# An integer from 100 through 999, inclusive, is to be chosen at random. What is the probability that the number chosen will have 0 as at least 1 digit? Hello!

Probability is (the quantity of numbers having zero) / (the total quantity of numbers).

It is obvious that the total quantity of numbers is  `(999 - 100 + 1) = 900.`  The quantity of numbers among these having at least one zero have to be computed.

Zero may be at...

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Hello!

Probability is (the quantity of numbers having zero) / (the total quantity of numbers).

It is obvious that the total quantity of numbers is  `(999 - 100 + 1) = 900.`  The quantity of numbers among these having at least one zero have to be computed.

Zero may be at the ones position or tens position, not hundreds. It is obvious that one-tenth of these `900` numbers has zero at ones position, this is `90` numbers. Also `90` numbers have zero at the tens position.

But the total quantity of numbers with zeros is NOT 90+90, because some numbers have zeros at both ones and tens. There are 9 such numbers (100, 200, 300, ... , 900).

So there are `90+90-9=9*19` numbers with zeros, and the probability is

`(9*19)/900=19/100=0.19` or 19%. This is the answer.

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