What is the probability the number contains at least one digit 2?

Answer 1

The probability is 1. And the same is true for any other digit.

This counter-intuitive result can be proved as follows:

Ignore numbers with leading zeros.

There are 9 1-digit numbers of which 8 do not contain any 2s;

There are 90 2-digit numbers of which 8*9 do not contain any 2s;

There are 900 3-digit numbers of which 8*9*9 do not contain any 2s;

and so on.

The total number of k digit numbers is 9 + 90 + 900 + ... + 9*10^(k-1)

= 9*[10^k-1]/[10 - 1] = 10^k - 1

Of these, the count of numbers which do not contain any 2s = 8 + 8*9 + 8*9^2 + ... + 8*9^(k-1)

= 8*[9^k-1]/[9 - 1] = 9^k - 1

So the probability that the numbers do not contain any 2s is [9^k - 1]/[10^k - 1] which, as k gets larger, approaches 9^k/10^k = (0.9)^k which tends to 0.

So, the probability that the number DOES contain at least one 2 = 1 - Pr(It contains none) = 1 - 0 = 1.