suppose you select a 3 digit number at random from the set of all positive 3 digit nimbers.
First we need to determine the number of 3 digit numbers.
The largest 3 digit number is 999. Among the first 999
counting numbers, the first 99 of them have less that 3 digits,
so the number of 3 digit counting numbers is 999-99 or 900.
1) odds in favor of a multiple of 10
Now we have to find the number of 3-digit numbers which are multiplse
of 10. They are
100, 110, 120, ... 990.
They are all the 2-digit
numbers
10, 11, 12, ... 99
with a zero added on the end. So we only
need to find the number of two digit numbers. Among the first 99
counting numbers, the first 9 of them have less that 2 digits,
so the number of 2 digit counting numbers is 99-9 or 90. And
that's the same as the number of 3-digit multiples of 10.
So out of the 900 3-digit numbers, 90 of them are multiples of 10.
The other 900-90 or 810 are not multiples of 10.
So the odds in favor of a multiple of 10 is
90:810 which reduces to 1:9
2) odds against 243 or 244
Out of the 900 3-digit numbers, 2 of them are 243 or 244, and the other
900-2 or 898 are neither of those, so the odds against those two are
898:2 which reduces to 449:1
3) odds against a number greater than 400.
Among the first 999 counting numbers, the first 400 of them are not
greater than 400, and the other 999-400 or 599 counting number are
3-digit numbers greater than 400. The other 900-599 or 301 3-digit
numbers are not greater than 400. So the odds against a number
greater than 400 are
301:599
Edwin
