 6 and 10 separately, their intersections, and the union to find probabilities in each case.

Key concepts

Divisibility Rules

Divisibility rules help us quickly determine if a number can be divided by another number without a remainder. For this exercise, we focused on 6 and 10.

To check if a number is divisible by **6**, it must meet two conditions: it has to be divisible by 2 and by 3. For divisibility by 2, a number just needs to be even. For divisibility by 3, the sum of the digits of the number must be divisible by 3. If both conditions are satisfied, the number is divisible by 6.

For example, consider the number 18. It is even, and the sum of its digits is 1 + 8 = 9, which is divisible by 3. Hence, 18 is divisible by 6.

Next, to check for divisibility by **10**, a number must end in 0. Simple right? Numbers like 10, 20, 30, all the way to 90 are examples of numbers from 1 to 99 that are divisible by 10.

Understanding these rules sets the foundation for identifying numbers that meet these criteria within the given range.

Multiple Inclusion-Exclusion Principle

The inclusion-exclusion principle helps us consider combined probabilities without double-counting.

In our problem, we need to account for numbers divisible by 6 or 10 or both. To do this, we:

• First count numbers divisible by 6
• Then count numbers divisible by 10
• Subtract numbers divisible by both 6 and 10 (to avoid double-counting)

From 1 to 99, the multiples of 6 are 6, 12, 18, up to 96. Count them to get 16 numbers.

The multiples of 10 are 10, 20, up to 90. Count these to get 9.

Multiples of both 6 and 10, i.e., of 30 (the LCM of 6 and 10), in the range are 30, 60, 90, giving us 3 counts.

By adding the number of multiples of 6 and 10 and subtracting those of both, we avoid double-counting. The principle ensures our counts are accurate and precise.

Probability Calculation

Probability allows us to measure how likely an event is to happen. It is calculated by dividing the number of favorable outcomes by the total possible outcomes.

To find how likely it is for a number between 1 and 99 to be divisible by 6 and 10 or both:

• Count the total numbers divisible by 6, 10, and both (which we calculated using inclusion-exclusion in the previous section).
• This count signifies the favorable outcomes.

For our exercise, numbers divisible by 6: 16, by 10: 9, and their intersection: 3.

Total unique favorable outcomes = 16 + 9 - 3 = 22.

Total possible outcomes are the numbers from 1 to 99, giving us a count of 99.

Therefore, the probability is \[ P = \frac{22}{99} \approx 0.222 \] or about 22.2%

Breaking down steps and understanding each concept makes tackling such problems simpler and clearer.