Question

If a number is picked at random from the first `100` natural numbers, then what is the probability that it is divisible by `3` ?

Answer 1

Assuming the choice is uniformly distributed, either `33/100` or `34/100` according to whether `0` is counted as a Natural number.

In either case, close to `1/3`.

Explanation:

Unfortunately, the term "Natural number" is used by different people to mean different things.

Sometimes the set of Natural numbers `NN` is taken to include `0` and sometimes not.

If you want to be precise, you can the slightly more verbose terms "positive integer" to start at `1` or "non-negative integer" to start at `0`.

Write:

`NN_0 = { n in ZZ : 0 <= n } = { 0, 1, 2, 3,... }`

`NN_1 = { n in ZZ : 0 < n } = { 1, 2, 3,...}`

Natural numbers starting at 0 (non-negative integers)

`H_0 = { n in ZZ : 0 <= n < 100 } = { 0, 1, 2, 3,...,99 }`

`T_0 = { n in H_0 : EE k in ZZ : n = 3k } = { 0, 3, 6, ... , 99 }`

So `H_0 sub NN_0` is the set of the first `100` `NN_0` numbers and `T_0 sub H_0` the subset of `H_0` of numbers divisible by `3`.

Then the probability of picking a member of `T_0` at random from `H_0` is:

`abs(T_0)/abs(H_0) = 34/100`

Natural numbers starting at 1 (positive integers)

`H_1 = { n in ZZ : 0 < n <= 100 } = { 1, 2, 3,...,100 }`

`T_1 = { n in H_1 : EE k in ZZ : n = 3k } = { 3, 6, ..., 99 }`

So `H_1 sub NN_1` is the set of the first `100` `NN_1` numbers and `T_1 sub H_1` the subset of `H_1` of numbers divisible by `3`.

Then the probability of picking a member of `T_1` at random from `H_1` is:

`abs(T_1)/abs(H_1) = 33/100`

Practical note

In practice, if you just ask some people to pick a number less than `100`, then the results you get will not be anywhere near evenly distributed.