The Probability of Randomly Selecting a Divisible by Three Number from an Infinite Set of Natural Numbers

The question of determining the probability of randomly selecting a number that is divisible by three from an infinite set of natural numbers can be a fascinating exploration in the realm of probability theory and set theory. The answer, as we will delve into, depends on the set and the probability measure in use.

Understanding the Problem

When considering the infinite set of natural numbers, the notion of selecting a number with equal probability presents several mathematical challenges. Conventionally, each number cannot be assigned a non-zero probability due to the fact that the sum of probabilities over an infinite set of numbers must equal 1. However, by using specific probability measures, we can explore the possible probabilities for such a selection.

Geometric Probability Measure

Consider a geometric probability assignment, where the probability of selecting a number (n) is given by (p(n) frac{1}{2^n}). For the set of all natural numbers, starting with 0, the probabilities are as follows:

0: (frac{1}{2}) 1: (frac{1}{4}) 2: (frac{1}{8}) 3: (frac{1}{16}) …

The cumulative sum of these probabilities is:

(frac{1}{2} frac{1}{4} frac{1}{8} frac{1}{16} … 1)

Now, we need to calculate the probability of selecting a number that is divisible by three. In this geometric distribution, the numbers divisible by three are 0, 3, 6, 9, etc., with respective probabilities ( frac{1}{2^0}, frac{1}{2^3}, frac{1}{2^6}, frac{1}{2^9} ) and so on. The sum of these probabilities is:

(frac{1}{2^0} frac{1}{2^3} frac{1}{2^6} frac{1}{2^9} … 1 frac{1}{8} frac{1}{64} frac{1}{512} … frac{1}{1 - frac{1}{8}} frac{8}{7})

However, this sum exceeds 1, indicating a non-standard probability measure. To correct this, we can scale down by a factor of ( frac{7}{8} ), giving a probability of ( frac{4}{7} ) of selecting a number divisible by three.

Brute Force Solution

A practical approach to verify this concept can be seen in the J programming language, where:

avg 031 to 1e70.333333

This brute force solution indicates that the average frequency of numbers divisible by three among the first one million natural numbers is approximately 33.3333, which is consistent with the theoretical probability of ( frac{1}{3} ).

Probability in Different Sets

The question of probability further depends on the specific set of numbers under consideration. Here are a few scenarios:

Set of all natural numbers: The probability is ( frac{1}{3} ). Set of all natural numbers not divisible by three: The probability is 0. Set of all multiples of three: The probability is 1.

Unequal Probability Distributions

Using different probability distributions, the probability can vary. If the probability is non-zero only for multiples of three, the probability of selecting a divisible by three number is 1. Conversely, if the probability is non-zero for a set of numbers that avoids multiples of three, the probability is 0.

Conclusion

The probability of randomly selecting a number that is divisible by three from an infinite set of natural numbers is not a straightforward 1/3, but rather depends on the specific probability measure and the set of numbers under consideration. The examples and calculations provided here illustrate the complexity and the various possibilities in determining such probabilities.